Category: ML Aggarwal Solutions

The availability of ML Aggarwal Class 12 ISC Solutions Chapter 9 Differential Equations Ex 9.1 encourages students to tackle difficult exercises.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.1

Very short answer type questions :

Determine the order and the degree (when defined) of each of the following (1 to 20) differential equations:

Question 1.
(i) \(\frac{d y}{d x}\) – cos x = 0 (NCERT)
(ii) \(\left(\frac{d y}{d x}\right)^2+\frac{d y}{d x}\) – sin y = 0.
Solution:
Given differential equation be,
\(\frac{d y}{d x}\) – cos x = 0
Which is polynomial in derivatives.
The highest ordered derivative existing in the given diff. eqn. be \(\frac{d y}{d x}\) so its order is 1.
Also, the exponent of \(\frac{d y}{d x}\) is 1.
∴ degree of given differential equation be 1.

(ii) Given differential equation be,
\(\left(\frac{d y}{d x}\right)^2+\frac{d y}{d x}\) – sin y = 0
which is polynomial in derivatives.
The higher order derivative existing in given the given diff. eqn. be \(\frac{d y}{d x}\) so its order is 1.
and the highest power of \(\frac{d y}{d x}\) in given diff. eqn. be 2 so its degree is 2.

Question 2.
(i) y’ + y = e^x
(ii) \(\frac{d^2 y}{d x^2}\) = sin 3x + cos 3x. (NCERT)
Solution:
Given differential eqn. be
y’ + y = e^x i.e. \(\frac{d y}{d x}\) + y = e^x
which is polynomial in derivatives.
The highest order derivative present in given differential eqn. be \(\frac{d y}{d x}\) and its exponent be 1.
So order of given diff. eqn. be 1 and its degree be also 1.

(ii) Given diff. eqn. be,
\(\frac{d^2 y}{d x^2}\) = sin 3x + cos 3x
which is polynomial in derivatives.
The highest order derivative existing in given diff. eqn. be \(\frac{d^2 y}{d x^2}\).
So its order be 2.
The highest power of \(\frac{d^2 y}{d x^2}\) be 1.
∴ degree of given differential eqn. be 1.

Question 3.
(i) (x²y – 3x) dy + (x³ – 3y²) dx = 0
(ii) \(\sqrt{1-y^2} d x+\sqrt{1-x^2} d y\) = 0
Solution:
(i) Given differential eqn. can be written as,
\(\frac{d y}{d x}+\frac{x^3-3 y^2}{x^2 y-3 x}\) = 0
which is polynomial in derivatives.
The highest order derivative present in given differential eqn. be \(\frac{d y}{d x}\).
so its order be 1.
Also the degree of given diff. eqn. be the highest exponent of \(\frac{d y}{d x}\) which is 1.
Thus its degree be 1.

(ii) Given differential eqn. can be written as,
\(\frac{d y}{d x}+\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}\) = 0
which is polynomial in derivatives.
The highest order derivatives present in given diff. eqn. be \(\frac{d y}{d x}\) and its order be 1.
The degree of given diff. eqn. be the highest power of \(\frac{d y}{d x}\) which is 1.
So its degree be 1.

Question 4.
(i) \(\frac{1}{x} \cdot \frac{d^2 y}{d x^2}+5 x \frac{d y}{d x}\) = sin 2x
(ii) \(\left(\frac{d y}{d x}\right)^4+3 y \frac{d^2 y}{d x^2}\) = 0
Solution:
(i) Given differential eqn. be,
\(\frac{1}{x} \cdot \frac{d^2 y}{d x^2}+5 x \frac{d y}{d x}\) = sin 2x
which is polynomial in derivatives.
The highest ordered derivative present in given diff. eqn. be \(\frac{d^2 y}{d x^2}\)
So its order is 2.
The degree of given diff. eqn. be the highest exponent of \(\frac{d^2 y}{d x^2}\) which is 1.
So its degree be 1.

Question 5.
(i) (y’)² + y² – 1 = 0
(ii) y” + 5x (y’)² – 6y = log x
Solution:
(i) Given diff. eqn. be,
y’² + y² – 1 = 0
which is polynomial in derivatives.
The highest ordered derivative present in given diff. eqn. be \(\frac{d y}{d x}\) i.e. y’.
so its order be 1.
The degree of the given diff. eqn. is the highest exponent of which is \(\frac{d y}{d x}\).
So its degree be 2.

(ii) Given diff. eqn. be,
y” + 5xy’² – 6y = log x
which is polynomial in derivatives.
The highest ordered derivative in the given diff. eqn. be y”.
so its order be 2.
The degree of the given diff. eqn. be the highest exponent of \(\frac{d^2 y}{d x^2}\) which is 1.
Thus its degree be 1.

Question 6.
(i) x³ (\(\frac{d^2 y}{d x^2}\))² + x (\(\frac{d y}{d x}\))⁴ = 0
(ii) y^(iv) + sin y⁽ⁱ⁾ = 0 (NCERT)
Solution:
(i) Here the highest ordered derivative existing in given differential eqn. be \(\frac{d^2 y}{d x^2}\)
so its order be 2.
Further given differential eqn. can be expressed as polynomial in derivatives.
So the exponent of highest ordered derivative \(\frac{d^2 y}{d x^2}\) which is 2 gives the degree of given differential eqn.

(ii) Here the highest ordered derivative existing in given differential equation be y^(iv) so its order be 4.
Since the given differential eqn. can’t be expressed as polynomial in derivatives as it contains term like sin y⁽¹⁾ which contains infinite number of terms.
So degree of given differential eqn. is not defined.

Question 7.
(i) \(\left(\frac{d^2 y}{d x^2}\right)^3+\frac{d^2 y}{d x^2}+\sin \left(\frac{d y}{d x}\right)\) = 2x
(ii) \(\left(\frac{d^2 y}{d x^2}\right)^3+2 y \frac{d y}{d x}+\sin y=5 x^{\frac{2}{3}}+\log x\)
Solution:
(i) Here, the highest ordered derivative existing in given diff. eqn. be \(\frac{d^2 y}{d x^2}\) and its order be 2.
Clearly given diff. eqn. cannot be expressed as polynomial in \(\frac{d y}{d x}\)
so its degree is not defined [as it contains terms like sin (\(\frac{d y}{d x}\))]

(ii) Clearly the highest ordered derivative existing in given differential eqn. be \(\frac{d^2 y}{d x^2}\)
so its order is 2.
Since each term in derivatives is a polynomial.
So degree of given differential eqn. be highest exponent of \(\frac{d^2 y}{d x^2}\) i.e. 3.

Question 8.
(i) 2x² \(\frac{d^2 y}{d x^2}\) – 3 \(\frac{d y}{d x}\) + y = 0 (NCERT)
(ii) (y”)³ + (y’)² + sin y’ + 1 = 0. (NCERT)
Solution:
(i) Given differential eqn. be
2x² \(\frac{d^2 y}{d x^2}\) – 3 \(\frac{d y}{d x}\) + y = 0,
which is polynomial in derivatives.
The highest ordered derivative present in \(\frac{d^2 y}{d x^2}\) given diff. eqn. be \(\frac{d^2 y}{d x^2}\).
So its order be 2.
The degree of given diff. eqn. is the highest power of \(\frac{d^2 y}{d x^2}\) which is 1.
Thus, its degree be 1.

(ii) Given differential eqn. be
(y”)³ + (y’)² + sin y’ + 1 = 0
The highest order derivative existing in given diff. eqn. be y” and its order be 2.
Thus, the order of given diff. eqn. be 2.
Since the term sin y’ is not polynomial in y’.
∴ The degree of given diff. eqn. is not defined.

Question 9.
(i) x \(\frac{d^2 y}{d x^2}\) = (1 + (\(\frac{d y}{d x}\))²)⁴
(ii) \(\left(\frac{d^4 y}{d x^4}\right)^2=\left(x+\left(\frac{d y}{d x}\right)^2\right)^3\)
Solution:
(i) Clearly the highest ordered derivative existing in given diff. eqn. be \(\frac{d^2 y}{d x^2}\).
so its order be 2.
Clearly the given diff. eqn. can be expressed as polynomial in derivatives.
Hence degree be the highest exponent of \(\frac{d^2 y}{d x^2}\) which is 1.

(ii) Clearly the highest ordered derivative existing in given diff. eqn. be \(\frac{d^4 y}{d x^4}\).
so its order be 4.
Here given differential eqn. can be expressed as polynomial in derivatives.
Hence degree of given diff. eqn. be the highest exponent of \(\frac{d^4 y}{d x^4}\) which is 2.

Question 10.
(i) 3x \(\frac{d y}{d x}\) + \(\frac{5}{\frac{d y}{d x}}\) = y³
(ii) y = x \(\frac{d y}{d x}\) + a \(\sqrt{1+\left(\frac{d y}{d x}\right)^2}\)
Solution:
(i) Given differential eqn. be,
3x \(\frac{d y}{d x}\) + \(\frac{5}{\frac{d y}{d x}}\) = y³
⇒ 3x (\(\frac{d y}{d x}\))² + 5 = y³ \(\frac{d y}{d x}\)
which is polynomial in derivatives.
The highest ordered derivative present in given diff. eqn. be \(\frac{d y}{d x}\).
so its order be 1.
The degree of given diff. eqn. be the highest exponent of \(\frac{d y}{d x}\) which is 2.
Thus its degree be 2.

(ii) Given differential eqn. be,
y = x \(\frac{d y}{d x}\) + a \(\sqrt{1+\left(\frac{d y}{d x}\right)^2}\)
⇒ y – x \(\frac{d y}{d x}\) = a \(\sqrt{1+\left(\frac{d y}{d x}\right)^2}\)
On squaring, we have
(y – x \(\frac{d y}{d x}\))² = a² [1 + (\(\frac{d y}{d x}\))²]
The highest ordered derivative exiosting in given diff. eqn. be \(\frac{d y}{d x}\) and its power 2.
∴ it is of order 1 and degree 2.
Clearly it is a non-linear differential equation.

Question 11.
(i) \(\sqrt{1+\left(\frac{d y}{d x}\right)^2}\) = 3x – \(\frac{d y}{d x}\)
(ii) \(5 \frac{d^2 y}{d x^2}=\left(1+\left(\frac{d y}{d x}\right)^2\right)^{\frac{1}{4}}\)
Solution:
(i) Given differential eqn. be,
\(\sqrt{1+\left(\frac{d y}{d x}\right)^2}\) = 3x – \(\frac{d y}{d x}\)
On squaring both sides ;
1 + (\(\frac{d y}{d x}\))² = 9x² + (\(\frac{d y}{d x}\))² – 6x \(\frac{d y}{d x}\)
⇒ 6x \(\frac{d y}{d x}\) = 9x² + 1 = 0
The highest ordered derivative existing in given diff. eqn. be \(\frac{d y}{d x}\) so its order be 1.
Further given duff. eqn. can be expressed as polynomial in derivative.
So degree be the highest exponent of which is 1.

(ii) Given differential eqn. can be written as
\(5 \frac{d^2 y}{d x^2}=\left(1+\left(\frac{d y}{d x}\right)^2\right)^{\frac{1}{4}}\)
Clearly the highest ordered derivative existing in given diff. eqn. be \(\frac{d^2 y}{d x^2}\) and
hence its order be 2.
Here, given diff. eqn. can be expressed as polynomial in derivatives.
So degree of given diff. eqn. be the highest power of which is 4.

Question 11 (old).
\(\left(\frac{d^2 x}{d t^2}\right)^4-7 t\left(\frac{d x}{d t}\right)^3\) = log t.
Solution:
Given differential eqn. be,
\(\left(\frac{d^2 x}{d t^2}\right)^4-7 t\left(\frac{d x}{d t}\right)^3\) = log t
which is polynomial in derivatives.
The highest ordered derivative present in given diff. eqn. be \(\frac{d^2 y}{d x^2}\) and its order be 2.
The degree of given diff. eqn. be the highest exponent of \(\frac{d^2 y}{d x^2}\) which is 4.
Thus its degree be 4.

Question 12.
(i) y” + (y’)² + 2y = 0 (NCERT)
(ii) y” + 2y’ + sin y = 0
Solution:
(i) Given differential eqn. be,
y” + (y’)² + 2y = 0
which is polynomial in derivatives.
The highest order derivative present in given diff. eqn. be y” and its order be 2.
The degree of given differential equation is the highest exponent of y” which is 1.
Thus its degree be 1.

(ii) Given differential eqn. be,
y” + 2y’ + sin y = 0
which is polynomial in derivatives.
The highest order derivative present in given diff. eqn. be y” and its order is 2.
The degree of given diff. eqn. be the highest exponent of y” which is 1.
Thus its degree be 1.

Question 12 (old).
y^(iv) + sin y”’ = 0 (NCERT)
Solution:
Given diff. eqn. be,
y^(iv) + sin y”’ = 0
The order of given diff. eqn. be the highest order derivative present in given diff. eqn. which is y^(iv) and its order be 4.
Here the term sin y”’ is not polynomial in y”’.
Thus, the degree of given differential eqn. is not defined.

Question 13.
(i) (y”’)² + (y”)³ + (y’)⁴ + y⁵ = 0 (NCERT)
(ii) (1 – (y’)²)^3/2 = ky”
Solution:
(i) Given diff. eqn. be,
(y”’)² + (y”)³ + (y’)⁴ + y⁵ = 0
which is polynomial in derivatives.
The highest order derivative present in given diff. eqn. be y”’ so its order be 3.
The degree of given diff. eqn. is the highest exponent of y”’, which is 2.
Thus degree of given diff. eqn. be 2.

(ii) Given diff. eqn. can be written as
(1 – (y’)²)³ = (ky”)²
Clearly the highest ordered derivative existing in given duff. eqn. be y” so its order 2.
Here the given duff. eqn. can be expressed as polynomial in derivatives.
Thus its degree be the highest exponent of y” which is 2.

Question 14.
Write the sum of the order and the degree of the differential equation \(\left(\frac{d y}{d x}\right)^5+3 x y\left(\frac{d^3 y}{d x^3}\right)^2+y\left(\frac{d^2 y}{d x^2}\right)^4\) = 0.
Solution:
Clearly the highest ordered derivative existing in given diff. eqn. be \(\frac{d^3 y}{d x^3}\).
so its order be 3.
Clearly the given duff. eqn. can be expressed as polynomial in derivatives.
So degree of given diff. eqn. be the highest exponent of \(\frac{d^3 y}{d x^3}\) which is 2.
∴ required sum = 3 + 2 = 5.

Question 15.
Find the product of the order and degree of the following differential equation :
x (\(\frac{d^2 y}{d x^2}\))² + (\(\frac{d y}{d x}\))² + y² = 0.
Solution:
The given differential eqn. be
x (\(\frac{d^2 y}{d x^2}\))² + (\(\frac{d y}{d x}\))² + y² = 0
which is polynomial in derivatives.
The highest order derivative present in given diff. eqn. be \(\frac{d^2 y}{d x^2}\).
So its order be 2.
The degree of given diff. eqn. be the highest exponent of \(\frac{d^2 y}{d x^2}\) which is 2.
Thus, its degree be 2.
∴ required product of order and degree of given differential equation 2 × 2 = 4.

Question 20 (old).
y = px + \(\sqrt{a^2 p^2+b^2}\), where p = \(\frac{d y}{d x}\).
Solution:
Given diff. eqn. can be written as
(y – px)² = a²p² + b²; where p = \(\frac{d y}{d x}\)
Clearly it is a polynomial in p.
The highest order derivative existing in diff. eqn. be p i.e. \(\frac{d y}{d x}\) and its power 2.
Thus given diff. eqn. is of order 1 and degree 2.
Clearly it is a non-linear differential equation.

Question 21 (old).
Write the sum of the order and the degree of the differential equation \(\left(\frac{d^2 y}{d x^2}\right)^2-\left(\frac{d y}{d x}\right)^3\) = y³.
Solution:
Given differential equation be,
\(\left(\frac{d^2 y}{d x^2}\right)^2-\left(\frac{d y}{d x}\right)^3\) = y³
which is polynomial in derivatives.
The highest order derivative present in given
diff. eqn. be \(\frac{d^2 y}{d x^2}\).
so its order be 2.
The degree of given diff. eqn. be the highest power of \(\frac{d^2 y}{d x^2}\) which is 2.
So its degree be 2 and required sum = 2 + 2 = 4.

Regular engagement with ML Aggarwal Class 12 Solutions ISC Chapter 8 Integrals MCQs can boost students’ confidence in the subject.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals MCQs

Choose the correct answer from the given four options in questions (1 to 55) :

Question 1.
∫ sin \(\frac{x}{2}\) cos \(\frac{x}{2}\) cos x dx is equal to
(a) – \(\frac{1}{4}\) cos 2x + C
(b) – \(\frac{1}{8}\) cos 2x + C
(c) \(\frac{1}{8}\) cos 2x + C
(d) – \(\frac{1}{8}\) cos 2x + C
Solution:
(b) – \(\frac{1}{8}\) cos 2x + C

∫ sin \(\frac{x}{2}\) cos \(\frac{x}{2}\) cos x dx
= \(\frac{1}{2}\) ∫ (2 sin \(\frac{x}{2}\) cos \(\frac{x}{2}\)) cos x dx
= \(\frac{1}{2}\) ∫ sin x cos x dx
= \(\frac{1}{4}\) ∫ sin 2x dx
= \(\frac{1}{4}\left(-\frac{\cos 2 x}{2}\right)\) + C
= – \(\frac{1}{8}\) cos 2x + C

Question 2.
∫ \(\frac{d x}{\sin ^2 x \cos ^2 x}\) is equal to
(a) tan x + cot x + C
(b) (tan x + cot x)² + C
(c) tan x – cot x + C
(d) (tan x – cot x)² + C
Solution:
∫ \(\frac{d x}{\sin ^2 x \cos ^2 x}\)
= ∫ \(\frac{\sin ^2 x+\cos ^2 x}{\sin ^2 x \cos ^2 x}\) dx
= ∫ \(\left[\frac{1}{\cos ^2 x}+\frac{1}{\sin ^2 x}\right]\) dx
= ∫ [sec² x + cosec² x] + C
= tan x – cot x + C

Question 3.
∫ \(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\) dx is equal to
(a) 2 (sin x + x cos α) + C
(b) 2 (sin x – x cos α) + C
(c) 2 (sin x + 2x cos α) + C
(d) 2 (sin x – 2x cos α) + C
Solution:
(a) 2 (sin x + x cos α) + C

∫ \(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\) dx
= 2 ∫ \(\frac{2 \cos ^2 x-1-\left(2 \cos ^2 \alpha-1\right)}{\cos x-\cos \alpha}\) dx
= 2 ∫ \(\frac{(\cos x-\cos \alpha)(\cos x+\cos \alpha)}{\cos x-\cos \alpha}\) dx
= 2 [sin x + cos x] + C

Question 4.
∫ \(\frac{x}{4+x^4}\) dx is equal to
(a) \(\frac{1}{4}\) tan^-1 x² + C
(b) \(\frac{1}{2}\) tan^-1 \(\frac{x^2}{2}\) + C
(c) \(\frac{1}{4}\) tan^-1 \(\left(\frac{x^2}{2}\right)\) + C
(d) \(\frac{1}{2}\) tan^-1 2x² + C
Solution:
(c) \(\frac{1}{4}\) tan^-1 \(\left(\frac{x^2}{2}\right)\) + C

∫ \(\frac{x}{4+x^4}\) dx ;
put x² = t
⇒ 2x dx = dt

Question 5.
∫ \(\frac{\cos \sqrt{x}}{\sqrt{x}}\) is equal to
(a) 2 cos √x + C
(b) 2 sin √x + C
(c) – 2 sin √x + C
(d) sin √x + C
Solution:
(b) 2 sin √x + C

put √x = t
⇒ \(\frac{1}{2 \sqrt{x}}\) dx = dt
∴ ∫ \(\frac{\cos \sqrt{x}}{\sqrt{x}}\) dx = ∫ cos t . 2t dt
= 2 sin t + C
= 2 sin √x + C

Question 6.
If ∫ x e^kx2 dx = \(\frac{1}{4}\) e^2x2 + C, then the value of k is,
(a) 4
(b) – 2
(c) 2
(d) 1
Solution:
(c) 2

Given ∫ x e^kx2 dx = \(\frac{1}{4}\) e^2x2 + C
put x² = t
⇒ 2x dx = dt
⇒ ∫ e^kt \(\frac{dt}{2}\) = \(\frac{1}{4}\) e^2x2 + C
⇒ \(\frac{1}{2} \frac{e^{k t}}{k}\) = \(\frac{1}{4}\) e^2x2 + C
⇒ \(\frac{1}{2 k}\) e^2x2 + C = \(\frac{1}{4}\) e^2x2 + C
∴ 2k = 4
⇒ k = 2

Question 7.
If ∫ x⁶ sin (5x⁷) dx = \(\frac{k}{5}\) cos (5x⁷) + C, then the value of k is
(a) – \(\frac{1}{7}\)
(b) \(\frac{1}{7}\)
(c) \(\frac{1}{35}\)
(d) – \(\frac{1}{35}\)
Solution:
(a) – \(\frac{1}{7}\)

∫ x⁶ sin (5x⁷) dx = \(\frac{k}{5}\) cos (5x⁷) + C
put x⁷ = t
⇒ 7x⁶ dx = dt

Question 8.
If ∫ \(\frac{2^x}{\sqrt{1-4^x}}\) dx = k sin^-1 (2^x) + C, then the value of k is
(a) log 2
(b) \(\frac{1}{\log 2}\)
(c) \(\frac{2}{\log 2}\)
(d) \(\frac{\log 2}{2}\)
Solution:
(b) \(\frac{1}{\log 2}\)

Given ∫ \(\frac{2^x}{\sqrt{1-4^x}}\) dx = k sin^-1 (2^x) + C
put 2^x = t
⇒ 2^x log 2 dx = dt
⇒ ∫ \(\frac{d t}{\log 2 \sqrt{1-t^2}}\) = k sin^-1 (2^x) + C
⇒ \(\frac{1}{\log 2}\) sin^-1 + C = k sin^-1(2^x + C
⇒ \(\frac{1}{\log 2}\) sin^-1 + C = k sin^-1 2^x + C
∴ k = \(\frac{1}{\log 2}\)

Question 9.
If ∫ |x| dx = kx |x| + C, x ≠ 0, then the value of k is
(a) 2
(b) – 2
(c) – \(\frac{1}{2}\)
(d) \(\frac{1}{2}\)
Solution:
(d) \(\frac{1}{2}\)

Let I = ∫ |x| . 1 dx
= |x| . x – ∫ \(\frac{x}{|x|}\) . x dx
= x |x| – ∫ \(\frac{|x|^2}{|x|}\) dx
[∵ x² = |x|²]
= x |x| – ∫ |x| dx
[∵ x ≠ 0]
⇒ I = x |x| – I
⇒ 2I = x |x|
⇒ I = \(\frac{x|x|}{2}\) + C
Now I = kx |x| + C
⇒ k = \(\frac{1}{2}\).

Question 10.
∫ cot x log (sin x) dx is equal to
(a) \(\frac{1}{2}\) (log (sin x)² + C
(b) log (sin x) + C
(c) \(\frac{1}{2}\) (log (cos x))² + C
(d) none of these
Solution:
(a) \(\frac{1}{2}\) (log (sin x)² + C

Let I = ∫ cot x log (sin x) dx
put log sin x = t
⇒ \(\frac{1}{\sin x}\) cos x dx = dt
⇒ cot x dx = dt
= ∫ t dt
= \(\frac{t^2}{2}\) + C
= \(\frac{1}{2}\) [log (sin x)]² + C

Question 11.
∫ \(\frac{x+\sin x}{1+\cos x}\) dx is equal to
(a) log |1 + cos x| + C
(b) log |x + sin x| + C
(c) x – tan \(\frac{x}{2}\) + C
(d) x tan \(\frac{x}{2}\) + C
Solution:
(d) x tan \(\frac{x}{2}\) + C

Question 12.
∫ \(\left(\frac{1-x}{1+x^2}\right)^2\) e^x dx is equal to
(a) \(\frac{e^x}{1+x^2}\) + C
(b) – \(\frac{e^x}{1+x^2}\) + C
(c) \(\frac{e^x}{\left(1+x^2\right)^2}\) + C
(d) – \(\frac{e^x}{\left(1+x^2\right)^2}\) + C
Solution:
(a) \(\frac{e^x}{1+x^2}\) + C

Question 13.
∫ (x – 1)e^-x dx is equal to
(a) (x – 2) e^-x + C
(b) xe^-x + C
(c) – xe^-x + C
(d) (x + 1)e^-x + C
Solution:
(c) – xe^-x + C

∫ (x – 1)e^-x dx = (x – 1) \(\frac{e^{-x}}{-1}\) – ∫ 1 . \(\frac{e^{-x}}{-1}\) dx +C
= – (x – 1) e^-x – e^-x + C
= – (x – 1 + 1) e^-x + C
= – x e^-x + C

Question 14.
∫ e^x (1 – cot x + cot² x) dx is equal to
(a) e^x cosec x + C
(b) – e^x cosec x + C
(c) e^x cot x + C
(d) – e^x cot x + C
Solution:
(d) – e^x cot x + C

Let I = ∫ e^x (1 – cot x + cot² x) dx
= ∫ e^x (1 – cot x + cosec² x – 1) dx
= ∫ e^x (cosec² x) dx – ∫ e^x cot x dx
= e^x (- cot x) – ∫ e^x (- cot x) dx
= e^x (- cot x) – ∫ e^x (- cot x) dx – ∫ e^x cot x dx + C
= – e^x cot x + C

Question 15.
If ∫ \(\frac{1+\cos 4 x}{\cot x-\tan x}\) dx = k cos 4x + C, then the value of k is
(a) \(\frac{1}{4}\)
(b) – \(\frac{1}{2}\)
(c) – \(\frac{1}{8}\)
(d) – \(\frac{1}{4}\)
Solution:
(c) – \(\frac{1}{8}\)

Let I = ∫ \(\frac{1+\cos 4 x}{\cot x-\tan x}\) dx
= ∫ \(\frac{2 \cos ^2 2 x}{\left(\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}\right)}\) dx
= ∫ \(\frac{\frac{2 \cos ^2 2 x d x}{\cos ^2 x-\sin ^2 x}}{\sin x \cos x}\) dx
= ∫ \(\frac{2 \cos ^2 2 x d x}{\frac{\cos 2 x}{\sin x \cos x}}\) dx
= ∫ cos 2x (2 sin x cos x) dx
= ∫ cos 2x sin 2x dx
= \(\frac{1}{2}\) ∫ sin 4x dx
= – \(\frac{1}{2} \frac{\cos 4 x}{4}\) + C
= – \(\frac{1}{8}\) cos 4x + C
Also, I = k cos 4x + C
∴ k = – \(\frac{1}{8}\)

Question 16.
∫ \(\frac{d x}{e^x+e^{-x}+2}\) is equal to
(a) \(\frac{1}{e^x+1}\) + C
(b) \(\frac{1}{1+e^{-x}}\) + C
(c) – \(\frac{1}{e^x+1}\) + C
(d) none of these
Solution:
(c) – \(\frac{1}{e^x+1}\) + C

Question 17.
∫ \(\frac{(\log x)^5}{x}\) dx is equal to
(a) \(\frac{\log x^6}{6}\) + C
(b) \(\frac{(\log x)^6}{6}\) + C
(c) \(\frac{(\log x)^6}{3 x^2}\) + C
(d) none of these
Solution:
(b) \(\frac{(\log x)^6}{6}\) + C

∫ \(\frac{(\log x)^5}{x}\) dx ;
put log x = t
⇒ \(\frac{1}{x}\) dx = dt
= ∫ t⁵ dt
= \(\frac{t^6}{6}\) + C
= \(\frac{(\log x)^6}{6}\) + C

Question 18.
∫ \(\frac{d x}{\sqrt{2 x-x^2}}\) is equal to
(a) sin^-1 (x – 1) + C
(b) sin^-1 (x – 1) + C
(c) – \(\sqrt{2 x-x^2}\) + C
(d) none of these
Solution:
(a) sin^-1 (x – 1) + C

Question 19.
∫ \(\frac{x^2+1}{x^2-1}\) dx is equal to
(a) x + log \(\left|\frac{x+1}{x-1}\right|\) + C
(b) x + log \(\left|\frac{x-1}{x+1}\right|\) + C
(c) log |(x – 1) (x + 1)| + C
(d) log |x² + 1| + C
Solution:
(b) x + log \(\left|\frac{x-1}{x+1}\right|\) + C

Question 20.
∫ (sin⁶ x + cos⁶ x + 3 sin² x cos² x) dx is equal to
(a) x + C
(b) \(\frac{3}{2}\) sin 2x + C
(c) – \(\frac{3}{2}\) cos 2x + C ‘
(d) none of these
Solution:
(a) x + C

∫ (sin⁶ x + cos⁶ x + 3 sin² x cos² x) dx
= ∫ [(sin² x)³ + (cos² x)³ + 3 sin² x cos² x (sin² x + cos² x)] dx
= ∫ (sin² x+ cos² x)³ dx
= ∫ dx
= x + C

Question 21.
∫ \(\frac{d x}{x\left(x^7+1\right)}\) is equal to
(a) log \(\left|\frac{x^7}{x^7+1}\right|\) + C
(b) \(\frac{1}{7} \log \left|\frac{x^7}{x^7+1}\right|\) + C
(c) log \(\left|\frac{x^7+1}{x^7}\right|\) + C
(d) \(\frac{1}{7} \log \left|\frac{x^7+1}{x^7}\right|\) + C
Solution:
(b) \(\frac{1}{7} \log \left|\frac{x^7}{x^7+1}\right|\) + C

Let I = ∫ \(\frac{d x}{x\left(x^7+1\right)}\) ;
put x⁷ = t
⇒ 7x⁶ dx = dt

Question 22.
∫ (sin⁴ x – cos⁴ x) dx is equal to
(a) \(\frac{1}{2}\) cos 2x + C
(b) – \(\frac{1}{2}\) cos 2x + C
(c) \(\frac{1}{2}\) sin 2x + C
(d) – \(\frac{1}{2}\) sin 2x + C
Solution:
(d) – \(\frac{1}{2}\) sin 2x + C

∫ (sin⁴ x – cos⁴ x) dx = ∫ (sin² x – cos² x) (sin² x + cos² x) dx
= – ∫ cos 2x dx
= – \(\frac{\sin 2 x}{2}\) + C

Question 23.
∫ \(\frac{\left(\tan ^{-1} x\right)^3}{1+x^2}\) dx is equal to
(a) 3 (tan^-1 x)² + C
(b) (tan^-1 x)⁴ + C
(c) (tan^-1 x)⁴ + C
(d) none of these
Solution:
(b) (tan^-1 x)⁴ + C

put tan^-1 x = t
⇒ \(\frac{1}{1+x^2}\) dx = dt
∴ I = ∫ \(\frac{\left(\tan ^{-1} x\right)^3}{1+x^2}\) dx
= ∫ t³ dt
= \(\frac{t^4}{4}\) + C
= \(\frac{\left(\tan ^{-1} x\right)^4}{4}\) + C

Question 24.
∫ e^{3 log x} (x⁴ + 1)^-1 dx is equal to
(a) \(\frac{1}{4}\) log (x⁴ + 1) + C
(b) – \(\frac{1}{4}\) (x⁴ + 1) + C
(c) log (x⁴ + 1) + C
(d) none of these
Solution:
(a) \(\frac{1}{4}\) log (x⁴ + 1) + C

∫ e^{3 log x} (x⁴ + 1)^-1 dx
= ∫ e^{log x3} \(\frac{1}{x^4+1}\) dx
= ∫ \(\frac{x^3 d x}{x^4+1}\) ;
put x⁴ + 1 = t
⇒ 4x³ dx = dt
= \(\frac{1}{4} \int \frac{d t}{t}\)
= \(\frac{1}{4}\) log |t| + C
= \(\frac{1}{4}\) log |x⁴ + 1| + C

Question 25.
∫ (sin (log x) + cos (log x)) dx is equal to
(a) sin (log x) + C
(b) cos (log x) + C
(c) x cos (log x) + C
(d) x sin (log x) + C
Solution:
(d) x sin (log x) + C

Let I = ∫ (sin (log x) + cos (log x)) dx
= sin (log x) . x – ∫ cos (log x) . \(\frac{1}{x}\) . x dx + ∫ cos (log x) dx
= x sin (log x) + C

Question 26.
∫ \(\frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}}\) dx is equal to
(a) \(\frac{1}{2} \sqrt{1+x}\) + C
(b) \(\frac{1}{2}\) (1 + x)^3/2 + C
(c) 2 (1 + x)^3/2 + C
(d) \(\sqrt{1 + x}\) + C
Solution:
(b) \(\frac{1}{2}\) (1 + x)^3/2 + C

Question 27.
∫ \(\frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}}\) dx is equal to
(a) f(x) log (f(x)) + C
(b) log (log (f(x)) + C
(c) \(\frac{f(x)}{\log (f(x))}\) + C
(d) \(\frac{1}{\log (\log (f(x)))}\) + C
Solution:
(b) log (log (f(x)) + C

Let I = ∫ \(\frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}}\) dx
put log (f(x)) = t
⇒ \(\frac{f^{\prime}(x)}{f(x)}\) dx = dt
= ∫ \(\frac{1}{t}\) dt
= |log |t| + C
= log |log f(x)| + C

Question 28.
∫ x³ log x dx is equal to
(a) \(\frac{x^4 \log x}{4}\) + C
(b) \(\frac{x^4}{8}\left(\log x-\frac{4}{x^2}\right)\) + C
(c) \(\frac{x^4}{16}\) (4 log x – 1) + C
(d) \(\frac{x^4}{16}\) (4 log x + 1) + C
Solution:
(c) \(\frac{x^4}{16}\) (4 log x – 1) + C

∫ x³ log x dx

Question 29.
∫ e^{x log a} e^x dx is equal to
(a) \(\frac{a^x}{\log a e}\) + C
(b) \(\frac{e^x}{1+\log a}\)
(c) (ae)^x + C
(d) \(\frac{(a e)^x}{\log a e}\) + C
Solution:
(d) \(\frac{(a e)^x}{\log a e}\) + C

∫ e^{x log a} e^x dx = ∫ e^{log ax} . e^x dx
= ∫ a^x . e^x dx
= ∫ (ae)^x dx
= \(\frac{(a e)^x}{\log a e}\) + C

Question 30.
∫ \(\left(\frac{1-\sin x}{1-\cos x}\right)\) e^x dx is equal to
(a) – e^x tan \(\frac{x}{2}\) + C
(b) – e^x cot \(\frac{x}{2}\) + C
(c) – \(\frac{1}{2}\) e^x tan \(\frac{x}{2}\) + C
(d) – \(\frac{1}{2}\) e^x cot \(\frac{x}{2}\) + C
Solution:
(b) – e^x cot \(\frac{x}{2}\) + C

Question 31.
∫ \(\left(\frac{1+x+x^2}{1+x^2}\right)\) e^{tan-1 x} dx is equal to
(a) x + e^{tan-1 x} + C
(b) e^{tan-1 x} – x + C
(c) e^{tan-1 x} + C
(d) x e^{tan-1 x} + C
Solution:
(d) x e^{tan-1 x} + C

Question 32.\(\)
If \(\int_0^{40} \frac{d x}{2 x+1}\) = log k, then value of k is
(a) 3
(b) \(\frac{9}{2}\)
(c) 9
(d) none of these
Solution:
(c) 9

Given \(\int_0^{40} \frac{d x}{2 x+1}\) = log k
⇒ \(\left.\frac{\log |2 x+1|}{2}\right]_0^{40}\) = log k
⇒ \(\frac{1}{2}\)[log 81 – log 1] = log k
⇒ \(\frac{1}{2}\) log 9² = log k
⇒ log 9 = log k
∴ k = 9

Question 33.
\(\int_0^{\sqrt{3}} \frac{d x}{1+x^2}\) is equal to
(a) \(\frac{\pi}{3}\)
(b) \(\frac{\pi}{4}\)
(c) \(\frac{\pi}{6}\)
(d) \(\frac{\pi}{12}\)
Solution:
(d) \(\frac{\pi}{12}\)

\(\left.\int_0^{\sqrt{3}} \frac{d x}{1+x^2}=\tan ^{-1} x\right]_1^{\sqrt{3}}\)
= tan^-1 √3 – tan^-1 1
= \(\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}\)

Question 34.
If \(\int_0^k \frac{1}{9 x^2+1} d x=\frac{\pi}{12}\), then k is equal to
(a) \(\frac{\pi}{4}\)
(b) \(\frac{1}{3}\)
(c) 3
(d) none of these
Solution:
(b) \(\frac{1}{3}\)

Question 35.
\(\int_0^{\pi / 2} \frac{\sin x \cos x}{1+\sin x}\) dx is equal to
(a) 3 – log 12
(b) 1 – log 2
(c) 1 + log 2
(d) none of these
Solution:
(b) 1 – log 2

\(\int_0^{\pi / 2} \frac{\sin x \cos x}{1+\sin x}\) dx ;
put sin x = t
⇒ cos x dx = dt
When x = 0 ⇒ t = 0 ;
When x = \(\frac{\pi}{2}\) ⇒ t = 1
= \(\int_0^1 \frac{t d t}{1+t}\)
= \(\int_0^1\left[\frac{1+t-1}{1+t}\right]\) dt
= \(\int_0^1\left[1-\frac{1}{1+t}\right]\) dt
= t – log |1 + t|\(\)
= 1 – log 2 – 0 – 0
= 1 – log 2

Question 36.
The value of \(\int_{\pi / 6}^{\pi / 3} \frac{1}{\sin 2 x}\) dx is
(a) log 3
(b) \(\frac{1}{2}\) log 2
(c) \(\frac{1}{2}\) log 3
(d) none of these
Solution:
(c) \(\frac{1}{2}\) log 3

Question 37.
\(\int_{-1}^0 \frac{d x}{x^2+2 x+2}\) is equal to
(a) 0
(b) \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{4}\)
(d) – \(\frac{\pi}{4}\)
Answer:
(c) \(\frac{\pi}{4}\)

\(\int_{-1}^0 \frac{d x}{x^2+2 x+2}\) = \(\int_{-1}^0 \frac{d x}{(x+1)^2+1^2}\)
= tan^-1 x\(]_{-1}^0\)
= tan^-1 1 – tan^-1 0
= \(\frac{\pi}{4}\) – 0 = \(\frac{\pi}{4}\)

Question 38.
\(\int_0^1 \frac{\tan ^{-1} x}{1+x^2}\) dx is equal to
(a) \(\frac{\pi}{4}\)
(b) \(\frac{\pi^2}{32}\)
(c) 1
(d) \(\frac{\pi^2}{16}\)
Solution:
(b) \(\frac{\pi^2}{32}\)

\(\int_0^1 \frac{\tan ^{-1} x}{1+x^2}\) dx ;
put tan^-1 x = t
⇒ \(\frac{1}{1+x^2}\) dx = dt
When x = 0
⇒ t = tan^-1 0 = 0
When x = 1
⇒ t = tan^-1 1 = \(\frac{\pi}{4}\)
= \(\left.\int_0^{\pi / 4} t d t=\frac{t^2}{2}\right]_0^{\pi / 4}\)
= \(\frac{\pi^2}{32\)

Question 39.
\(\int_0^1 \frac{d x}{e^x+e^{-x}}\) is equal to
(a) \(\frac{\pi}{4}\)
(b) \(\frac{\pi}{2}\)
(c) tan^-1 \(\left(\frac{e-1}{e+1}\right)\)
(d) tan^-1 \(\left(\frac{1-e}{1+e}\right)\)
Solution:
(c) tan^-1 \(\left(\frac{e-1}{e+1}\right)\)

\(\int_0^1 \frac{d x}{e^x+e^{-x}}=\int_0^1 \frac{e^x d x}{e^{2 x}+1}\) ;
put e^x = t
⇒ e^x dx = dt
When x = 0 ⇒ t = 1
When x = 1 ⇒ t = e
= \(\int_1^e \frac{d t}{t^2+1}\)
= tan^-1 t\(]_1^e\)
= tan^-1 e – tan^-1 1
= tan^-1 e – \(\frac{\pi}{4}\)
= tan^-1 \(\left(\frac{e-1}{e+1}\right)\)

Question 40.
\(\int_0^1 \frac{x^4+1}{x^2+1}\) dx is equal to
(a) \(\frac{1}{6}\) (3 – 4π)
(b) \(\frac{1}{6}\) (3π + 4)
(c) \(\frac{1}{6}\) (3 + 4π)
(d) \(\frac{1}{6}\) (3π – 4)
Solution:
(d) \(\frac{1}{6}\) (3π – 4)

Question 41.
\(\int_{a+c}^{b+c}\) f(x) dx is equal to
(a) \(\int_a^b\int_a^b\) f(x – c) dx
(b) \(\int_a^b\) f(x + c) dx
(c) \(\int_a^b\) f(x) dx
(d) \(\int_{a-c}^{b-c}\) f(x) dx
Solution:
(b) \(\int_a^b\) f(x + c) dx

put x = t + c
⇒ dx = dt
When x = a + c
⇒ t = a
When x = b + c
⇒ t = b
∴ I = \(\int_a^b\) f(t + c) dt
= \(\int_a^b\) f(x + c) dx
[∵ \(\int_a^b\) f(x) dx = \(\int_a^b\) f(t) dt]

Question 42.
\(\int_{-\pi / 2}^{\pi / 2}\) cos x dx is equal to
(a) 0
(b) 1
(c) 2
(d) 4
Solution:
(c) 2

\(\int_{-\pi / 2}^{\pi / 2}\) cos x dx = + sin x\(]_{-\pi / 2}^{\pi / 2}\)
= sin \(\frac{\pi}{2}\) – sin (- \(\frac{\pi}{2}\))
= 1 + sin \(\frac{\pi}{2}\)
= 1 + 1 = 2

Question 43.
\(\int_{-\pi / 4}^{\pi / 4} \frac{d x}{1+\cos 2 x}\) is equla to
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(a) 1

Question 44.
\(\int_0^{\pi / 2}\) log |tan x| dx is equal to
(a) π – log 2
(b) – π log 2
(c) 0
(d) none of these
Solution:
(c) 0

Let I = \(\int_0^{\pi / 2}\) log |tan x| dx
I = \(\int_0^{\pi / 2}\) log tan x dx ………………(1)
[when 0 ≤ x < \(\frac{\pi}{2}\) ⇒ tan x > 0
⇒ |tan x| = tan x]
We know that
\(\int_0^a\) f(x) dx = \(\int_0^a\) f(a – x) dx
∴ I = \(\int_0^{\pi / 2}\) log tan (\(\frac{\pi}{2}\) – x) dx
= \(\frac{\pi}{2}\) log cot x dx ………………(2)
On adding (1) and (2) ; we have
2I = \(\int_0^{\pi / 2}\) (log tan x + log cot x) dx
= \(\int_0^{\pi / 2}\) log (tan x . cot x) dx
= \(\int_0^{\pi / 2}\) log 1 dx = 0
⇒ I = 0

Question 45.
\(\int_{-\pi / 2}^{\pi / 2} \sqrt{\frac{1-\cos 2 x}{2}}\) dx is equal to
(a) 0
(b) \(\frac{1}{2}\)
(c) 1
(d) 2
Solution:
(d) 2

\(\int_{-\pi / 2}^{\pi / 2} \sqrt{\frac{1-\cos 2 x}{2}}\) dx
= \(\int_{-\pi / 2}^{\pi / 2} \sqrt{\frac{2 \sin ^2 x}{2}}\)
= \(\int_{-\pi / 2}^{\pi / 2}\) |sin x| dx
= \(\int_{-\pi / 2}^0\) |sin x| dx + \(\int_0^{\pi / 2}\) |sin x| dx
When – \(\frac{\pi}{2}\) ≤ x < 0
⇒ sin x ≥ 0
⇒ |sin x| = – sin x
When 0 ≤ x ≤ \(\frac{\pi}{2}\)
⇒ sin x ≥ 0
⇒ |sin x| = sin x
= \(\int_{-\pi / 2}^0\) – sin x dx + \(\int_0^{\pi / 2}\) sin x dx
= \(\left.\cos x]_{-\pi / 2}^0+(-\cos x)\right]_0^{\pi / 2}\)
= (1 – 0) + (- 0 + 1) = 2

Question 46.
\(\int_{-\pi}^\pi \frac{e^{\sin x}}{e^{\sin x}+e^{-\sin x}}\) dx is equal to
(a) 2π
(b) π
(c) \(\frac{\pi}{2}\)
(d) \(\frac{\pi}{4}\)
Solution:
(b) π

Question 47.
\(\int_a^b \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a+b-x}}\) dx is equal to
(a) \(\frac{\pi}{2}\)
(b) π
(c) \(\frac{1}{2}\) (b – a)
(d) b – a
Solution:
(c) \(\frac{1}{2}\) (b – a)

Question 48.
\(\int_0^{\pi / 2} \frac{\cos x-\sin x}{2+\sin x \cos x}\) dx is equal to
(a) 0
(b) \(\frac{\pi}{6}\)
(c) \(\frac{\pi}{4}\)
(d) \(\frac{\pi}{2}\)
Solution:
(a) 0

Let I = \(\int_0^{\pi / 2} \frac{\cos x-\sin x}{2+\sin x \cos x}\) dx ……………..(1)
We know that
\(\int_0^a\) f(x) dx = \(\int_0^a\) f(a – x) dx
∴ I = \(\int_0^{\pi / 2} \frac{\cos \left(\frac{\pi}{2}-x\right)-\sin \left(\frac{\pi}{2}-x\right)}{2+\sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)}\)
I = \(\int_0^{\pi / 2} \frac{\sin x-\cos x}{2+\cos x \sin x}\) ………………….(2)
On adding (1) and(2) ; we have
2I = \(\int_0^{\pi / 2} \frac{\cos x-\sin x+\sin x-\cos x}{2+\cos x \sin x}\) dx = 0
⇒ I = 0

Question 49.
The value of \(\int_0^2\left|\cos \frac{\pi}{2} t\right|\) dt is equal to
(a) \(\frac{3}{4 \pi}\)
(b) \(\frac{4}{\pi}\)
(c) \(\frac{\pi}{2}\)
(d) 2π
Solution:
(b) \(\frac{4}{\pi}\)

Since, 0 ≤ t ≤ 2
⇒ 0 ≤ \(\frac{\pi}{2}\) t ≤ π
⇒ cos \(\frac{\pi}{2}\) t > 0
and 1 ≤ t ≤ 2
⇒ \(\frac{\pi}{2}\) ≤ \(\frac{\pi}{2}\) t ≤ π
⇒ cos \(\frac{\pi}{2}\) t < 0

Question 50.
\(\int_0^{\pi / 2} \sqrt{1-\sin 2 x}\) dx is equal to
(a) 2√2
(b) 2 (√2 + 1)
(c) 2
(d) 2 (√2 – 1)
Solution:
(d) 2 (√2 – 1)

Let I = \(\int_0^{\pi / 2} \sqrt{1-\sin 2 x}\) dx
= \(\int_0^{\pi / 2} \sqrt{\sin ^2 x+\cos ^2 x-2 \sin x \cos x}\)
= \(\int_0^{\pi / 2} \sqrt{(\sin x-\cos x)^2}\) dx
= \(\int_0^{\pi / 2}\) |sin x – cos x| dx
When 0 ≤ x < \(\frac{\pi}{4}\) ⇒ cos x > sin x
⇒ sin x – cos x < 0
∴ |sin x – cos x| = – (sin x – cos x)
and \(\frac{\pi}{4}\) ≤ x ≤ \(\frac{\pi}{2}\)
⇒ sin x > cos x
⇒ sin x – cos x > 0
|sin x – cos x| = sin x – cos x
= \(\int_0^{\pi / 4}\) – (sin x -cos x) dx + \(\int_{\pi / 4}^{\pi / 2}\) (sin x – cos x) dx
= – [- cos x – sin x\(]_0^{\pi / 4}\) + (- cos x- sin x)\(]_{\pi / 4}^{\pi / 2}\)
= \(\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-(1+0)\right]+\left(0-1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)\)
= √2 – 1 – 1 + √2
= 2 (√2 – 1)

Question 51.
\(\int_0^{\pi / 2}\) sin 2x log cot x dx is equal to
(a) π
(b) 2π
(c) 0
(d) \(\frac{\pi}{2}\)
Solution:
(c) 0

Let I = \(\int_0^{\pi / 2}\) sin 2x log cot x dx ………..(1)
We know that
\(\int_0^a\) f(x) dx = \(\int_0^a\) f (a – x) dx
∴ I = \(\int_0^{\pi / 2}\) sin 2 (\(\frac{\pi}{2}\) – x) log cot (\(\frac{\pi}{2}\) – x) dx
= \(\int_0^{\pi / 2}\) sin (π – x) log cot (\(\frac{\pi}{2}\) – x) dx
∴ I = \(\int_0^{\pi / 2}\) sin x log tan x dx ……………..(2)
On adding (1) and (2) ; we have
2I = \(\int_0^{\pi / 2}\) sin 2x [log cot x + log tan x] dx
= \(\int_0^{\pi / 2}\) sin 2x log (cot x . tan x) dx
= \(\int_0^{\pi / 2}\) sin 2x log 1 dx = 0
[∵ log 1 = 0 ]
⇒ I = 0

Question 52.
The value of \(\int_{-\pi / 2}^{\pi / 2}\) (x⁵ + x sin² x + 2 tan^-1 x – 1) dx is
(a) π
(b) 2
(c) 1
(d) 0
Solution:
(a) π

Let I = \(\int_{-\pi / 2}^{\pi / 2}\) (x⁵ + x sin² x + 2 tan^-1 x – 1) dx
= \(\int_{-\pi / 2}^{\pi / 2}\) (x⁵ + x sin² x + 2 tan^-1 x) dx + \(\int_{-\pi / 2}^{\pi / 2}\) dx
= I₁ + x\(]_{-\pi / 2}^{\pi / 2}\)
= I₁ + \(\left(\frac{\pi}{2}+\frac{\pi}{2}\right)\)
= I₁ + π …………..(1)
Let f(x) = x⁵ + x sin² x + 2 tan^-1 x
∴ f(- x) = (- x)⁵ – x (sin (- x))² + 2 tan^-1 (- x)
= – x⁵ – x sin² x – 2 tan^-1 x
= – f(x)
Thus f(x) is an odd function.
∴ I₁ = \(\int_{-\pi / 2}^{\pi / 2}\) f(x) = 0
∴ from (1) ;
I = 0 + π = π

Question 53.
If f (a + b – x) = f(x), then \(\int_a^b\) x f(x) dx is equal to
(a) \(\frac{a+b}{2} \int_a^b\) f (b – x) dx
(b) \(\frac{a+b}{2} \int_a^b\) f(a – x) dx
(c) \(\frac{b-a}{2} \int_a^b\) f(x) dx
(d) \(\frac{a+b}{2} \int_a^b\) f(x) dx

Solution:
(d) \(\frac{a+b}{2} \int_a^b\) f(x) dx

Let I = \(\int_a^b\) x f(x) dx ………………(1)
We know that
\(\int_a^b\) f(x) dx = \(\int_a^b\) f(a + b – x) dx
I = \(\int_a^b\) (a + b – x) f(a + b – x) dx
I = \(\int_a^b\) (a + b – x) f(x) dx ………………(2)
[∵ f(a + b – x) = f(x)]
On adding (1) and (2) ; we get
2I = \(\int_a^b\) (a + b – x + x) f(x) dx
⇒ I = \(\frac{a+b}{2} \int_a^b\) f(x) dx

Question 54.
If a is a real number such that \(\int_0^a\) x dx ≤ a + 4, then
(a) 0 ≤ a ≤ 4
(b) – 2 ≤ a ≤ 0
(c) a ≤ – 2 or a ≥ 4
(d) – 2 ≤ a ≤ 4
Solution:
(d) – 2 ≤ a ≤ 4

\(\int_0^a\) x dx ≤ a + 4
⇒ \(\left.\frac{x^2}{2}\right]_0^a\) ≤ a + 4
⇒ \(\frac{a^2}{2}\) ≤ a + 4
⇒ a² ≤ 2a + 8
⇒ (a² – 2a – 8) ≤ 0
⇒ (a + 2) (a – 2) ≤ 0
⇒ – 2 ≤ a ≤ 4
[if (x – a) (x – b) ≤ 0 and a < b. Then a ≤ x ≤ b]

Question 55.
\(\int_0^{\pi / 8}\) tan² 2x dx is equal to
(a) \(\frac{4-\pi}{8}\)
(b) \(\frac{4+\pi}{8}\)
(c) \(\frac{4-\pi}{4}\)
(d) \(\frac{4-\pi}{2}\)
Solution:
(a) \(\frac{4-\pi}{8}\)

Let I = \(\int_0^{\pi / 8}\) tan² 2x dx
put 2x = t
⇒ 2 dx = dt
When x = 0 ⇒ t = 0 ;
When x = \(\frac{\pi}{8}\)
⇒ t = \(\frac{\pi}{4}\)

Peer review of ML Aggarwal Class 12 Solutions Chapter 8 Integrals Chapter Test can encourage collaborative learning.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Chapter Test

Very Short answer type questions (1 to 7):

Evaluate the following (1 to 4) :

Question 1.
If f'(x) = √x and f(1) = 2, then find the f(x).
Solution:
Since f(x) = ∫ f'(x) dx + C
⇒ f(x) = ∫ √x dx + C
[∵ f'(x) = √x]
⇒ f(x) = \(\frac{2}{3}\) x^3/2 + C ………..(1)
Since f(1) = 2
⇒ When x = 1 ; f(x) = 2
∴ from (1);
2 = \(\frac{2}{3}\) + C
⇒ C = \(\frac{4}{3}\)
∴ f(x) = \(\frac{2}{3}\) x^3/2 + \(\frac{4}{3}\)

Question 1 (old).
(i) ∫ \(\frac{x}{\sqrt{1-x^2}}\) dx
(ii) ∫ x² e^x3 dx
(iii) ∫ e^ex e^x dx
Solution:
(i) Let I = ∫ \(\frac{x}{\sqrt{1-x^2}}\) dx
put x² = t
⇒ 2x dx = dt
= ∫ \(\frac{d t}{2 \sqrt{1-t}}\)
= \(\frac{1}{2}\) ∫ (1 – t)^{– \(\frac{1}{2}\)}
= \(\frac{1}{2} \frac{(1-t)^{-\frac{1}{2}+1}}{(-1)\left(-\frac{1}{2}+1\right)}\) + C
= – \(\sqrt{1-x^2}\) + C

(ii) Let I = ∫ x² e^x3 dx
put x³ = t
⇒ 3x² dx = dt
= ∫ e^t \(\frac{d t}{3}\)
= \(\frac{e^{x^3}}{3}\) + C

(iii) put e^x = t
e^x dx = dt
∴ I = ∫ e^ex e^x dx
= ∫ e^t dt
= e^t + C
= e^ex + C

Question 2.
If f'(x) = 4x³ – \(\frac{3}{x^4}\) and f(- 1) = 0, find f(x).
Solution:
Given f'(x) = 4x³ – \(\frac{3}{x^4}\)
On integrating both sides w.r.t x, we have
∫ f'(x) dx = ∫ (4x³ – \(\frac{3}{x^4}\)) dx + C
⇒ f(x) = \(\frac{4 x^4}{4}-3 \frac{x^{-4+1}}{-4+1}\) + C
⇒ f(x) = x⁴ + \(\frac{1}{x^3}\) + C …………….(1)
Given f(- 1) = 0 i.e.
When x = – 1 ;
f(x) = 0
0 = 1 – 1 + C
⇒ C = 0
∴ from (1) ;
f(x) = x⁴ + \(\frac{1}{x^3}\)

Question 2 (old).
(i) ∫ \(\frac{d x}{\left(1+x^2\right) \tan ^{-1} x}\)
(ii) ∫ \(\sqrt{2+\sin 3 x}\) cos 3x dx
(iii) ∫ \(\frac{d x}{4 \cos ^3 x-3 \cos x}\)
Solution:
(i) put tan^-1 x = t
⇒ \(\frac{1}{1+x^2}\) dx = dt
∴ ∫ \(\frac{d x}{\left(1+x^2\right) \tan ^{-1} x}\) = ∫ \(\frac{d t}{t}\)
= log |t| + C
= log |tan^-1 x| + C

(ii) Let I = ∫ \(\sqrt{2+\sin 3 x}\) cos 3x dx
put sin 3x = t
⇒ 3 cos 3x dx = dt
∴ I = ∫ \(\sqrt{2+t} \frac{d t}{3}\)
= \(\frac{1}{3} \frac{(2+t)^{3 / 2}}{3 / 2}\) + C
= \(\frac{2}{9}\) (2 + sin 3x)^3/2 + C

(iii) Let I = ∫ \(\frac{d x}{4 \cos ^3 x-3 \cos x}\)
= ∫ \(\frac{d x}{\cos 3 x}\)
= ∫ sec 3x dx
= \(\frac{1}{3}\) log |sec 3x + tan 3x| + C

Question 3.
(i) ∫ sec x \(\sqrt{\frac{1-\sin x}{1+\sin x}}\) dx
(ii) ∫ \(\frac{1}{\ {cosec} x-1}\) dx
Solution:
(i) ∫ sec x \(\sqrt{\frac{1-\sin x}{1+\sin x}}\) dx
= ∫ sec x \(\sqrt{\frac{1-\sin x}{1+\sin x} \times \frac{1-\sin x}{1-\sin x}}\) dx
= ∫ \(\frac{\sec x(1-\sin x)}{\cos x}\) dx
= ∫ \(\frac{(1-\sin x)}{\cos ^2 x}\) dx
= ∫ sec² x dx – ∫ tan x sec x dx + C
= tan x – sec x + C

(ii) ∫ \(\frac{1}{\ {cosec} x-1}\) dx
= ∫ \(\frac{\sin x d x}{1-\sin x}\) dx
= ∫ \(\frac{\sin x(1+\sin x)}{\cos ^2 x}\) dx
= ∫ tan x sec x dx + ∫ (sec² x – 1) dx
= sec x + tan x – x + C

Question 3 (old).
(i) \(\int_1^2 \frac{x}{x^2+1}\) dx
(ii) \(\int_1^e \frac{1+\log x}{2 x}\) dx
(iii) \(\int_0^{\pi / 4} \frac{\cos 3 x}{\cos x}\) dx
Solution:
(i) Let I = \(\int_1^2 \frac{x}{x^2+1}\) dx
put x² + 1 = t
⇒ 2x dx = dt
When x = 1 ⇒ t = 2 ;
When x = 2 ⇒ t = 5
∴ I = \(\int_2^5 \frac{d t}{2 t}\)
= \(\left.\frac{1}{2} \log |t|\right]_2^5\)
= \(\frac{1}{2}\) (log 5 – log 2)
= \(\frac{1}{2}\) log \(\frac{5}{2}\).

(ii) put 1 + log x = t
⇒ \(\frac{1}{x}\) dx = dt
When x = 1
⇒ t = 1 + log 1 = 1
When x = e
⇒ t = 1 + log e = 1 + 1 = 2
∴ I = \(\int_1^2 \frac{1+\log x}{2 x}\) dx
= \(\frac{1}{2} \int_1^2\) t dt
= \(\left.\frac{t^2}{4}\right]_1^2\)
= \(\frac{1}{4}\) [4 – 1]
= \(\frac{3}{4}\)

(iii) \(\int_0^{\pi / 4} \frac{\cos 3 x}{\cos x}\) dx

Question 4.
(i) ∫ \(\frac{1+\sin ^2 x}{1+\cos x}\) dx
(ii) ∫ (4 cot x – 5 tan x)² dx
Solution:
(i) Let I = ∫ \(\frac{1+\sin ^2 x}{1+\cos x}\) dx

(ii) Let I = ∫ (4 cot x – 5 tan x)² dx
= ∫ (16 cot² x + 25 tan² x – 40) dx
= ∫ [16 (cosec² x – 1) + 25 (sec² x – 1) – 40] dx
= ∫ 16 cosec² x dx + 25 ∫ sec² x dx – 81 ∫ dx
= – cot x + 25 tan x – 81 x + C

Question 4 (old).
(ii) \(\int_{0}^{\pi / 2} \frac{\sin ^n x}{\sin ^n x+\cos ^n x}\) dx
Solution:
Let I = \(\int_{0}^{\pi / 2} \frac{\sin ^n x}{\sin ^n x+\cos ^n x}\) dx

Question 5.
(i) ∫ \(\frac{d x}{\sqrt{1-3 x}-\sqrt{5-3 x}}\)
(ii) ∫ (1 – x) \(\sqrt{1+x}\) dx
Solution:
(i) Let I = \(\frac{d x}{\sqrt{1-3 x}-\sqrt{5-3 x}}\)

(ii) Let I = ∫ (1 – x) \(\sqrt{1+x}\) dx
= ∫ – (x + 1 – 2) \(\sqrt{1+x}\) dx
= – ∫ [(x + 1)^3/2 – 2 (1 + x)^1/2] dx
= – \(\frac{2}{5}\) (x + 1)^5/2 + \(\frac{4}{3}\) (1 + x)^3/2 + C

Question 5 (old).
If ∫ |x| dx = k x |x| + C, then what is the value of k ?
Solution:
Let I = ∫ |x| dx
= ∫ |x| . 1 dx
= |x| . x – ∫ \(\frac{x}{|x|}\) . x dx
∴ I = x |x| – ∫ \(\frac{|x|^2}{|x|}\) dx + 2C
⇒ I = x |x| – ∫ |x| dx + 2C
⇒ 2I = x |x| + 2C
⇒ I = \(\frac{x|x|}{2}\) + C ………….(1)
Also I = kx |x| + C …………….(2)
∴ from (1) and (2) ;
we have k = \(\frac{1}{2}\).

Question 6.
(i) ∫ (x + 1) (2x – 1)^3/2 dx
(ii) ∫ sec² 2x cos 4x dx
Solution:
(i) Let I = ∫ (x + 1) (2x – 1)^3/2 dx
= \(\frac{1}{2}\) ∫ (2x – 1 + 3) (2x – 1)^3/2 dx
= \(\frac{1}{2} \int(2 x-1)^{5 / 2} d x+\frac{3}{2} \int(2 x-1)^{3 / 2}\) dx
= \(\frac{1}{2} \frac{(2 x-1)^{7 / 2}}{\frac{7}{2} \times 2}+\frac{3}{2} \frac{(2 x-1)^{5 / 2}}{\frac{5}{2} \times 2}\) + C
= \(\frac{1}{14}\) (2x – 1)^3/2 + \(\frac{3}{10}\) (2x – 1)^5/2 + C

(ii) ∫ sec² 2x cos 4x dx
= ∫ sec² 2x cos (2 × 2x) dx
= ∫ sec² 2x (2 cos² 2x – 1) dx
= ∫ [2 – sec² 2x] dx
= 2x – \(\frac{\tan 2 x}{2}\) + C

Question 7.
(i) ∫ \(\frac{10 x^9+10^x \log 10}{x^{10}+10^x}\) dx (NCERT)
(ii) ∫ \(\frac{d x}{x \sqrt{a x-x^2}}\) dx (NCERT)
Solution:
(i) put x¹⁰ + 10^x = t
⇒ (10x⁹ + 10^x log 10) dx = dt
∴ ∫ \(\frac{10 x^9+10^x \log 10}{x^{10}+10^x}\) dx = ∫ \(\frac{d t}{t}\)
= log |t| + C
= log |x¹⁰ + 10^x| + C

(ii) Let I = ∫ \(\frac{d x}{x \sqrt{a x-x^2}}\) dx

Question 8.
(i) ∫ \(\frac{d x}{\cos (x+a) \cos (x+b)}\)
(ii) ∫ \(\frac{d x}{3 \cos x+4 \sin x}\)
Solution:
(i) Let I = ∫ \(\frac{d x}{\cos (x+a) \cos (x+b)}\)

(ii) Let I = ∫ \(\frac{d x}{3 \cos x+4 \sin x}\)
put 3 = r cos α …………..(1)
and 4 = r sin α …………..(2)
On squaring (1) and (2) ; we have
r = \(\sqrt{9+16}\) = 5
On dividing (2) by (1) ; we have
tan α = \(\frac{4}{3}\)
⇒ α = tan^-1 \(\frac{4}{3}\)
∴ I = ∫ \(\frac{d x}{r(\cos \alpha \cos x+\sin \alpha \sin x)}\)
= ∫ \(\frac{d x}{5 \cos (x-\alpha)}\)
= \(\frac{1}{5}\) log |sec (x – α) + tan (x – α)| + C
where α = tan^-1 \(\frac{4}{3}\)

Question 9.
(i) ∫ \(\frac{\sin 3 x}{\sin x}\) dx
(ii) ∫ \(\frac{\sec x}{\sec 2 x}\) dx
Solution:
(i) ∫ \(\frac{\sin 3 x}{\sin x}\) dx
= ∫ \(\left[\frac{3 \sin x-4 \sin ^3 x}{\sin x}\right]\) dx
= ∫ [3 – 4 sin² x] dx
= ∫ [3 – 4 \(\left(\frac{1-\cos 2 x}{2}\right)\)] dx
= \(\frac{1}{2}\) ∫ [2 + 4 cos 2x] dx
= ∫ (1 + 2 cos 2x) dx
= x + \(\frac{2 \sin 2 x}{2}\) + C
= x + sin 2x + C

(ii) ∫ \(\frac{\sec x}{\sec 2 x}\) dx
= ∫ \(\frac{\cos 2 x}{\cos x}\) dx
= ∫ \(\left(\frac{2 \cos ^2 x-1}{\cos x}\right)\) dx
= 2 ∫ cos x dx – ∫ sec x dx
= 2 sin x – log |sec x + tan x| + C

Question 10.
(i) ∫ \(\frac{e^x}{\sqrt{e^{2 x}-4}}\) dx
(ii) ∫ e^{cot x} cosec² x dx
Solution:
(i) put e^x = t
⇒ e^x dx = dt
∴ I = ∫ \(\frac{e^x}{\sqrt{e^{2 x}-4}}\) dx
= ∫ \(\frac{d t}{\sqrt{t^2-2^2}}\)
= log |t + \(\sqrt{t^2-2^2}\)| + C
= log |e^x + \(\sqrt{e^{2 x}-4}\)| + C

(ii) Let I = ∫ e^{cot x} cosec² x dx
put cos x = t
⇒ – cosec² x dx = dt
∴ I = ∫ e^t (- dt)
= – e^t + C
= – e^{cot x} + C

Question 11.
(i) ∫ \(\frac{d x}{\sqrt{\sin ^3 x \cos x}}\)
(ii) ∫ \(\frac{\sin 2 x}{(a+b \cos 2 x)^2}\) dx
Solution:
(i) Let I = ∫ \(\frac{d x}{\sqrt{\sin ^3 x \cos x}}\)

(ii) Let I = ∫ \(\frac{\sin 2 x}{(a+b \cos 2 x)^2}\) dx
put cos 2x = t
⇒ – 2 sin 2x dx = dt
= ∫ \(\frac{d t}{-2(a+b t)^2}\)
= – \(\frac{1}{2} \frac{(a+b t)^{-2+1}}{(-2+1) b}\) + C
= \(\frac{1}{2 b} \frac{1}{(a+b \cos 2 x)}\) + C

Question 12.
(i) ∫ \(\frac{x+\left(\cos ^{-1} 3 x\right)^2}{\sqrt{1-9 x^2}}\) dx
(ii) ∫ \(\frac{x^3}{\sqrt{1+x^2}}\) dx
Solution:
(i) Let I = ∫ \(\frac{x+\left(\cos ^{-1} 3 x\right)^2}{\sqrt{1-9 x^2}}\) dx

(ii) put x² = t
⇒ 2x dx = dt

Question 13.
(i) ∫ \(\frac{\sec x}{\log (\sec x+\tan x)}\) dx
(ii) ∫ \(\frac{\cos x-\sin x}{1-\sin 2 x}\) dx
Solution:
(i) put log (sec x + tan x) = t
⇒ \(\frac{1}{\sec x+\tan x}\) (sec x tan x + sec² x) dx = dt
⇒ \(\frac{\sec x(\tan x+\sec x)}{\sec x+\tan x}\) dx = dt
⇒ sec x dx = dt
∴ I = ∫ \(\frac{\sec x d x}{\log (\sec x+\tan x)}\)
= ∫ \(\frac{d t}{t}\)
= log |t| + C

(ii) Let I = ∫ \(\frac{\cos x-\sin x}{1-\sin 2 x}\) dx

Question 14.
(i) ∫ \(\sqrt{1+2 \tan x(\tan x+\sec x)}\) dx
(ii) ∫ \(\frac{4 x+1}{\sqrt{2 x^2+x-3}}\) dx
Solution:
(i) Let I = ∫ \(\sqrt{1+2 \tan x(\tan x+\sec x)}\) dx

(ii) put 2x² + x – 3 = t
⇒ (4x + 1) dx = dt
∴ I = ∫ \(\frac{(4 x+1) d x}{\sqrt{2 x^2+x-3}}\)
= ∫ \(\frac{d t}{\sqrt{t}}\)
= ∫ t^-1/2 dt
= \(\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\) + C
= 2 \(\sqrt{2 x^2+x-3}\) + C

Question 15.
∫ \(\frac{\sin 2 x}{\sqrt{a^2 \sin ^2 x+b^2 \cos ^2 x}}\) dx, a ≠ ± b. What happens if a = ± b ?
Solution:
Case – I: When a ≠ ± b
Let I = ∫ \(\frac{\sin 2 x}{\sqrt{a^2 \sin ^2 x+b^2 \cos ^2 x}}\) dx
put a² sin² x + b² cos² x = t
⇒ [2a² sin x cos x + 2b² cos x (- sin x)] dx = dt
⇒ (a² – b²) sin 2x dx = dt

Question 16.
(i) ∫ \(\frac{x d x}{\sqrt{x^2+a^2}+\sqrt{x^2-a^2}}\)
(ii) ∫ \(\frac{x e^{\sqrt{x^2+3}}}{\sqrt{x^2+3}}\) dx
Solution:
(i) Let I = ∫ \(\frac{x d x}{\sqrt{x^2+a^2}+\sqrt{x^2-a^2}}\)

(ii) Let I = ∫ \(\frac{x e^{\sqrt{x^2+3}}}{\sqrt{x^2+3}}\) dx
put \(\sqrt{x^2+3}\) = t
⇒ \(\frac{1}{2}\left(x^2+3\right)^{-\frac{1}{2}}\) 2x dx = dt
⇒ \(\frac{x}{\sqrt{x^2+3}}\) dx = dt
⇒ I = ∫ e^t dt
= e^t + C
Thus, I = \(e^{\sqrt{x^2+3}}\) + C

Question 17.
(i) ∫ tan x sec² x \(\sqrt{1-\tan ^2 x}\) dx
(ii) ∫ sin³ x cos⁵ x dx
Solution:
(i) Let I = ∫ tan x sec² x \(\sqrt{1-\tan ^2 x}\) dx
put tan² x dx = t
⇒ 2 tan x sec² x dx = dt
∴ I = \(\frac{1}{2}\) ∫ \(\sqrt{1-t}\) dt
= \(\frac{1}{2} \frac{(1-t)^{3 / 2}}{\frac{3}{2}(-1)}\) + C
= – \(\frac{1}{3}\) (1 – tan² x)^3/2 + C

(ii) Let I = ∫ sin³ x cos⁵ x dx
= ∫ sin² x cos⁵ x sin x dx
= ∫ (1 – cos² x) cos⁵ x (sin x) dx
put cos x = t
⇒ – sin x dx = dt
= ∫ (1 – t²) t⁵ (- dt)
= – \(\left[\frac{t^6}{6}-\frac{t^8}{8}\right]\) + C
= \(-\frac{\cos ^6 x}{6}+\frac{\cos ^8 x}{8}\) + C

Question 18.
(i) ∫ \(\frac{(a+\sqrt{x})^n}{\sqrt{x}}\) dx, n ≠ – 1
(ii) ∫ \(\frac{d x}{\cos ^2 x \sqrt{\tan x-1}}\)
Solution:
(i) Let I = ∫ \(\frac{(a+\sqrt{x})^n}{\sqrt{x}}\) dx, n ≠ – 1
put √x = t
⇒ \(\frac{1}{2 \sqrt{x}}\) dx = dt
∴ I = ∫ (a + t)ⁿ (2 dt)
= \(\frac{2(a+t)^{n+1}}{n+1}\) + C
= \(\frac{2}{n+1}\) (a + √x)^{n + 1} + C ; n ≠ – 1

(ii) Let I = ∫ \(\frac{d x}{\cos ^2 x \sqrt{\tan x-1}}\)

Question 19.
(i) ∫ \(\frac{d x}{(2 \sin x+3 \cos x)^2}\)
(ii) ∫ \(\frac{d x}{x^{2 / 3} \sqrt{x^{2 / 3}-4}}\)
Solution:
(i) Let I = ∫ \(\frac{d x}{(2 \sin x+3 \cos x)^2}\)
Divide Numerator and deno. by cos² x ; we have
= ∫ \(\frac{\sec ^2 x d x}{(2 \tan x+3)^2}\)
put tan x = t
⇒ sec² x dx = dt
∴ I = ∫ \(\frac{d t}{(2 t+3)^2}\)
= ∫ (2t + 3)^-2 dt
= \(\frac{(2 t+3)^{-2+1}}{2 \cdot(-2+1)}\) + C
= \(\frac{1}{2(2 t+3)}\) + C
= – \(\frac{1}{2(2 \tan x+3)}\) + C

(ii) Let I = ∫ \(\frac{d x}{x^{2 / 3} \sqrt{x^{2 / 3}-4}}\)

Question 20.
(i) ∫ x \(\sqrt[3]{2 x+1}\) dx
(ii) ∫ \(\frac{2 x}{x^2+3 x+2}\) dx (NCERT)
Solution:
(i) Let I = ∫ x \(\sqrt[3]{2 x+1}\) dx
put \(\sqrt[3]{2 x+1}\) = t
⇒ 2x + 1 = t³
⇒ 2 dx = 3t² dt

(ii) Let I = ∫ \(\frac{2 x}{x^2+3 x+2}\) dx

Question 21.
(i) ∫ \(\frac{\sqrt{x}}{\sqrt{x}+2}\) dx
(ii) ∫ √x (log x)² dx
Solution:
(i) put √x + 2 = t
⇒ √x = t – 2
⇒ x = (t – 2)²
⇒ dx = 2 (t – 2) dt

= t² – 8t + 8 log |t| + C
= (√x + 2)² – 8 (√x + 2) + 8 log |t| + C
= x – 4√x + 8 log |√x + 2| + C’

(ii) Let I = ∫ √x (log x)² dx
put log x = t
⇒ x = e^t
⇒ dx = e^t dt
∴ I = ∫ e^t/2 t² e^t dt

Question 22.
(i) ∫ \(\frac{1}{x-x^3}\) dx (NCERT)
(ii) ∫ \(\frac{\cos x}{2+\cos ^2 x}\) dx
Solution:
(i) Let I = ∫ \(\frac{1}{x-x^3}\) dx

(ii) Let I = ∫ \(\frac{\cos x}{2+\cos ^2 x}\) dx

Question 23.
(i) ∫ \(\frac{x^2}{\left(x^2+a^2\right)\left(x^2+b^2\right)}\) dx (NCERT Exemplar)
(ii) ∫ \(\frac{e^x}{e^{2 x}+6 e^x+5}\) dx
Solution:
(i) put x² = y

(ii) Let I = ∫ \(\frac{e^x}{e^{2 x}+6 e^x+5}\) dx ;
put e^x = t
⇒ e^x dx = dt

Question 24.
(i) ∫ \(\frac{\sin x}{(2+\cos x)(5+\cos x)}\) dx
(ii) ∫ \(\frac{x^4}{(x-1)\left(x^2+1\right)}\) dx
Solution:
(i) I = ∫ \(\frac{\sin x}{(2+\cos x)(5+\cos x)}\) dx
put cos x = t
⇒ – sin x dx = dt

(ii) Let I = ∫ \(\frac{x^4}{(x-1)\left(x^2+1\right)}\) dx

Question 25
(i) ∫ cos 2x log (1 + tan x) dx
(ii) ∫ \(\frac{e^{\log \left(1+\frac{1}{x^2}\right)}}{x^2+\frac{1}{x^2}}\) dx
Solution:
(i) Let I = ∫ cos 2x log (1 + tan x) dx

sin x = l (sin x + cos x) + m \(\frac{d}{d x}\) (sin x + cos x)
⇒ sin x = l (sin x + cos x) + m (cos x – sin x)
Coeff. of sin x ;
1 = l – m ;
Coeff. of cos x ;
0 = l + m
On solving these eqn.’s ;
l = \(\frac{1}{2}\) ; m = – \(\frac{1}{2}\)

(ii) Let I = ∫ \(\frac{e^{\log \left(1+\frac{1}{x^2}\right)}}{x^2+\frac{1}{x^2}}\) dx

Question 25 (old).
(ii) ∫ \(\frac{1-\cos x}{\cos x(1+\cos x)}\) dx
Solution:
Let I = ∫ \(\frac{1-\cos x}{\cos x(1+\cos x)}\) dx

Question 26.
(i) ∫ \(\frac{x^3 \sin ^{-1}\left(x^4\right)}{\sqrt{1-x^4}}\) dx
(ii) ∫ sin⁴ x cos⁵ x dx
Solution:
(i) Let I = ∫ \(\frac{x^3 \sin ^{-1}\left(x^4\right)}{\sqrt{1-x^4}}\) dx ;
put x⁴ = t
⇒ 4x³ dx = dt

(ii) Let I = ∫ sin⁴ x cos⁵ x dx
= ∫ sin⁴ x cos⁴ x cos x dx
= ∫ sin⁴ x (1 – sin² x)⁴ cos x dx
put sin x = t
⇒ cos x dx = dt
∴ I = ∫ t⁴ (1 – t²)² dt
= ∫ t⁴ (t⁴ – 2t² + 1) dt
= \(\frac{t^9}{9}-2 \frac{t^7}{7}+\frac{t^5}{5}\) + C
= \(\frac{\sin ^9 x}{9}-\frac{2}{7} \sin ^7 x+\frac{1}{5} \sin ^5 x\) + C

Question 27.
(i) ∫ \(\frac{d x}{x^{1 / 2}-x^{1 / 4}}\)
(ii) ∫ tan^-1 \(\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)\) dx
Solution:
(i) Let I = \(\frac{d x}{x^{1 / 2}-x^{1 / 4}}\)
put x^1/4 = t
⇒ x = t⁴
⇒ dx = 4t³ dt

= 2t² + 4t + 4 log |t – 1| + C
= 2√x + 4x^1/4 + 4 log |x^1/4 – 1| + C

(ii) Let I = ∫ tan^-1 \(\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)\) dx
put x = cos 2t
⇒ dx = – 2 sin 2t dt

Question 28.
(i) ∫ \(\frac{\log (x+1)-\log x}{x(x+1)}\) dx
(ii) ∫ \(\frac{x^2-x+1}{\left(x^2+1\right)^{3 / 2}}\) e^x dx
Solution:
(i) Let I = ∫ \(\frac{\log (x+1)-\log x}{x(x+1)}\) dx ;
put log (x + 1) – log x = t

(ii) Let I = ∫ \(\frac{x^2-x+1}{\left(x^2+1\right)^{3 / 2}}\) e^x dx

Question 29.
(i) ∫ x³ e^{– x2} dx
(ii) ∫ sin^-1 \(\left(\frac{2 x+2}{\sqrt{4 x^2+8 x+13}}\right)\) dx
Solution:
(i) Let I = ∫ x³ e^{– x2} dx ;
put x² = t
⇒ 2x dx = dt
∴ I = ∫ t e^-t \(\frac{d t}{2}\)
= \(\frac{1}{2}\left[t \frac{e^{-t}}{(-1)}-\int 1 \times \frac{e^{-t}}{(-1)} d t\right]\) + C
∴ I = \(\frac{1}{2}\) [- t e^-t – e^-t] + C
= – \(\frac{1}{2}\) (t + 1) e^-t + C
= – \(\frac{1}{2}\) (x² + 1) e^{– x2} + C

(ii) Let I = ∫ sin^-1 \(\left(\frac{2 x+2}{\sqrt{4 x^2+8 x+13}}\right)\) dx
first of all, we convert sin^-1 to tan^-1 with p = 2x + 2;
h = \(\sqrt{4 x^2+8 x+3}\)

Question 30.
(i) ∫ f'(ax + b) (f (ax + b))ⁿ dx (NCERT)
(ii) ∫ \(\frac{\sqrt{1-x}}{x^{7 / 2}}\) dx
Solution:
(i) Case – I :
When n ≠ – 1
Let I = ∫ f'(ax + b) (f (ax + b))ⁿ dx
put f (ax + b) = t
⇒ f’ (ax + b) . a dx = dt
∴ I = ∫ \(\frac{1}{a}\) tⁿ dt
= \(\frac{1}{a} \frac{t^{n+1}}{n+1}\) + C ; n ≠ – 1
∴ I = \(\frac{(f(a x+b))^{n+1}}{(n+1) a}\) + C ; n ≠ – 1

Case – II :
When n = – 1
I = ∫ \(\frac{f^{\prime}(a x+b) d x}{f(a x+b)}\)
put f (ax + b) = t
⇒ f’ (ax + b) . a dx = dt
= ∫ \(\frac{d t}{a t}\)
= \(\frac{1}{a}\) log |t| + C
= \(\frac{1}{a}\) log |f (ax + b)| + C

(ii) Let I = ∫ \(\frac{\sqrt{1-x}}{x^{7 / 2}}\) dx

Question 31.
(i) ∫ \(\sqrt{4-x+x^2}\) dx
(ii) ∫ \(\frac{\tan ^{-1} x}{x^2}\) dx
(iii) ∫ \(\frac{\log |x|}{(x+1)^3}\) dx
Solution:
(i) Let I = ∫ \(\sqrt{4-x+x^2}\) dx

(ii) Let I = ∫ \(\frac{\tan ^{-1} x}{x^2}\) dx
put tan^-1 x = θ
⇒ x = tan θ
⇒ dx = sec² θ
= ∫ \(\frac{\theta}{\tan ^2 \theta}\) sec² θ dθ
= ∫ θ cosec² θ dθ
= θ (- cot θ) – ∫ (- cot θ) dθ + C
= – θ cot θ + log |sin θ| + C
= \(-\frac{\tan ^{-1} x}{x}+\log \left|\frac{x}{\sqrt{1+x^2}}\right|\) + C

(iii) Let I = ∫ \(\frac{\log |x|}{(x+1)^3}\) dx
= ∫ log |x| . \(\frac{1}{(x+1)^3}\) dx

Multiplying both sides of eqn. (2) by x(x + 1)²; we get
1 = A(x + 1)² + Bx (x + 1) + Cx ………….(3)
putting x = 0, – 1 successively in eqn. (3); we have
1 = A
and 1= – C
⇒ C = – 1
Coeff. of x²;
0 = A + B,
⇒ B = – 1
∴ I₁ = ∫ \(\left[\frac{1}{x}-\frac{1}{x+1}-\frac{1}{(x+1)^2}\right]\) dx
= log |x| – log |x + 1| + \(\frac{1}{x+1}\)
∴ from (1); we have
= \(-\frac{\log |x|}{2(x+1)^2}+\frac{1}{2} \log \left|\frac{x}{x+1}\right|+\frac{1}{2(x+1)}\) + C

Question 32.
Evaluate the following definite integrals as the limit of sums :
(i) \(\int_1^3\) (x² + 5x) dx
(ii) \(\int_0^1\) e^{2 – 3x} dx (NCERT)
Solution:
(i) On comparing \(\int_1^3\) (x² + 5x) dx with \(\int_a^b\) f(x) dx
Here f(x) = x² + 5x ;
a = 1 ; b = 3
∴ nh = b – a
= 3 – 1 = 2
∴ f(a) = f(1)
= 1² + 5
f(a + h) = f(1 + h)
= (1 + h)² + 5 (1 + h)
f(a + 2h) = f(1 + 2h)
= (1 + 2h)² + 5 (1 + 2h)
……………………………………
……………………………………

(ii) On comparing\(\int_0^1\) e^{2 – 3x} dx with \(\int_a^b\) f(x) dx
Here, f(x) = e^2-3x;
a = 0;
b = 1;
nh = b – a = 1 – 0 = 1
∴ f(a) = f(0) = e²
f(a + h) = f(h) = e^{2 – 3h}
f(a + 2h) = f(2h) = e^{2 – 6h}
……………………………
……………………………
\(f(a+\overline{n-1} h)\) = f((n – 1)h)
= e^{2 – 3 (n – 1) h}
∴ \(\int_0^1\) e^{2 – 3h} dx = \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) h[f(a) + f(a + h) + f(a + 2h) ………..+ \(f(a+\overline{n-1} h)\)]
= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) h [e² + e^{2 – 3h} + e^{2 – 6h} + ……. + e^{2 – 3(n – 1) h}]
= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) h[1 + e^{– 3h} + e^{– 6h} ……… n terms]

Evaluate the following (33 to 36) definite integrals:

Question 33.
(i) \(\int_0^{\pi / 4}\) tan² x dx
(ii) \(\int_0^2 \frac{5 x+1}{x^2+4}\) dx
Solution:
(i) \(\int_0^{\pi / 4}\) tan² x dx
= \(\int_0^{\pi / 4}\) (sec² – 1) dx
= tan x – x\(]_0^{\pi / 4}\)
= (tan \(\frac{\pi}{4}\) – \(\frac{\pi}{4}\)) – (0 – 0)
= 1 – \(\frac{\pi}{4}\)

(ii) \(\int_0^2 \frac{5 x+1}{x^2+4}\) dx

Question 34.
(i) \(\int_0^{\pi / 6}\) (2 + 3x²) cos 3x dx
(ii) \(\int_0^1\) (cos^-1 x)² dx
Solution:
(i) \(\int_0^{\pi / 6}\) (2 + 3x²) cos 3x dx

(ii) Let I = \(\int_0^1\) (cos^-1 x)² dx
put cos^-1 x = t
⇒ – \(\frac{1}{\sqrt{1-x^2}}\) dx = dt
⇒ x = cos t
⇒ dx = – sin t dt
When x = 0
⇒ t = \(\frac{\pi}{2}\) ;
When x = 1
⇒ t = 0
∴ I = \(\int_{\pi / 2}^0\) t² (- sin t) dt
= – [- t² cos t\(\}_{\pi / 2}^0\) + \(\int_{\pi / 2}^0\) t cos t dt]
∴ I = (0 × 1 – \(\frac{\pi^2}{4}\) × 0) – 2 [t sin t\(\}_{\pi / 2}^0\) – \(\int_{\pi / 2}^0\) sin t dt]
∴ I = – 2 [t sin t + cos t\(]_{\pi / 2}^0\)
= + 2 [t sin t + cos t\(t]_0^{\pi / 2}\)
= + 2 [\(\frac{\pi}{2}\) × 1 + 0 – 0 – 1]
= + 2 (\(\frac{\pi}{2}\) – 1)
= π – 2

Question 35.
(i) \(\int_0^\pi\) x sin x cos² x dx (NCERT Exemplar)
(ii) \(\int_0^{\pi / 2}\) \(\sqrt{sin x}\) cos⁵ x dx (NCERT)
Solution:
(i) Let I = \(\int_0^\pi\) x sin x cos² x dx ……………(1)
∴ I = \(\int_0^\pi\) (π – x) sin (π – x) cos² (π – x) dx
I = \(\int_0^\pi\) (π – x) sin x cos² x dx ……………(2)
[∵ \(\int_0^a\) f(x) dx = \(\int_0^a\) f(a – x) dx]
On adding (1) and (2) ; we have
2I = \(\int_0^\pi\) π sin x cos² x dx
put cos x = t
⇒ – sin x dx = dt
When x = 0 ⇒ t = 1 ;
When x = π ⇒ t = – 1
⇒ 2I = π \(\int_1^{-1}\) t² (- dt)
= – π \(\left.\frac{t^3}{3}\right]_1^{-1}\)
= – \(\frac{\pi}{3}\) (- 1 – 1)
= \(\frac{2 \pi}{3}\)
Thus, I = \(\frac{\pi}{3}\)

(ii) Let I = \(\int_0^{\pi / 2}\) \(\sqrt{sin x}\) cos⁵ x dx
= \(\int_0^{\pi / 2}\) \(\sqrt{sin x}\) (1 – sin² x)² cos x dx
put sin x = t ⇒ cos x dx = dt
When x = 0 ⇒ t = 0 ;
When x = \(\frac{\pi}{2}\) ⇒ t = 1
∴ I = \(\int_0^1\) √t (1 – t²)² dt
= \(\int_0^1\) √t (t⁴ – 2t² + 1) dt
= \(\left.\frac{2 t^{11 / 2}}{11}-\frac{2 t^{7 / 2}}{7 / 2}+\frac{t^{3 / 2}}{3 / 2}\right]_0^1\)
= \(\left[\frac{2}{11}-\frac{4}{7}+\frac{2}{3}\right]\)
= \(\frac{42-132+154}{231}=\frac{64}{231}\).

Question 36.
(i) \(\int_0^{\pi / 4} \frac{\sin x \cos x}{\cos ^2 x+\sin ^4 x}\) dx
(ii) \(\int_0^{\pi / 4} \frac{\sin x+\cos x}{\cos ^2 x+\sin ^4 x}\) dx
Solution:
(i) I = \(\int_0^{\pi / 4} \frac{\sin x \cos x}{\cos ^2 x+\sin ^4 x}\) dx

(ii) Let I = \(\int_0^{\pi / 4} \frac{\sin x+\cos x}{\cos ^2 x+\sin ^4 x}\) dx

put sin x – cos x = t
⇒ (cos x + sin x) dx = dt
On squaring; we have
(sin x – cos x)² = t²
⇒ sin² x + cos² x – sin 2x = t²
⇒ 1 – sin 2x = t²
⇒ sin 2x = 1 – t²
When x = 0 ⇒ t = – 1 ;
When x = \(\frac{\pi}{4}\) ⇒ t = 0
∴ I = 4 \(\int_{-1}^0 \frac{d t}{4-\left(1-t^2\right)^2}\) ………………(1)
= 4 \(\int_{-1}^0 \frac{d t}{\left(2+1-t^2\right)\left(2-1+t^2\right)}\)
⇒ I = 4 \(\int_{-1}^0 \frac{d t}{\left(3-t^2\right)\left(1+t^2\right)}\)
put t² = y
Then \(\frac{1}{\left(3-t^2\right)\left(1+t^2\right)}=\frac{1}{(3-y)(1+y)}=\frac{\mathrm{A}}{3-y}+\frac{\mathrm{B}}{1+y}\) …………….(2)
Multiply both sides of eqn. (2) by (3 – y) (1 + y) ; we get
1 = A (1 + y) + B (3 – y) ………….(3)
putting y = – 1, 3 successively in eqn. (3) ; we have
1 = 4B
⇒ B = \(\frac{1}{4}\)
and 1 = 4A
⇒ A = \(\frac{1}{4}\)
∴ from (2) ; we get ;

By using properties of definite integrals, evaluate the following (37 to 39) :

Question 37.
\(\int_0^4\) (|x| + |x – 2| + |x – 4|
Solution:
Let f(x) = |x| + |x – 2| + |x – 4|
The critical points of f(x) are given by x = 0,2, 4
When 0 ≤ x < 2
⇒ |x| = x ;
x – 2 < 0
⇒ |x – 2| = – (x – 2)
and x – 4< 0
⇒ |x – 4| = – (x – 4)
∴ f(x) = x – (x – 2) – (x – 4) = – x + 6
When 2 ≤ x < 4
Then x – 2 ≥ 0;
x – 4 < 0; x > 0
∴ f(x) = x + x – 2 – (x – 4) = x+ 2
∴ \(\int_0^4\) f(x) dx = \(\int_0^2\) f(x) dx + \(\int_2^4\) f(x) dx
= \(\int_0^2\) (- x + 6) dx + \(\int_2^4\) (x + 2) dx
= \(\left.\left[-\frac{x^2}{2}+6 x\right]_0^2+\frac{(x+2)^2}{2}\right]_2^4\)
= [(- 2 + 12) – (0 + 0)] + \(\frac{1}{2}\) [36 – 16]
= 10 + 10 = 20.

Question 38.
(i) \(\int_{1 / e}^e\) |log x| dx
(ii) \(\int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}\)
(iii) \(\int_0^{\pi / 2} \frac{\sqrt{\cot x}}{\sqrt{1+\cot x}}\) dx
Solution:
(i) When \(\frac{1}{e}\) ≤ x ≤ 1 ;
log x ≤ 0
∴ |log x| = – log x
When 1 ≤ x ≤ e ;
log x ≥ 0
∴ |log x| = + log x

(ii) Let I = \(\int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}\) …………….(1)

(iii) Let I = \(\int_0^{\pi / 2} \frac{\sqrt{\cot x}}{\sqrt{1+\cot x}}\) dx

Question 39.
\(\int_0^1\) x (1 – x)⁵ dx
Solution:
Let I = \(\int_0^1\) x (1 – x)⁵ dx
∴ I = \(\int_0^1\) (1 – x) (1 – (1 – x))⁵ dx
[∵ \(\int_0^a\) f(x) dx = \(\int_0^a\) f (a – x) dx]
I = \(\int_0^1\) (1 – x) x⁵ dx
= \(\left[\frac{x^6}{6}-\frac{x^7}{7}\right]_0^1\)
= \(\left(\frac{1}{6}-\frac{1}{7}-0-0\right)\)
= \(\frac{1}{42}\)

(ii) Let I = \(\int_0^{\pi / 2} \frac{\sin ^2 x}{1+\sin x \cos x}\) dx

Question 40.
(i) \(\int_{-\pi / 2}^{\pi / 2}\) (x³ + x cos x + tan⁵ x + 1) dx
(ii) \(\int_0^{\pi / 2} \frac{\sin ^2 x}{1+\sin x \cos x}\) dx
Solution:
(i) Here f(x) = x³ + x cos x + tan⁵ x + 1
∴ f (- x) = (- x)³ + (- x) cos (- x) + [tan (- x)⁵] + 1
= – x³ – x cos x – tan⁵ x + 1
So f(- x) is neither equal to f(x) nor equal to – f(x)
Let I = \(\int_{-\pi / 2}^{\pi / 2}\) [x³ + x cos x + tan⁵ x + 1] dx
= \(\int_{-\pi / 2}^{\pi / 2} x^3 d x+\int_{-\pi / 2}^{\pi / 2} x \cos x+\int_{-\pi / 2}^{\pi / 2} \tan ^5 x+\int_{-\pi / 2}^{\pi / 2} 1 d x\) …………..(1)
Let f(x) = x³ ;
f(- x) = (- x)³
= – x³
= – f(x)
∴ \(\int_{-\pi / 2}^{\pi / 2}\) f(x) = \(\int_{-\pi / 2}^{\pi / 2}\) x³ = 0 ……………….(2)
[∵ \(\int_{-a}^a\) f(x) = 0 if f(- x) = – f(x)]
Let g(x) = x cos x
∴ g(- x) = – x cos (- x)
= – x cos x
= – g(x)
∴ g(x) be an odd function.
\(\int_{-\pi / 2}^{\pi / 2}\) g(x) dx = 0
⇒ \(\int_{-\pi / 2}^{\pi / 2}\) x cos x dx = 0 ……………(3)
Let h(x) = tan⁵ x
⇒ h (- x) = [tan (- x)]⁵
= – tan⁵ x
= – h(x)
∴ \(\int_{-\pi / 2}^{\pi / 2}\) h(x) dx = 0
⇒ \(\int_{-\pi / 2}^{\pi / 2}\) tan⁵ x dx = 0 ……………….(4)
Using eqn. (2), (3) and (4) in eqn. (1) ; we get
I = 0 + 0 + 0 + x\(]_{-\pi / 2}^{\pi / 2}\)
= \(\frac{\pi}{2}+\frac{\pi}{2}\)
= π

(ii) Let I = \(\int_0^{\pi / 2} \frac{\sin ^2 x}{1+\sin x \cos x}\) dx …………….(1)

Question 45 (old).
(i) \(\int_0^1 \tan ^{-1}\left(\frac{2 x-1}{1+x-x^2}\right)\) dx
Solution:
Let I = \(\int_0^1 \tan ^{-1}\left(\frac{2 x-1}{1+x-x^2}\right)\) dx
= \(\int_0^1 \tan ^{-1}\left(\frac{x+(x-1)}{1-x(x-1)}\right)\) dx
= \(\int_0^1\) [tan^-1 x + tan^-1 (x – 1)] dx
I = \(\int_0^1\) [tan^-1 x + tan^-1 (x – 1)] dx …………….(1)
∴ I = \(\int_0^1\) [tan^-1 (1 – x) + tan^-1 (1 – x – 1)] dx
[∵ \(\int_0^a\) f(x) dx = \(\int_0^a\) f (a – x) dx]
⇒ I = \(\int_0^1\) [tan^-1 (1 – x) + tan^-1 (- x)] dx
⇒ I = \(\int_0^1\) [- tan^-1 (x – 1) – tan^-1 x] dx …………….(2)
[∵ tan^-1 (- x) = – tan^-1 x]
On adding (1) and (2) ; we have
2I = 0
⇒ I = 0.

Continuous practice using Class 12 ISC Maths Solutions Chapter 8 Integrals Ex 8.19 can lead to a stronger grasp of mathematical concepts.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.19

Evaluate the following (1 to 9) integrals :

Question 1.
(i) ∫ \(\frac{x^5}{\sqrt{1+x^3}}\) dx
(ii) ∫ \(\frac{d x}{\sqrt{2 e^x-1}}\)
Solution:
(i) Let I = ∫ \(\frac{x^5}{\sqrt{1+x^3}}\) dx
put \(\sqrt{1+x^3}\) = t
⇒ 1 + x³ = t²
⇒ x³ = t² – 1
⇒ 3x² dx = 2t dt
∴ I = ∫ \(\frac{\left(t^2-1\right) 2 t d t}{3 t}\)
= \(\frac{2}{3}\) ∫ (t² – 1) dt
= \(\frac{2}{3}\left[\frac{t^3}{3}-t\right]\)
= \(\frac{2}{9}\) (1 + x³)^3/2 – \(\frac{2}{3} \sqrt{1+x^3}\) + C

(ii) Let I = ∫ \(\frac{d x}{\sqrt{2 e^x-1}}\)
put \(\sqrt{2 e^x-1}\) = t
⇒ 2 e^x – 1 = t²
⇒ 2 e^x dx = 2t dt
∴ I = ∫ \(\frac{t d t}{e^x \cdot t}\)
= ∫ \(\frac{\frac{d t}{t^2+1}}{\frac{2}{2}}\)
= 2 ∫ \(\frac{d t}{t^2+1^2}\)
= 2 tan^-1 t + C
= 2 tan^-1 \(\left(\sqrt{2 e^x-1}\right)\) + C

Question 2.
(i) ∫ \(\frac{d x}{\tan x+\cot x+\sec x+\ {cosec} x}\)
(ii) ∫ \(\frac{d x}{\sec x+\ {cosec} x}\)
Solution:
(i) ∫ \(\frac{d x}{\tan x+\cot x+\sec x+\ {cosec} x}\)

(ii) ∫ \(\frac{d x}{\sec x+\ {cosec} x}\)

Question 3.
(i) ∫ \(\frac{\cos 7 x-\cos 8 x}{1+2 \cos 5 x}\) dx
(ii) ∫ \(\frac{\cos 5 x+\cos 4 x}{1-2 \cos 3 x}\) dx (NCERT Exemplar)
Solution:
(i) Let I = ∫ \(\frac{\cos 7 x-\cos 8 x}{1+2 \cos 5 x}\) dx

(ii) Let I = ∫ \(\frac{\cos 5 x+\cos 4 x}{1-2 \cos 3 x}\) dx

Question 4.
(i) ∫ \(\frac{d x}{2 \sin x+3 \sec x}\)
(ii) ∫ \(\frac{d x}{\sin ^3 x+\cos ^3 x}\)
Solution:
(i) Let I = ∫ \(\frac{d x}{2 \sin x+3 \sec x}\)

(ii) Let I = ∫ \(\frac{d x}{\sin ^3 x+\cos ^3 x}\)

Question 5.
(i) ∫ \(\frac{x^2-3}{x^3-2 x^2-x+2}\) dx
(ii) ∫ x tan^-1 (2x + 3) dx
Solution:
(i) Let I = ∫ \(\frac{x^2-3}{x^3-2 x^2-x+2}\) dx
= ∫ \(\frac{\left(x^2-3\right) d x}{(x-1)\left(x^2-x-2\right)}\)
= ∫ \(\frac{\left(x^2-3\right) d x}{(x-1)(x+1)(x-2)}\)
Let \(\frac{x^2-3}{(x-1)(x+1)(x-2)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x-2}\) …………..(1)
Multiply both sides ofeqn. (1) by (x- 1) (x + 1) (x – 2); we get
x² – 3 = A (x + 1) (x – 2) + B (x – 1) (x – 2) + C (x – 1) (x + 1) …………..(2)
putting x = – 1, 1, 2 successively in eqn. (2) ; we have
– 2 = B (- 2) (- 3)
⇒ B = – \(\frac{1}{3}\)
– 2 = A (2) (- 1)
⇒ A = 1
and1 = 3C
⇒ C = \(\frac{1}{3}\)
∴ from (1) ; we get

(ii) Let I = ∫ x tan^-1 (2x + 3) dx

Question 6.
(i) ∫ \(\frac{\tan ^{-1} x}{x^2}\) dx
(ii) ∫ \(\frac{\log |x|}{(x+1)^3}\) dx
Solution:
(i) Let I = ∫ tan^-1 x . \(\frac{1}{x^2}\) dx

(ii) Let I = ∫ \(\frac{\log |x|}{(x+1)^3}\) dx

Multiplying eqn.(2) by x (x + 1)²;
we have 1 = A (x + 1)² + B x (x + 1) + Cx ………….(3)
puuing x = 0, – 1 successively in eqn.(3);
we have 1 = A and 1 = – C
⇒ C = – 1
Coeff. of x² ;
0 = A + B
⇒ B = – 1
∴ from eqn. (2); we get

Question 7.
(i) ∫ e^x (log x + \(\frac{1}{x^2}\)) dx
(ii) ∫ cos 2x log \(\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\) dx
Solution:
(i) Let I = ∫ e^x (log x + \(\frac{1}{x^2}\)) dx

(ii) Let I = ∫ cos 2x log \(\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\) dx [by parts]

Question 8.
(i) ∫ \(\frac{x^2-x+1}{\left(x^2+1\right)^{3 / 2}}\) e^x dx
(ii) ∫ \(\frac{x^2-1}{\left(x^4+3 x^2+1\right) \tan ^{-1}\left(x+\frac{1}{x}\right)}\) dx
Solution:
(i) Let I = ∫ \(\frac{x^2-x+1}{\left(x^2+1\right)^{3 / 2}}\) e^x dx

(ii) Let I = ∫ \(\frac{x^2-1}{\left(x^4+3 x^2+1\right) \tan ^{-1}\left(x+\frac{1}{x}\right)}\) dx
Divide numerator and denominator by x² ; we have

Question 9.
\(\int_0^\pi\) |cos x – sin x| dx
Solution:
Let I = \(\int_0^\pi\) |cos x – sin x| dx
When 0 ≤ x ≤ \(\frac{\pi}{4}\) ;
cos x ≥ sin x
⇒ cos x – sin x ≥ 0
∴ |cos x – sin x| = cos x – sin x
When \(\frac{\pi}{4}\) ≤ x ≤ 7 ;
sin x ≥ cos x
⇒ cos x – sin x ≤ 0
∴ |cos x – sin x| = – (cos x – sin x)
∴ I = \(\int_0^{\pi / 4}\) |cos x – sin x| dx + \(\int_{\pi / 4}^\pi\) |cos x – sin x| dx
= \(\int_0^{\pi / 4}\) (cos x – sin x) dx + \(\int_{\pi / 4}^\pi\) – (cos x – sin x) dx
= [sin x + cos x\(]_0^{\pi / 4}\) – [sin x + cos x\(]_{\pi / 4}^\pi\)
= \(\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-0-1\right]-\left[0-1-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right]\)
= (√2 – 1) – (- 1 – √2)
= 2√2

Question 10.
Prove that \(\int_0^\pi \frac{x}{1-\cos \alpha \sin x} d x=\frac{\pi(\pi-\alpha)}{\sin \alpha}\).
Solution:
Let I = \(\int_0^\pi \frac{x}{1-\cos \alpha \sin x}\) dx ……………(1)

Question 11.
Prove that \(\int_0^{\pi / 2} \frac{f(\sin x)}{f(\sin x)+f(\cos x)} d x=\frac{\pi}{4}\).
Solution:
Let I = \(\int_0^{\pi / 2} \frac{f(\sin x)}{f(\sin x)+f(\cos x)} d x=\frac{\pi}{4}\)

Question 12.
Evaluate \(\int_0^2 \frac{d x}{\left(17+8 x-4 x^2\right)\left(e^{6(1-x)}+1\right)}\).
Solution:
Let I = \(\int_0^2 \frac{d x}{\left(17+8 x-4 x^2\right)\left(e^{6(1-x)}+1\right)}\) ……………(1)

Question 13.
If p (x) is a polynomial of least degree that has maximum value equal to 6 at x = 1, and a minimum value equal to 2 at x = 3, then show that \(\int_0^1\) p(x) dx = \(\frac{19}{4}\).
Solution:
Since p (x) be a polynomial of least degree
and it has maximum and minimum values.
∴ p’ (x) must be atleast polynomial of degree 2.
Thus p (x) must be a polynomial of degree 3.
Let p(x) = ax³ + bx² + cx + d ……………….(1)
∴ p’(x) = 3ax² + 2bx + c
Since p (x) has maximum values equal to 6 at x = 1
and a minimum value equal to 2 at x = 3.
∴ p'(1) = 0 – p'(3)
⇒ 3a + 2b + c = 0 …………..(2)
and 27a + 6b + c = 0 ……………(3)
also p(1) 6
⇒ a + b + c + d= 6 …………(4)
and p (3) = 2
⇒ 27a + 9b + 3c + d = 2 …………….(5)
eqn. (5) – eqn. (4) gives:
26a + 8b + 2c = – 4
⇒ 13a + 4b + c = – 2 ……………..(6)
eqn. (3) – eqn. (2) gives;
24a + 4b = 0
⇒ 6a + b = 0 …………….(7)
eqn. (3) – eqn. (6) gives
14a + 2b = 2
⇒ 7a + b = 1 ……………(8)
eqn. (8) – eqn. (7) gives;
a = 1
∴ from (7);
b = – 6
∴ from eqn.(6);
13 – 24 + c = – 2
⇒ c = 9
from eqn (4) ;
1 – 6 + 9 + d = 6
⇒ d = 2
∴ from (1) ;
p(x) = x³ – 6x² + 9x + 2
∴ \(\int_0^1\) p(x) dx = \(\int_0^1\) (x³ – 6x² + 9x + 2) dx
= \(\left[\frac{x^4}{4}-2 x^3+\frac{9 x^2}{2}+2 x\right]_0^1\)
= \(\left[\frac{1}{4}-2+\frac{9}{2}+2\right]=\frac{19}{4}\)

Effective ISC Maths Class 12 Solutions Chapter 8 Integrals Ex 8.18 can help bridge the gap between theory and application.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.18

Very Short answer type questions:

Evaluate the following (1 to 5) definite integrals :

Question 1.
(i) \(\int_{-1}^1\) |x| dx
(ii) \(\int_0^3\) |x – 2| dx
(iii) \(\int_0^2\) dx
(iv) \(\int_{-1}^1 \frac{|x+2|}{x+2}\) dx
(v) \(\int_{-3}^6 \frac{x+3}{|x+3|}\) dx
(vi) \(\int_0^3\) [x] dx
Solution:
(i) When – 1 ≤ x ≤ 0 ⇒ |x| = – x
When 1 ≥ x ≥ 0 ⇒ |x| = x
∴ \(\int_{-1}^1\) |x| dx = \(\int_{-1}^0\) |x| dx + \(\int_0^1\) |x| dx
= \(\int_{-1}^0\) – x dx + \(\int_0^1\) x dx
= \(\left.\left.-\frac{x^2}{2}\right]_{-1}^0+\frac{x^2}{2}\right]_0^1\)
= – \(\frac{1}{2}\) (0 – 1) + \(\frac{1}{2}\) (1 – 0)
= \(\frac{1}{2}+\frac{1}{2}\) = 1

(ii) \(\int_0^3\) |x – 2| dx
= \(\int_0^2|x-2| d x+\int_2^3|x-2| d x\)
When 0 ≤ x ≤ 2
⇒ x – 2 ≤ 0
When 2 ≤ x ≤ 3, x – 2 ≥ 0
⇒ |x – 2| = x – 2
∴ \(\int_0^3\) |x – 2| dx = \(\int_0^2\) – (x – 2) dx + \(\int_2^3\) (x – 2) dx
= \(\left.\left.-\frac{(x-2)^2}{2}\right]_0^2+\frac{(x-2)^2}{2}\right]_2^3\)
= – \(\frac{1}{2}\) (0 – 4) + \(\frac{1}{2}\) (1 – 0)
= 2 + \(\frac{1}{2}\)
= \(\frac{5}{2}\)

(iii) When 0 ≤ x ≤ 1
⇒ x – 1 ≤ 0
⇒ |x – 1| = – (x + 1)
When 1 ≤ x ≤ 2
⇒ x – 1 ≥ 0
⇒ |x – 1| = x – 1

= – (1 – 0) + (2 – 1)
= – 1 + 2 – 1 = 0

(iv) Let I = \(\int_{-1}^1 \frac{|x+2|}{x+2}\) dx
When – 1 ≤ x ≤ 1
⇒ 1 ≤ x + 2 ≤ 3
⇒ x + 2 > 0
⇒ |x + 2| = x + 2
= \(\left.\int_{-1}^1 \frac{x+2}{x+2} d x=x\right]_{-1}^1\)
= 1 – (- 1) = 2

(v) Let I = \(\int_{-3}^0 \frac{x+3}{|x+3|}\) dx
When – 3 ≤ x ≤ 0
⇒ x + 3 ≥ 0
⇒ |x + 3| = x + 3
= \(\left.\int_{-3}^0 \frac{x+3}{x+3} d x=x\right]_{-3}^0\)
= 0 – (- 3)
= 0 + 3 = 3

(vi) Let I = \(\int_0^3\) [x] dx
[x] = \(\left\{\begin{array}{lll}
0 & ; & 0 \leq x<1 \\
1 & ; & 1 \leq x<2 \\
2 & ; & 2 \leq x<3
\end{array}\right.\)
= \(\int_0^1[x] d x+\int_1^2[x] d x+\int_2^3[x] d x\)
= \(\int_0^1 0 d x+\int_1^2 d x+\int_2^3 2 d x\)
= 0 + 1 (2 – 1) + 2 (3 – 2) = 3.

Question 2.
(i) \(\int_{-3}^1 \frac{|x+2|}{x+2}\) dx
(ii) \(\int_{-1}^1\) |x| dx
(iii) \(\int_0^{1.5}\) |x| dx
Solution:
(i) When – 3 ≤ x ≤ – 2, x + 2 ≤ 0
∴ |x + 2| = – (x + 2)
When – 2 ≤ x ≤ 1, x + 2 ≥ 0
∴ |x + 2| = x + 2

= – (- 2 + 3) + (1 + 2)
= – 1 + 3
= 2

(ii) When – 1 ≤ x ≤ 0
∴ [x] = – 1
When 0 ≤ x < 1
∴ [x] = 0
Since [x] is discontinuous at x = 0
∴ \(\int_{-1}^1[x] d x=\int_{-1}^0[x] d x+\int_0^1[x] d x\)
= \(\int_{-1}^0-1 d x+\int_0^1 0 \cdot d x\)
= – 1 (0 + 1) + 0 (1 – 0) = – 1

(iii) When 0 ≤ x < 1
∴ [x] = 0
and When 1 ≤ x < 1.5
∴ [x] = 1
and [x] is discontinuous at x = 1
∴ \(\int_0^{1.5}[x] d x=\int_0^1[x] d x+\int_1^{1.5}[x] d x\)
= \(\int_0^1 0 d x+\int_1^{1 \cdot 5} 1 \cdot d x\)
= 0 (1 – 0) + (1.5 – 1) = 0.5

Question 3.
(i) \(\int_{-\pi / 2}^{\pi / 2}\) sin⁵ x dx = 0
(ii) \(\int_{-\pi / 2}^{\pi / 2}\) x cos² x dx
(iii) \(\int_{-\pi / 3}^{\pi / 3}\) (x² sin x + tan³ x) dx
Solution:
(i) Here f(x) = sin⁵ x
∴ f(- x) = sin⁵ (- x)
= [sin (- x)]⁵
= [- sin x]⁵
= – sin⁵ x
= – f(x)
Thus f(x) is an odd function
∴ \(\int_{-\pi / 2}^{\pi / 2}\) sin⁵ x dx = 0

(ii) Let I = \(\int_{-\pi / 2}^{\pi / 2}\) x cos² x dx
Here f(x) = x cos² x dx
∴ f(- x) = – x (cos (- x))²
= – x cos² x
= – f(x)
∴ \(\int_{-a}^a\) f(x) dx = 0
⇒ \(\int_{-\pi / 2}^{\pi / 2}\) x cos² x dx = 0

(iii) Let f(x) = x² sin x + tan³ x
∴ f(- x) = (- x)² sin (- x) + [tan (- x)]³
= – x² sin x – tan³ x
= – f(x)
∴ \(\int_{-a}^a\) f(x) dx = 0
⇒ \(\int_{-\pi / 3}^{\pi / 3}\) (x² sin x + tan³ x) dx = 0

Question 3 (old).
(ii) \(\int_{-\pi}^\pi\) (x⁵ + x cos x) dx
Solution:
Let f(x) = x⁵ + x cos x
∴ f(- x) = (- x)⁵ + (- x) cos (- x)
= – [x⁵ + x cos x]
= – f(x)
Thus f(x) be an odd function.
∴ \(\int_{-a}^a\) f(x) dx = 0
⇒ \(\int_{-\pi}^\pi\) (x⁵ + x cos x) dx = 0.

Question 4.
(i) \(\int_{-1}^1 \log \left(\frac{2-x}{2+x}\right)\) dx
(ii) \(\int_{-\pi}^\pi \frac{2 x}{1+\cos x}\) dx
(iii) \(\int_0^\pi\) e^{cos2 x} cos x dx
Solution:
(i) Here f(x) = log \(\left(\frac{2-x}{2+x}\right)\)
∴ f(- x) = log \(\left(\frac{2-x}{2+x}\right)\)
= log \(\left(\frac{2-x}{2+x}\right)\)^-1
= – log \(\left(\frac{2-x}{2+x}\right)\)
= – f(x)
[∵ a log b = log b^a]
Thus f(x) is an odd function.
∴ \(\int_{-1}^1\) f(x) dx = 0
⇒ \(\int_{-1}^1 \log \left(\frac{2-x}{2+x}\right)\) dx = 0

(ii) Let I = \(\int_{-\pi}^\pi \frac{2 x}{1+\cos x}\) dx

(iii) Here, f(x) = e^{cos2 x} cos x dx
Also f(π – x) = e^{cos2 (π – x)} cos (π – x)
= e^{cos2 x} (- cos x)
= – f(x)
∴ \(\int_0^\pi\) e^{cos2 x} cos x dx = 0
[∵ \(\int_0^{2 a}\) f(x) dx = 0 if (2a – x) = – f(x)]

Question 5.
(i) \(\int_0^{\pi / 2}\) log (tan x) dx (ISC 2009)
(ii) \(\int_0^{\pi / 2} \frac{\sin x-\cos x}{\sqrt{1-\sin 2 x}}\) dx
(iii) \(\int_0^{2 \pi}\) cos² x sin x dx
Solution:
(i) Let I = \(\int_0^{\pi / 2}\) log (tan x) dx ………………(1)
∴ I = \(\int_0^{\pi / 2}\) log tan (\(\frac{\pi}{2}\) – x) dx
[∵ \(\int_0^a\) f(x) dx = \(\int_0^a\) f(a – x) dx]
∴ I = \(\int_0^{\pi / 2}\) log cot x dx ……………(2)
On adding (1) and (2) ; we have
2I = \(\int_0^{\pi / 2}\) log (tan x . cot x) dx
∴ 2I = \(\int_0^{\pi / 2}\) log 1 . dx = 0
⇒ I = 0

(ii) Let I = \(\int_0^{\pi / 2} \frac{\sin x-\cos x}{\sqrt{1-\sin 2 x}}\) dx ………………(1)

On adding (1) and (2) ; we have
2I = 0
⇒ I = 0

(iii) Here, f(x) = cos² x sin x
∴ f(2π – x) = cos² (2π – x) sin (2π – x)
= cos² x (- sin x)
= – f(x)
∴ \(\int_0^{2 \pi}\) f(x) dx = 0
\(\int_0^{2 \pi}\) cos² x sin x dx = 0
[∵ \(\int_0^{2 a}\) f(x) dx = 0 if f(2a – x) = – f(x)]

Question 6.
(i) \(\int_0^\pi\) sin² x cos³ x dx
(ii) \(\int_0^{2 \pi}\) cos⁵ x dx
(iii) \(\int_{\log \frac{1}{2}}^{\log 2} \sin \left(\frac{e^x+1}{e^x-1}\right)\) dx
Solution:
(i) Let f(x) = sin² x cos³ x
∴ f(π – x) = sin² (π – x) cos³ (π – x)
= [sin (π – x)]² [cos (π – x)]³
= (sin x)² (- cos x)³
= – sin² x cos³ x
= – f(x)
Thus, \(\int_0^\pi\) f(x) dx = 0
⇒ \(\int_0^\pi\) sin² x cos³ x dx
[∵ \(\int_0^{2 a}\) f(x) dx = 0 if f(2a – x) = – f(x)]

(ii) f(x) = cos⁵ x
∴ f(2π – x) = cos⁵ (2π – x)
= [cos (2π – x)]⁵
= cos⁵ x
= f(x)
\(\int_0^{2 \pi}\) f(x) dx = 2 \(\int_0^\pi\) f(x) dx
⇒ \(\int_0^{2 \pi}\) cos⁵ x dx = 2 \(\int_0^\pi\) cos⁵ x dx …………….(1)
[∵ \(\int_0^{2 a}\) f(x) dx = 2 \(\int_0^{a}\) f(x) dx
if f(2a – x) = f(x)]
Since f(π – x) = cos⁵ (π – x)
= (- cos x)⁵
= – cos⁵ x
= – f(x)
Thus, \(\int_0^\pi\) f(x) dx = 0
⇒ \(\int_0^\pi\) cos⁵ x dx = 0
[∵ \(\int_0^{2 a}\) f(x) dx = 0 if f(2a – x) = – f(x)]
∴ from (1) ;
\(\int_0^{2 \pi}\) cos⁵ x dx = 2 × 0 = 0

(iii) Let I = \(\int_{\log \frac{1}{2}}^{\log 2} \sin \left(\frac{e^x+1}{e^x-1}\right)\) dx

Question 7.
\(\int_0^{2 a}\) f(x) dx = \(\int_0^{2 a}\) f(2a – x) dx
Solution:
Let I = \(\int_0^{2 a}\) f(2a – x) dx
put 2a – x = t
⇒ – dx = dt
When x = 0 ⇒ t = 2a
When x = 2a ⇒ t = 0

Question 8.
(i) Prove that \(\int_0^a\) f(x) dx = \(\int_0^a\) f(a – x) dx and hence prove that \(\int_0^{\pi / 2} \frac{\sin x}{\sin x+\cos x} d x=\frac{\pi}{4}\).
(ii) Prove that \(\int_0^a\) f(x) dx = \(\int_0^a\) f (a – x) dx. Hence evaluate \(\int_0^{\pi / 2} \frac{d x}{1+\tan x}\).
Solution:
(i) Let I = \(\int_0^a\) f(a – x) dx
put a – x = t
⇒ dx = – dt
When x = 0 ⇒ t = a
When x = a ⇒ t = 0
∴ I = \(\int_0^a\) f(t) (- dt)
= \(\int_0^a\) f(t) dt
[∵ \(\int_a^b\) f(x) dx = – \(\int_a^b\) f(x) dx]

(ii) Let I = \(\int_0^{\pi / 2} \frac{d x}{1+\tan x}\) …………….(1)

By using properties of definite integrals:
(Valuate the following (9 to 21))

Question 9.
(i) \(\int_0^4\) |x – 1| dx (NCERT)
(ii) \(\int_2^8\) |x – 5| dx (NCERT)
(iii) \(\int_{-5}^5\) |x + 2| dx (NCERT)
Solution:
(i) \(\int_0^4\) |x – 1| dx

(ii) \(\int_2^8\) |x – 5| dx

(iii) \(\int_{-5}^5\) |x + 2| dx
= \(\int_{-5}^{-2}\) |x + 2| dx + \(\int_{-2}^5\) |x + 2| dx
= \(\int_{-5}^{-2}\) – (x + 2) dx + \(\int_{-2}^5\) (x + 2) dx
[∵ – 5 ≤ x < – 2 ⇒ x + 2 < 0
⇒ |x + 2| = – (x + 2)
and – 2 ≤ x < 5 ⇒ x + 2 ≥ 0
⇒ |x + 2| = x + 2]
= \(\left.\left.\frac{-(x+2)^2}{2}\right]_{-5}^{-2}+\frac{(x+2)^2}{2}\right]_{-2}^5\)
= – \(\frac{1}{2}\) [0 – 9] + \(\frac{1}{2}\) [49 – 0] = 29

Question 10.
(i) \(\int_0^1\) |2x – 1| dx
(ii) \(\int_{-2}^2\) |2x + 3| dx
(iii) \(\int_1^3\) |x² – 2x| dx
Solution:
(i) When 0 ≤ x < \(\frac{1}{2}\)
⇒ 2x – 1 < 0
∴ |2x – 1| = – (2x – 1)
When \(\frac{1}{2}\) ≤ x ≤ 1
⇒ 2x – 1 ≥ 0
∴ |2x – 1| = 2x – 1

(ii) Since |2x + 3| = \(\left\{\begin{aligned}
2 x+3 & \text { if } 2 x+3 \geq 0 \text { i.e. } x \geq-3 / 2 \\
-(2 x+3) & \text { if } 2 x+3<0 \text { i.e. } x<-3 / 2
\end{aligned}\right.\)

(iii) When 1 ≤ x < 2, x ≥ 0 ;
x – 2 ≤ 0
∴ |x| = x ;
|x – 2| = – (x – 2)
Thus, |x² – 2x| = |x| |x – 2|
= – x (x – 2)
= – x² + 2x
When 2 ≤ x ≤ 3, x ≥ 0 ;
x – 2 ≥ 0
∴ |x| = x ;
|x – 2| = x – 2
Thus, |x² – 2x| = |x| |x – 2|
= x (x – 2)
= x² – 2x

Question 11.
(i) \(\int_{-\pi / 4}^{\pi / 4}\) |sin x| dx
(ii) \(\int_0^\pi\) |cos x| dx
(iii) \(\int_0^{\pi / 2}\) |cos 2x| dx
Solution:
(i) When – \(\frac{\pi}{4}\) ≤ x < 0
⇒ sin x < 0
∴ |sin x| = – sin x
When 0 ≤ x < \(\frac{\pi}{4}\) then sin x > 0
∴ |sin x| = sin x

(ii) \(\int_0^\pi\) |cos x| dx

(iii) We see that x ∈ [0, \(\frac{\pi}{2}\)]
⇒ 2x ∈ [0, π]

Question 12.
(i) \(\int_0^{2 \pi}\) |sin x| dx
(ii) \(\int_{-1}^1 \sqrt{|x|-x}\) dx
(iii) \(\int_0^{\pi / 2} \sqrt{1-\sin 2 x}\) dx
Solution:
(i) |sin x| = \(\left\{\begin{array}{c}
\sin x \text { if } 0 \leq x<\pi \\
-\sin x \text { if } \pi \leq x<2 \pi
\end{array}\right.\)
When 0 ≤ x < π, sin x > 0
and when π ≤ x < 2π, sin x < 0
∴ I = \(\int_0^{2 \pi}\) |sin x| dx
= \(\int_0^\pi\) |sin x| dx + \(\int_\pi^{2 \pi}\) |sin x| dx
= \(\int_0^\pi\) sin x dx + \(\int_\pi^{2 \pi}\) – sin x dx
= \(\left.-\cos x]_0^\pi+\cos x\right]_\pi^{2 \pi}\)
= – (cos π – cos 0) + (cos 2π – cos π)
= – (- 1 – 1) + (1 – (- 1))
= – (- 2) + (2)
= 2 + 2 = 4

(ii) When – 1 ≤ x < 0
∴ |x| = – x
When 0 ≤ x ≤ 1
∴ |x| = x

(iii) Let I = \(\int_0^{\pi / 2} \sqrt{1-\sin 2 x}\) dx
= \(\int_0^{\pi / 2} \sqrt{\sin ^2 x+\cos ^2 x-2 \sin x \cos x}\) dx
= \(\int_0^{\pi / 2} \sqrt{(\sin x-\cos x)^2}\) dx
= \(\int_0^{\pi / 2}\) |sin x – cos x| dx

When 0 ≤ x< \(\frac{\pi}{4}\) ⇒ cos x > sin x
⇒ sin x – cos x < 0 ∴ |sin x – cos x| = – (sin x – cos x) and when \(\frac{\pi}{4}\) ≤ x ≤ sin x> cos x
⇒ sin x – cos x > 0
∴ |sin x – cos x| = (sin x – cos x)
= \(\int_0^{\pi / 4}\) – (sin x – cos x) dx + \(\int_{\pi / 4}^{\pi / 2}\) (sin x – cos x) dx
= – [- cos x – sin x\(]_0^{\pi / 4}\) + (- cos x – sin x)\(]_{\pi / 4}^{\pi / 2}\)
= \(\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-(1+0)\right]+\left(0-1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)\)
= √2 – 1 – 1 +√2
= 2 (√2 – 1).

Question 12 (old).
(v) \(\int_{-2}^3\) |x² – 1| dx
Solution:
I = \(\int_{-2}^3\) |x² – 1| dx
= \(\int_{-2}^3\) |x – 1| |x + 1| dx
When – 2 ≤ x < – 1
∴ x + 1 < 0
⇒ |x + 1| = – (x + 1)
and x – 1 < 0
⇒ |x – 1| = – (x – 1)
Thus |x² – 1| = {- (x + 1)} {- (x – 1)}
= x² – 1
When – 1 ≤ x < 1
∴ x + 1 ≥ 0
⇒ |x + 1| = x + 1
x – 1 < 0
⇒ |x – 1| = – (x – 1)
Thus |x² – 1| = – (x + 1) (x – 1)
= – (x² – 1)
When 1 ≤ x < 3 x + 1 > 0 ; x – 1 ≥ 0
∴ |x² – 1| = x² – 1

Question 13.
(i) \(\int_0^5\) f(x) dx where f(x) = |x + 2| + |x – 3|
(ii) \(\int_1^3\) (|x – 1| + |x – 2| + |x – 3|) dx
(iii) \(\int_0^3\) f(x) dx, where f(x) = |x| + |x – 1| + |x – 2|
(iv) \(\int_{-\pi / 2}^{\pi / 2}\) (sin |x| – cos |x|) dx
(v) \(\int_0^{3 / 2}\) |x sin πx| dx
Solution:
(i) When 0 ≤ x ≤ 3, x + 2 > 0 ;
x – 3 ≤ 0
∴ |x + 2| = x + 2 ;
|x – 3| = – (x – 3)
Thus f(x) = |x + 2| + |x – 3|
= x + 2 – (x – 3) = 5
When 3 ≤ x ≤ 5 ; x + 2 > 0 ; x – 3 ≥ 0
∴ |x + 2| = x + 2 ;
|x – 3| = x – 3
∴ f(x) = |x + 2| + |x – 3|
= x + 2 + x – 3
= 2x – 1

(ii) Given f(x) = |x – 1| + |x – 2| + |x – 3|
When 1 ≤ x ≤ 2, x – 1 > 0, x – 2 ≤ 0;
x – 3 ≤ 0
∴ |x – 1| = x – 1 ;
|x – 2| = – (x – 2);
| x — 3 | = – (x – 3)
Thus f(x) = x – 1 – (x – 2) – (x – 3) = – x + 4
When 2 ≤ x ≤ 3; x – 1 > 0, x – 2 ≥ 0
and x – 3 ≤ 0
∴ |x – 1| = x – 1;
|x – 2| = + (x – 2);
|x – 3| = – (x – 3)
Thus, f(x) = (x – 1) + (x – 2) – (x – 3) = x

(iii) \(\int_0^3\) f(x) dx
= \(\int_0^3\) [|x| + |x – 1| + |x – 2|] dx
= \(\int_0^3\) |x| dx + \(\int_0^3\) |x – 1| dx + \(\int_0^3\) |x – 2| dx
= I₁ + I₂ + I₃ …………….(1)

(iv) Let f(x) = sin |x| – cos |x|
∴ f(- x) = sin |- x| – cos |- x|
= sin |x| – cos |x|
= f(x)
∴ f(x) be an even function.
∴ I = \(\int_{-\pi / 2}^{\pi / 2}\) f(x) dx
= 2 \(\int_0^{\pi / 2}\) f(x) dx
= 2 \(\int_0^{\pi / 2}\) [sin |x| – cos |x|] dx
[when 0 ≤ x ≤ \(\frac{\pi}{2}\)
∴ |x| = x]
∴ I = 2 \(\int_0^{\pi / 2}\) (sin x – cos x) dx
= 2 [- cos x – sin x\(]_0^{\pi / 2}\)
Thus I = 2 [- 0 – 1 – (- 1 – 0)] = 0

(v) Let I = \(\int_0^{3 / 2}\) |x sin πx| dx
= π |x| |sin πx| dx
When 0 ≤ πx ≤ π
∴ sin πx ≥ 0
⇒ |sin πx| = sin πx
When 1 ≤ x ≤ \(\frac{3}{2}\)
∴ |x| = x
⇒ π ≤ πx ≤ \(\frac{3 \pi}{2}\)
∴ sin πx ≤ 0
⇒ |sin πx| = – sin πx

Question 14.
\(\int_1^4\) f(x) dx, where f(x) = \(\left\{\begin{array}{l}
4 x+3,1 \leq x \leq 2 \\
3 x+5,2 \leq x \leq 4
\end{array}\right.\).
Solution:
\(\int_1^4\) f(x) dx = \(\int_1^2\) f(x) dx + \(\int_2^4\) f(x) dx
= \(\int_1^2\) (4x + 3) dx + \(\int_2^4\) (3x + 5) dx
= \(\left.\left.\frac{(4 x+3)^2}{2 \cdot 4}\right]_1^2+\frac{(3 x+5)^2}{2 \cdot 3}\right]_2^4\)
= \(\frac{1}{8}\) [(11)² – 7²] + \(\frac{1}{6}\) [(17)² – (11)²]
= \(\frac{1}{8}\) [121 – 49] + \(\frac{1}{6}\) [289 – 121]
= \(\frac{72}{8}+\frac{1}{6}\) × 168
= 9 +28 = 37

Question 15.
\(\int_0^9\) f(x) dx, where f(x) = \(\left\{\begin{array}{cc}
\sin x, & 0 \leq x \leq \frac{\pi}{2} \\
1, & \frac{\pi}{2} \leq x \leq 5 \\
e^{x-5}, & 5 \leq x \leq 9
\end{array}\right.\) (ISC 2008)
Solution:
\(\int_0^9\) f(x) dx = \(\int_0^{\pi / 2} f(x) d x+\int_{\pi / 2}^5 f(x) d x+\int_5^9 f(x) d x\)
= \(\int_0^{\pi / 2} \sin x d x+\int_{\pi / 2}^5 d x+\int_5^9 e^{x-5} d x\)
= \(\left.\left.-\cos x]_0^{\pi / 2}+x\right]_{\pi / 2}^5+e^{x-5}\right]_5^9\)
= – (0 – 1) + (5 – \(\frac{\pi}{2}\)) + (e⁴ – 1)
= 5 – \(\frac{\pi}{2}\) + e⁴

Question 15 (old).
(i) \(\int_1^2\) [2x] dx
(ii) \(\int_0^3\) [x] dx
(iii) \(\int_{0 \cdot 2}^{3 \cdot 5}\) [x] dx
Solution:
(i) When 1 ≤ x ≤ 2
⇒ 2 ≤ 2x ≤ 4
Clearly [2x] is discontinuous at x = \(\frac{3}{2}\)
∴ \(\int_1^2\) [2x] dx = \(\int_1^{3 / 2}\) [2x] dx + \(\int_{3 / 2}^2\) [2x] dx
[when 1 ≤ x < \(\frac{3}{2}\)
⇒ 2 ≤ 2x < 3
∴ [2x] = 2
when \(\frac{3}{2}\) ≤ x < 2
⇒ 3 ≤ 2x < 4
∴ [2x] = 3]
= \(\int_1^{3 / 2}\) 2 dx + \(\int_{3 / 2}^2\) 3 dx
= 2 (\(\frac{3}{2}\) – 1) + 3 (2 – \(\frac{3}{2}\))
= 2 × \(\frac{1}{2}\) + 3 × \(\frac{1}{2}\)
= \(\frac{5}{2}\)

(ii) Clearly [x] is discontinuous at x = 1, 2
When 0 ≤ x < 1 ; [x] = 0
When 1 ≤ x <2 ; [x] = 1
When 2 ≤ x < 3 ; [x] = 2
∴ \(\int_0^3\) [x] dx = \(\int_0^1\) [x] dx + \(\int_1^2\) [x] dx + \(\int_2^3\) [x] dx
= \(\int_0^1\) 0 . dx + \(\int_1^2\) 1 . dx + \(\int_2^3\) 2 dx
= 0 + 1 (2 – 1) + 2 (3 – 2) = 3

(iii) Clearly [x] is discontinuous x = 1, 2, 3
When 0.2 ≤ x < 1
∴ [x] = 0
When 1 ≤ x < 2;
∴ [x] = 1
When 2 ≤ x < 3
∴ [x] = 2
When 3 ≤ x < 3.5
∴ [x] = 3
∴ \(\int_{0.2}^{3 \cdot 5}\) [x] dx = \(\int_{0.2}^1\) [x] dx + \(\int_1^2\) [x] dx + \(\int_2^3\) [x] dx + \(\int_3^{3 \cdot 5}\) [x] dx
= \(\int_{0.2}^1\) 0 . dx + \(\int_1^2\) 1 . dx + \(\int_2^3\) 2 dx + \(\int_3^{3 \cdot 5}\) 3 dx
∴ \(\int_{0.2}^{3 \cdot 5}\) [x] dx = 0 + 1 (2 – 1) + 2 (3 – 2) + 3 (3.5 – 3)
= 1 + 2 + 1.5 = 4.5.

Question 16.
(i) \(\int_0^{\pi / 2}\) cos² x dx (NCERT)
(ii) \(\int_{-\pi / 2}^{\pi / 2}\) sin² x dx (NCERT)
(iii) \(\int_{-\pi / 2}^{\pi / 2}\) cos⁴ x dx
Solution:
(i) Let I = \(\int_0^{\pi / 2}\) cos² x dx ……………(1)
∴ I = \(\int_0^{\pi / 2}\) cos² (\(\frac{\pi}{2}\) – x) dx
[∵ \(\int_0^a\) f(x) dx = \(\int_0^a\) f(a – x) dx]
I = \(\int_0^{\pi / 2}\) sin² x dx ………………..(2)
On adding (1) and (2) ; we have
2I = \(\int_0^{\pi / 2}\) (cos² x + sin² x) dx
= \(\int_0^{\pi / 2}\) dx = x\(]_0^{\pi / 2}\)
= \(\frac{\pi}{2}\)
∴ I = \(\frac{\pi}{4}\).

(ii) Here f(x) = sin² x is an even function.
∴ \(\int_{-a}^a\) f(x) dx = 2 \(\int_0^a\) f(x) dx

(iii) Let I = \(\int_{-\pi / 2}^{\pi / 2}\) cos⁴ x dx
= 2 \(\int_0^{\pi / 2}\) cos⁴ x dx
[Here f(x) = cos⁴ x dx
⇒ f(- x) = cos⁴ (- x)
= cos⁴ x
= f(x)
∴ f(x) be an even function]
Also \(\int_{-a}^a\) f(x) dx = 2 \(\int_0^a\) f(x) dx if f(- x) = f(x)]

Question 17.
(i) \(\int_0^1\) x (1 – x)^5/2 dx
(ii) \(\int_0^2 x \sqrt{2-x}\) dx
(iii) \(\int_0^3 x^2 \sqrt{3-x}\) dx
Solution:
(i) Let I = \(\int_0^1\) x (1 – x)^5/2 dx

(ii) Let I = \(\int_0^2 x \sqrt{2-x}\) dx

(iii) \(\int_0^3 x^2 \sqrt{3-x}\) dx

Question 18.
(i) \(\int_0^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\) dx (ISC 2014)
(ii) \(\int_0^{\pi / 2} \frac{d x}{1+\sqrt{\cot x}}\)
(iii) \(\int_0^{\pi / 2} \frac{\sin ^3 x}{\sin ^3 x+\cos ^3 x}\) dx
(iv) \(\int_0^{\pi / 2} \frac{\cos ^5 x}{\cos ^5 x+\sin ^5 x}\) dx
(v) \(\int_0^{\pi / 2} \frac{\tan ^7 x}{\cot ^7 x+\tan ^7 x}\) dx (NCERT Exemplar)
(vi) \(\int_0^{\pi / 2} \frac{\sin ^{3 / 2} x}{\sin ^{3 / 2} x+\cos ^{3 / 2} x}\) (NCERT)
Solution:
(i) Let I = \(\int_0^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\) dx ……………….(1)

(ii) Let I = \(\int_0^{\pi / 2} \frac{d x}{1+\sqrt{\cot x}}\)

(iii) \(\int_0^{\pi / 2} \frac{\sin ^3 x}{\sin ^3 x+\cos ^3 x}\) dx ……………….(1)

(iv) Let I = \(\int_0^{\pi / 2} \frac{\cos ^5 x d x}{\sin ^5 x+\cos ^5 x}\) …………….(1)

(v) Let I = \(\int_0^{\pi / 2} \frac{\tan ^7 x d x}{\tan ^7 x+\cot ^7 x}\) ………………(1)

(vi) Let I = \(\int_0^{\pi / 2} \frac{\sin ^{3 / 2} x}{\sin ^{3 / 2} x+\cos ^{3 / 2} x}\) dx ………………(1)

Question 19.
(i) \(\int_0^{\pi / 2} \frac{\cos ^2 x}{\sin x+\cos x}\) dx
(ii) \(\int_0^{\pi / 2} \frac{x}{\sin x+\cos x}\) dx
(iii) \(\int_0^\pi \frac{x \sin x}{1+3 \cos ^2 x}\) dx
Solution:
(i) Let I = \(\int_0^{\pi / 2} \frac{\cos ^2 x}{\sin x+\cos x}\) dx ………….. (1)

(ii) \(\int_0^{\pi / 2} \frac{x}{\sin x+\cos x}\) dx ……………..(1)

(iii) Let I = \(\int_0^\pi \frac{x \sin x}{1+3 \cos ^2 x}\) dx ……………(1)
We Know that
\(\int_0^a\) f(x) dx = \(\int_0^a\) f (a – x) dx
I = \(\int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{1+3[\cos (\pi-x)]^2}\) dx
= \(\int_0^\pi \frac{(\pi-x) \sin x}{1+3 \cos ^2 x}\) dx …………….(2)
On adding eqn. (1) and eqn. (2) ; we have
2I = \(\int_0^\pi \frac{\pi \sin x}{1+3 \cos ^2 x}\) dx
put cos x = t
⇒ – sin x dx = dt
When x = 0
⇒ t = 1 ;
When x = π
⇒ t = – 1

Question 19 (old).
(iii) \(\int_0^\pi \frac{4 x \sin x}{1+\cos ^2 x}\) dx
Solution:
Let I = \(\int_0^\pi \frac{4 x \sin x}{1+\cos ^2 x}\) dx ………………(1)
∴ I = \(\int_0^\pi \frac{4(\pi-x) \sin (\pi-x)}{1+\cos ^2(\pi-x)}\) dx
[∵ \(\int_0^a\) f(x) dx = \(\int_0^a\) f(a – x) dx]
I = \(\int_0^\pi \frac{4(\pi-x) \sin x}{1+\cos ^2 x}\) dx ………………(2)
On adding eqn. (1) and eqn. (2) ; we have
2I = \(\int_0^\pi \frac{4 \pi \sin x}{1+\cos ^2 x}\)
put cos x = t
⇒ – sin x dx = dt
When x = 0 ⇒ t = 1 ;
When x = π ⇒ t = – 1
2I = 4π \(\int_1^{-1} \frac{-d t}{1+t^2}\)
= – 4π \(\int_1^{-1} \frac{d t}{t^2+1^2}\)
= – \(\left.\frac{4 \pi}{1} \tan ^{-1} \frac{t}{1}\right]_1^{-1}\)
= – 4π [tan^-1 (- 1) – tan^-1 1]
= – 4π [- tan^-1 1 – tan^-1 1]
[∵ tan^-1 (- x) = – tan^-1 x]
= + 8π tan^-1 1
= 8 × \(\frac{\pi^2}{4}\)
= 2π²
∴ I = π²

Question 20.
(i) \(\int_2^3 \frac{\sqrt{x}}{\sqrt{x}+\sqrt{5-x}}\) dx
(ii) \(\int_2^8 \frac{\sqrt[3]{x+1}}{\sqrt[3]{x+1}+\sqrt[3]{11-x}}\) dx
(iii) \(\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\) dx
(iv) \(\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\tan x}}\) dx
Solution:
(i) Let I = \(\int_2^3 \frac{\sqrt{x}}{\sqrt{x}+\sqrt{5-x}}\) dx …………..(1)

(ii) Let I = \(\int_2^8 \frac{\sqrt[3]{x+1}}{\sqrt[3]{x+1}+\sqrt[3]{11-x}}\) dx …………….(1)

(iii) Let I = \(\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x} d x}{\sqrt{\sin x}+\sqrt{\cos x}}\) …………….(1)

(iv) Let I = \(\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\tan x}}\) ……………(1)

Question 21.
Prove that \(\int_0^a\) f(x) dx = \(\int_0^a\) f(a – x) dx. Hence, evaluate :
(i) \(\int_0^a \frac{x \sin x}{1+\cos ^2 x}\) dx
(ii) \(\int_0^{\pi / 2} \frac{x}{\sin x+\cos x}\) dx
(iii) \(\int_0^1\) x² (1 – x)ⁿ dx
Solution:
(i) Let I = \(\int_0^a \frac{x \sin x}{1+\cos ^2 x}\) dx …………….(1)

(ii) Let I = \(\int_0^{\pi / 2} \frac{x}{\sin x+\cos x}\) dx …………..(1)
We know that
\(\int_0^a\) f(x) dx = \(\int_0^a\) f(a – x) dx

(iii) Let I = \(\int_0^1\) x² (1 – x)ⁿ dx
We know that
\(\int_0^a\) f(x) dx = \(\int_0^a\) f(a – x) dx
∴ I = \(\int_0^1\) (1 – x)² [1 – (1 – x)]ⁿ dx
= \(\int_0^1\) (1 – x)² xⁿ dx
= \(\int_0^1\) [x² – 2x + 1] xⁿ dx
= \(\int_0^1\) [x^{n + 2} – 2x^{n + 1} + xⁿ] dx

Question 22.
Find the value of \(\int_0^1\) tan^-1 \(\left(\frac{1-2 x}{1+x-x^2}\right)\) dx.
Solution:
Let I = \(\int_0^1\) tan^-1 \(\left(\frac{1-2 x}{1+x-x^2}\right)\) dx
= \(\int_0^1 \tan ^{-1}\left[\frac{1-x-x}{1+x(-x)}\right]\) dx
= \(\int_0^1\) [tan^-1 (1 – x) – tan^-1 x] dx
= \(\int_0^1\) [tan^-1 (1 – (1 – x)] dx – \(\int_0^1\) tan^-1 x dx
[∵ \(\int_0^a\) f(x) dx = \(\int_0^a\) f(a – x) dx]
= \(\int_0^1\) (tan^-1 x – tan^-1 x) dx
= \(\int_0^1\) 0 dx = 0

Question 23.
Prove that \(\int_a^b\) f(x) dx = \(\int_a^b\) f(a + b – x) dx. Hence, evaluate \(\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\tan x}}\).
Solution:
Let I = \(\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\tan x}}\) …………….(1)

Question 24.
(i) If a function f is continuous on [0, a], then show that \(\int_0^a \frac{f(x)}{f(x)+f(a-x)} d x=\frac{a}{2}\).
(ii) If a function f is continuous on [a, b] then show that \(\int_a^b \frac{f(x)}{f(x)+f(a+b-x)} d x=\frac{b-a}{2}\).
Solution:
(i) Let I = \(\int_0^a \frac{f(x)}{f(x)+f(a-x)}\) dx ……………..(1)

(ii) Let I = \(\int_a^b \frac{f(x)}{f(x)+f(a+b-x)}\) dx ………………(1)

Question 25.
Prove that \(\int_0^{\pi / 2}\) log (sin x) dx = \(\int_0^{\pi / 2}\) log (cos x) dx = – \(\frac{\pi}{2}\) log 2.
Solution:
Let I = \(\int_0^{\pi / 2}\) log (sin x) dx …………..(1)
∴ I = \(\int_0^{\pi / 2}\) log (sin (\(\frac{\pi}{2}\) – x)) dx
[∵ \(\int_0^a\) f(x) dx = \(\int_0^a\) f(a – x) dx]
I = \(\int_0^{\pi / 2}\) log cos x dx …………..(2)
On adding (1) and (2) ; we have
2I = \(\int_0^{\pi / 2}\) (log sin x + log cos x) dx
= \(\int_0^{\pi / 2}\) log (sin x cos x) dx
[∵ log a + log b = log ab]
⇒ 2I = \(\int_0^{\pi / 2} \log \left(\frac{\sin 2 x}{2}\right)\) dx
⇒ 2I = \(\int_0^{\pi / 2}\) log (sin 2x) dx – \(\int_0^{\pi / 2}\) log 2 dx
⇒ 2I = \(\int_0^{\pi / 2}\) log sin 2x dx – log 2 [x\(]_0^{\pi / 2}\)
⇒ 2I = I₁ – \(\frac{\pi}{2}\) log 2 ……………(3)
where I₁ = \(\int_0^{\pi / 2}\) log (sin 2x) dx
put 2x = t
⇒ 2 dx = dt
When x = 0
⇒ t = 0
When x = \(\frac{\pi}{2}\)
⇒ t = π

Question 26.
Evaluate the following integrals:
(i) \(\int_0^{\pi / 4}\) log sin 2x dx
(ii) \(\int_0^1 \frac{\log x}{\sqrt{1-x^2}}\) dx
Solution:
(i) Let I = \(\int_0^{\pi / 4}\) log sin 2x dx
put 2x = t
⇒ 2 dx = dt
When x = 0
⇒ t = 0 ;
When x = \(\frac{\pi}{4}\)
⇒ t = \(\frac{\pi}{2}\)

(ii) Let I = \(\int_0^1 \frac{\log x}{\sqrt{1-x^2}}\) dx
put x = sin t
⇒ dx = cos t dt
When x = 0 ⇒ t = 0 ;
When x = 1 ⇒ t = \(\frac{\pi}{2}\)

⇒ 2I = \(\int_0^{\pi / 2}\) [log sin 2t – log 2] dt
⇒ 2I = \(\int_0^{\pi / 2}\) log sin 2t – log 2 \(\int_0^{\pi / 2}\) dt
⇒ 2I = \(\int_0^{\pi / 2}\) log sin 2t – \(\frac{\pi}{2}\) log 2
⇒ 2I = I₁ – \(\frac{\pi}{2}\) log 2 ………….(2)
where I₁ = \(\int_0^{\pi / 2}\) log sin 2t dt
put 2t = θ
⇒ 2 dt = dθ
When t = 0
⇒ θ = 0 ;
When t = \(\frac{\pi}{2}\)
⇒ θ = π
∴ I₁ = \(\int_0^{\pi}\) log sin θ . \(\frac{d \theta}{2}\)
= \(\frac{1}{2} \times 2 \int_0^{\pi / 2}\) log sin θ dθ
= \(\int_0^{\pi / 2}\) log sin θ dθ
[∵ \(\int_0^{2 a}\) f(θ) dθ = 2 \(\int_0^{a}\) f(θ) dθ if f (2a – θ) = f(θ)]
∴ I₁ = \(\int_0^{\pi / 2}\) log sin t dt = I
∴ from (2);
2I = I – \(\frac{\pi}{2}\) log 2
⇒ I = – \(\frac{\pi}{2}\) log 2.

Question 27.
Evaluate \(\int_0^\pi\) x log (sin x) dx.
Solution:
Let I = \(\int_0^\pi\) x log (sin x) dx ……….(1)
∴ I = \(\int_0^\pi\) (π – x) log sin (π – x) dx
[∵ \(\int_0^{a}\) f(x) dx = \(\int_0^{a}\) f(a – x) dx]
= \(\int_0^\pi\) (π – x) log sin x dx …………..(2)
On adding (1) and (2) ; we have
2I = \(\int_0^{\pi / 2}\) [π – x + x] log sin x dx
= π \(\int_0^\pi\) log sin x dx
⇒ 2I = 2π \(\int_0^{\pi / 2}\) log sin x dx
[∵ \(\int_0^{2 a}\) f(x) dx = 2 \(\int_0^{a}\) f(x) dx if f (2a – x) = f(x)]
⇒ I = π \(\int_0^{\pi / 2}\) log sin x dx
= πI₁ ……………(3)
where I₁ = \(\int_0^{\pi / 2}\) log sin x dx ………….(4)
∴ I₁ = \(\int_0^{\pi / 2}\) log sin (\(\frac{\pi}{2}\) – x) dx
[∵ \(\int_0^{a}\) f(x) dx = \(\int_0^{a}\) f(a – x) dx]
I₁ = \(\int_0^{\pi / 2}\) log cos x dx ………….(5)
On adding (4) and (5) ; we get
2I₁ = \(\int_0^{\pi / 2}\) [log sin x + log cos x] dx
= \(\int_0^{\pi / 2} \log \left(\frac{2 \sin x \cos x}{2}\right)\) dx
= \(\int_0^{\pi / 2}\) log son 2x dx – \(\int_0^{\pi / 2}\) log 2 dx
put 2x = t
⇒ 2 dx = dt
When x = 0 ⇒ t = 0 ;
When x = \(\frac{\pi}{2}\) ⇒ t = π
⇒ t = π =\(\frac{1}{2} \int_0^\pi\) (log sin t) dt – (log 2) \(\frac{\pi}{2}\)
= \(\frac{1}{2} \times 2 \int_0^{\pi / 2} \log \sin t d t-\frac{\pi}{2} \log 2\)
2I₁ = \(\int_0^{\pi / 2}\) log sin – \(\frac{\pi}{2}\) log 2
[∵ \(\int_a^b\) f(t) dt = \(\int_a^b\) f(x) dx]
⇒ 2I₁ = I₁ – \(\frac{\pi}{2}\) log 2
⇒ I₁ = – \(\frac{\pi}{2}\) log 2
∴ From (3) ;
I = – \(\frac{\pi^2}{2}\) log 2.

Students can cross-reference their work with ISC Mathematics Class 12 Solutions Chapter 8 Integrals Ex 8.17 to ensure accuracy.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.17

Evaluate the following (1 to 21) definite integrals :

Question 1.
(i) \(\int_{-1}^1\) x² (x³ + 1)³ dx
(ii) \(\int_0^1\) x e^x2 dx (NCERT)
Solution:
(i) Let I = \(\int_{-1}^1\) x² (x³ + 1)³ dx
put x³ = t
⇒ 3x² dx = dt
When x = – 1 ⇒ t = – 1
When x = 1 ⇒ t = 1
∴ I = \(\int_{-1}^1\) (t + 1)³ \(\frac{d t}{3}\)
= \(\left.\frac{1}{3} \frac{(t+1)^4}{4}\right]_{-1}^1\)
= \(\frac{1}{12}\) [2⁴ – 0⁴]
= \(\frac{16}{12}\)
= \(\frac{4}{3}\)

(ii) Let I = \(\int_0^1\) x e^x2 dx
put x² = t
⇒ 2x dx = dt
When x = 0 ⇒ t = 0 ;
When x = 1 ⇒ t = 1² = 1
∴ I = \(\int_0^1 e^t \frac{d t}{2}\)
= \(\frac{1}{2}\left[e^t\right]_0^1\)
= \(\frac{1}{2}\) (e – 1).

Question 2.
(i) \(\int_0^1 \frac{x}{x^2+1}\) dx
(ii) \(\int_0^{\pi / 3} \frac{\cos x}{3+4 \sin x}\) dx
Solution:
(i) Let I = \(\int_0^1 \frac{x}{x^2+1}\) dx
put x² = t
⇒ 2x dx = dt
When x = 0 ⇒ t = 0
and When x – 1 ⇒ t = 1
∴ I = \(\int_0^1 \frac{d t}{2(t+1)}\)
= \(\left.\frac{1}{2} \log |t+1|\right]_0^1\)
= \(\frac{1}{2}\) log 2

(ii) Let I = \(\int_0^{\pi / 3} \frac{\cos x}{3+4 \sin x}\) dx
put sin x = t
⇒ cos x dx =dt
When x = 0 ⇒ t = 0
When x = \(\frac{\pi}{3}\)
⇒ t = \(\frac{\sqrt{3}}{2}\)

Question 3.
(i) \(\int_1^2 \frac{3 x}{9 x^2-1}\) dx
(ii) \(\int_a^b \frac{\log x}{x}\) dx
(iii) \(\int_0^{\pi / 4} \frac{\sin x+\cos x}{16+9 \sin 2 x}\) dx
Solution:
(i) Let I = \(\int_1^2 \frac{3 x}{9 x^2-1}\) dx

(ii) Let I = \(\int_a^b \frac{\log x}{x}\) dx
put log x = t
⇒ \(\frac{1}{x}\) dx = dt
When x = a
⇒ t = log a
When x = b
⇒ t = log b

(iii) Let I = \(\int_0^{\pi / 4} \frac{\sin x+\cos x}{16+9 \sin 2 x}\) dx

Question 4.
(i) \(\int_0^2 \frac{6 x+3}{x^2+4}\) dx (NCERT)
(ii) \(\int_0^{\pi / 2} \frac{\sin x}{\sqrt{1+\cos x}}\) dx
Solution:
(i) Let I = \(\int_0^2 \frac{6 x+3}{x^2+4}\) dx
= \(\left.\int_0^2 \frac{6 x d x}{x^2+4}+\frac{3}{2} \tan ^{-1} \frac{x}{2}\right]_0^2\)
put x² = t in Ist integral
∴ 2x dx = dt
When x = 0 ⇒ t = 0
and When x = 2 ⇒ t = 4

(ii) Let I = \(\int_0^{\pi / 2} \frac{\sin x}{\sqrt{1+\cos x}}\) dx

Question 5.
(i) \(\int_0^1 \frac{x^5}{1+x^6}\) dx
(ii) \(\int_0^1 \frac{x}{\sqrt{1+x^2}}\) (NCERT Exemplar)
Solution:
(i) Let I = \(\int_0^1 \frac{x^5}{1+x^6}\) dx
put x⁶ = t
⇒ 6x⁵ dx = dt
When x = 0 ⇒ t = 0 ;
When x = 1 ⇒ t = 1
∴ I = \(\left.\int_0^1 \frac{d t}{6(1+t)}=\frac{1}{6} \log (1+t)\right]_0^1\)
= \(\frac{1}{6}\) [log 2 – log 1]
= \(\frac{1}{6}\) log 2

(ii) Let I = \(\int_0^1 \frac{x dx}{\sqrt{1+x^2}}\)

Question 6.
(i) \(\int_1^e \frac{1+\log x}{2 x}\) dx
(ii) \(\int_0^1 \frac{x}{1+x^4}\) dx
Solution:
(i) Let I = \(\int_1^e \frac{1+\log x}{2 x}\) dx
put log x = t
⇒ \(\frac{1}{x}\) dx = dt
When x = 1
⇒ t = log 1 = 0
When x = e
⇒ t = log e = 1
= \(\int_0^1 \frac{(1+t)}{2} d t\)
= \(\left.\frac{1}{2} \frac{(1+t)^2}{2}\right]_0^1\)
= \(\frac{1}{4}\) [2² – 1]
= \(\frac{3}{4}\)

(ii) Let I = \(\int_0^1 \frac{x}{1+x^4}\) dx
put x² dx = t
⇒ 2x dx = dt
When x = 0 ⇒ t = 0 ;
When x = 1 ⇒ t = 1
∴ I = \(\int_0^1 \frac{d t}{2\left(1+t^2\right)}\)
= \(\left.\frac{1}{2} \tan ^{-1} t\right]_0^1\)
= \(\frac{1}{2}\) [tan^-1 1 – tan^-1 0]
= \(\frac{1}{2} \times \frac{\pi}{4}=\frac{\pi}{8}\)

Question 6 (old).
(i) \(\int_0^4 \frac{d x}{\sqrt{x^2+2 x+3}}\)
(ii) \(\int_0^a \frac{d x}{\sqrt{a x-x^2}}\)
Solution:
(i) Let I = \(\int_0^4 \frac{d x}{\sqrt{x^2+2 x+3}}\)

(ii) Let I = \(\int_0^a \frac{d x}{\sqrt{a x-x^2}}\)

Question 7.
(i) \(\int_0^1 \frac{e^{2 x}}{1+e^{4 x}}\) dx
(ii) \(\int_0^{\pi / 2} \sqrt{\cos x}\) sin³ x dx
Solution:
(i) Let I = \(\int_0^1 \frac{e^{2 x}}{1+e^{4 x}}\) dx
put e^2x = t
⇒ 2 e^2x dx = dt
When x = 0
⇒ t = 1 ;
When x = 1
⇒ t = e²

(ii) Let I = \(\int_0^{\pi / 2} \sqrt{\cos x}\) sin³ x dx
= \(\int_0^{\pi / 2} \sqrt{\cos x}\) (1 – cos² x) sin x dx
put cos x = t
⇒ – sin x dx = dt
When x = 0 ⇒ t = 1 ;
When x = \(\frac{\pi}{2}\) ⇒ t = 0

i

Question 7 (old).
(i) \(\int_{-1}^1 \frac{d x}{x^2+2 x+5}\) (NCERT)
Solution:
Let I = \(\int_{-1}^1 \frac{d x}{x^2+2 x+5}\)

Question 8.
(i) \(\int_0^{\pi / 4}\) sin³ 2x cos 2x dx
(ii) \(\int_0^{\pi / 2} \sqrt{\cos x}\) sin⁵ x dx
Solution:
(i) Let I = \(\int_0^{\pi / 4}\) sin³ 2x cos 2x dx
put sin 2x = y
⇒ + 2 cos 2x dx = dy
When x = 0 ⇒ y = 0
and when x = \(\frac{\pi}{4}\)
⇒ y = sin \(\frac{\pi}{2}\) = 1
= \(\int_0^1 y^3\left(+\frac{d y}{2}\right)\)
= \(+\frac{1}{2}\left[\frac{y^4}{4}\right]_0^1\)
= + \(\frac{1}{2}\left(\frac{1}{4}-0\right)\)
= + \(\frac{1}{8}\)

(ii) Let I = \(\int_0^{\pi / 2} \sqrt{\cos x}\) sin⁵ x dx
= \(\int_0^{\pi / 2} \sqrt{\cos x}\) sin⁴ x . sin x dx
= \(\int_0^{\pi / 2} \sqrt{\cos x}\) (1 – cos² x)² . sin x dx
put cos x = t
⇒ – sin x dx = dt
When x = 0
⇒ t = 1
When x = \(\frac{\pi}{2}\)
⇒ t = 0

Question 9.
(i) \(\int_1^3 \frac{\cos (\log x)}{x}\) dx
(ii) \(\int_1^2 \frac{d x}{x(1+\log x)^2}\)
Solution:
(i) Let I = \(\int_1^3 \frac{\cos (\log x)}{x}\) dx
put log x = t
⇒ \(\frac{1}{x}\) dx = dt
When x = 1
⇒ t = log 1 = 0
and when x = 3
⇒ t = log 3
∴ I = \(\int_0^{\log 3}\) cos t dt
= sin t \(]_0^{\log 3}\)
= sin (log 3) – sin 0
= sin (log 3) – 0
= sin (log 3)

(ii) Let I = \(\int_1^2 \frac{d x}{x(1+\log x)^2}\)
put log x = t
⇒ \(\frac{1}{x}\) dx = dt
When x = 1 ⇒ t = 0
and When x = 2 ⇒ t = log 2
= \(\left.\int_0^{\log 2} \frac{d t}{(1+t)^2}=-\frac{1}{1+t}\right]_0^{\log 2}\)
= – \(\frac{1}{1+\log 2}\) + 1
= \(\frac{-1+1+\log 2}{1+\log 2}\)
= \(\frac{\log 2}{1+\log 2}\)

Question 10.
(i) \(\int_0^{\pi / 3} \frac{\sec x \tan x}{1+\sec ^2 x}\) dx
(ii) \(\int_0^1\) tan^-1 x dx
Solution:
(i) Let I = \(\int_0^{\pi / 3} \frac{\sec x \tan x}{1+\sec ^2 x}\) dx ;
put sec x = t
⇒ sec x tan x dx = dt
When x = 0 ⇒ t = 1
and When x = \(\frac{\pi}{3}\) ⇒ t = 2
∴ I = \(\int_1^2 \frac{d t}{\left(1+t^2\right)}\)
= tan^-1 t\(]_1^2\)
= tan^-1 2 – tan^-1 1
Thus, I = tan^-1 2 – \(\frac{\pi}{4}\)

(ii) Let I = \(\int_0^1\) tan^-1 x . 1 dx

Question 10 (old).
(ii) \(\int_4^8 x \sqrt[3]{x-4}\) dx
Solution:
Let I = \(\int_4^8 x \sqrt[3]{x-4}\) dx
put x – 4 = t
⇒ dx = dt
When x = 4 ⇒ t = 0 ;
When x = 8 ⇒ t = 4

Question 11.
(i) \(\int_0^{1 / 2} \frac{\sin ^{-1} x}{\sqrt{1-x^2}}\) dx
(ii) \(\int_0^1\) sin^-1 x dx (NCERT)
Solution:
(i) Let I = \(\int_0^{1 / 2} \frac{\sin ^{-1} x}{\sqrt{1-x^2}}\) dx
put sin^-1 x = t
⇒ x = sin t
⇒ dx = cos t dt
When x = 0 ⇒ t = 0 ;
When x = \(\frac{1}{2}\) ⇒ t = \(\frac{\pi}{6}\)

(ii) Let I = \(\int_0^1\) sin^-1 x dx
= \(\int_0^1\) sin^-1 x . 1 dx

Question 12.
(i) \(\int_0^4 \frac{d x}{\sqrt{x^2+2 x+3}}\)
(ii) \(\int_0^a \frac{d x}{\sqrt{a x-x^2}}\)
Solution:
(i) Let I = \(\int_0^4 \frac{d x}{\sqrt{x^2+2 x+3}}\)
= \(\int_0^4 \frac{d x}{\sqrt{(x+1)^2+(\sqrt{2})^2}}\)
= log |x + 1 + \(\sqrt{x^2+2 x+3}\)|\(]_0^4\)
= log |5 + \(\sqrt{16+8+3}\)| – log |1 + √3|
= log |5 + 3√3| – log |1 + √3|
= log \(\left|\frac{5+3 \sqrt{3}}{1+\sqrt{3}}\right|\)

(ii) Let I = \(\int_0^a \frac{d x}{\sqrt{a x-x^2}}\)

Question 13.
(i) \(\int_0^{\pi / 2} \frac{\cos x}{(1+\sin x)(2+\sin x)}\) dx
(ii) \(\int_0^1 \frac{2 x}{\left(x^2+1\right)\left(x^2+2\right)}\) dx
Solution:
(i) Let I = \(\int_0^{\pi / 2} \frac{\cos x}{(1+\sin x)(2+\sin x)}\) dx
put sin x = t
⇒ cos x dx = dt
When x = 0 ⇒ t = 0 and
When x = \(\frac{\pi}{2}\) ⇒ t = 1

(ii) Let I = \(\int_0^1 \frac{2 x}{\left(x^2+1\right)\left(x^2+2\right)}\) dx
put x² = t
⇒ 2x dx = dt
When x = 0 ⇒ t = 0
When x = 1 ⇒ t = 1

Question 14.
(i) \(\int_0^{\pi / 2} \frac{\sin x \cos x}{1+\sin ^4 x}\) dx
(ii) \(\int_0^1 \frac{d x}{2 e^x-1}\)
(iii) \(\int_0^{\pi / 4} \frac{\sin x+\cos x}{16+9 \sin 2 x}\) dx
(iv) \(\int_0^a \sin ^{-1}\left(\sqrt{\frac{x}{a+x}}\right)\) dx
Solution:
(i) Let I = \(\int_0^{\pi / 2} \frac{\sin x \cos x}{1+\sin ^4 x}\) dx
put sin² x = t
⇒ 2 sin x cos x dx = dt
When x = 0 ⇒ t = 0
and When x = \(\frac{\pi}{2}\)
⇒ t = sin² \(\frac{\pi}{2}\) = 1

(ii) Let I = \(\int_0^1 \frac{d x}{2 e^x-1}\)
put 2e^x – 1 = t
⇒ e^x = \(\frac{t+1}{2}\)
⇒ e^x dx = \(\frac{1}{2}\) dt
When x = 0 ⇒ t = 2 – 1 = 1
When x = 1 ⇒ t = 2e – 1

(iii) Let I = \(\int_0^{\pi / 4} \frac{\sin x+\cos x}{16+9 \sin 2 x}\) dx

(iv) Let I = \(\int_0^a \sin ^{-1}\left(\sqrt{\frac{x}{a+x}}\right)\) dx
put x = a tan² θ
⇒ dx = 2a tan θ sec² θ
When x = 0 ⇒ θ = 0 ;
When x = a ⇒ θ = \(\frac{\pi}{4}\)

Question 15.
(i) \(\int_0^1 \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\) dx
(ii) \(\int_0^1 \tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\) dx
Solution:
(i) put x = tan θ
⇒ dx = sec² θ dθ
When x = 0 ⇒ θ = 0
and When x = 1 ⇒ θ = \(\frac{\pi}{4}\)

(ii) Let I = \(\int_0^1 \tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\) dx
put x = tan θ
⇒ dx = sec² θ dθ
When x = 0 ⇒ tan θ = 0 ⇒ θ = 0
When x = 1 ⇒ tan θ = 1 ⇒ θ = \(\frac{\pi}{4}\)

Question 15 (old).
(i) \(\int_0^1\) tan^-1 x dx
(ii) \(\int_0^1\) x (tan^-1 x)² dx
Solution:
(i) Let I = \(\int_0^1\) tan^-1 x dx
= \(\int_0^1\) tan^-1 x . 1 dx

(ii) L.H.S. = \(\int_0^1\) x (tan^-1 x)² dx

Question 16.
(i) \(\int_0^1 \frac{x \tan ^{-1} x}{\left(1+x^2\right)^2}\) dx
(ii) \(\int_0^{\pi / 2} \frac{d x}{a^2 \sin ^2 x+b^2 \cos ^2 x}\)
Solution:
(i) Let I = \(\int_0^1 \frac{x \tan ^{-1} x}{\left(1+x^2\right)^2}\) dx
put tan^-1 x = t
⇒ x = tan t
⇒ dx = sec² t dt
When x = 0 ⇒ t = 0 and
When x = 1 ⇒ t = tan^-1 1 = \(\frac{\pi}{4}\)

(ii) Let I = \(\int_0^{\pi / 2} \frac{1}{a^2 \sin ^2 x+b^2 \cos ^2 x}\)
Divide Numerator and denominator by cos² x ; we have
I = \(\int_0^{\pi / 2} \frac{\sec ^2 x d x}{a^2 \tan ^2 x+b^2}\)
put tan x = t
⇒ sec² x dx = dt
When x = 0 ⇒ t = 0
and When x = \(\frac{\pi}{2}\)
⇒ t = tan \(\frac{\pi}{2}\) → ∞

Question 17.
(i) \(\int_0^1 \frac{5 x}{\left(4+x^2\right)^2}\) dx
(ii) \(\int_0^{\pi / 4} \frac{\sin x \cos x}{\sin ^4 x+\cos ^4 x}\) dx (NCERT)
Solution:
(i) Let I = \(\int_0^1 \frac{5 x}{\left(4+x^2\right)^2}\) dx
put x² = t
⇒ 2x dx = dt
When x = 0 ⇒ t = 0 ;
When x = 1 ⇒ t = 1

(ii) Let I = \(\int_0^{\pi / 4} \frac{\sin x \cos x d x}{\sin ^4 x+\cos ^4 x}\)
Divide Numerator and denominator by cos⁴ x ; we get
I = \(\int_0^{\pi / 4} \frac{\tan x \sec ^2 x d x}{\tan ^4 x+1}\)
put tan² x = t
⇒ 2 tan x sec² x dx = dt
When x = 0 ⇒ t = 0 ;
When x = \(\frac{\pi}{4}\) ⇒ t = 1
∴ I = \(\int_0^1 \frac{d t}{2\left(t^2+1\right)}\)
= \(\left.\frac{1}{2} \tan ^{-1} t\right]_0^1\)
= \(\frac{1}{2} \times \frac{\pi}{4}=\frac{\pi}{8}\)

Question 18.
(i) \(\int_0^{\pi / 2} \frac{d x}{5+4 \sin x}\)
(ii) \(\int_0^{\pi / 2} \frac{d x}{4 \cos x+2 \sin x}\)
Solution:
(i) Let I = \(\int_0^{\pi / 2} \frac{d x}{5+4 \sin x}\)
put tan \(\frac{x}{2}\) = t
⇒ sec² \(\frac{x}{2}\) \(\frac{1}{2}\) dx = dt
⇒ dx = \(\frac{2 d x}{1+\tan ^2 \frac{x}{2}}\)
= \(\frac{2 d x}{1+t^2}\)
and sin x = \(\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\)
= \(\frac{2 t}{1+t^2}\)
∴ When x = 0 ⇒ t = 0
and when x = \(\frac{\pi}{2}\) ⇒ t = 1

(ii) \(\int_0^{\pi / 2} \frac{d x}{4 \cos x+2 \sin x}\)

Question 19.
(i) \(\int_0^\pi \frac{d x}{5+3 \cos x}\)
(ii) \(\int_0^\pi \frac{d x}{6-\cos x}\)
Solution:
(i) Let I = \(\int_0^\pi \frac{d x}{5+3 \cos x}\)

(ii) put tan \(\frac{x}{2}\) = t
⇒ sec²\(\frac{x}{2}\) \(\frac{1}{2}\) dx = dt
⇒ dx = \(\frac{2 d t}{1+t^2}\)
and cos x = \(\frac{1-t^2}{1+t^2}\)
When x = 0 ⇒ t = 0
and When x = \(\frac{\pi}{2}\) ⇒ t = 1

Question 20.
(i) \(\int_0^1 \frac{1-x^2}{x^4+x^2+1}\) dx
(ii) \(\int_0^1 \frac{x^3}{\left(1+x^2\right)^4}\) dx
Solution:
(i) Let I = \(\int_0^1 \frac{1-x^2}{x^4+x^2+1}\) dx
Divide Numerator and denominator by x² ; we have
= \(\int_0^1 \frac{\left(\frac{1}{x^2}-1\right) d x}{x^2+1+\frac{1}{x^2}}\)

(ii) Let I = \(\int_0^1 \frac{x^3}{\left(1+x^2\right)^4}\) dx
put x² = t
⇒ 2x dx = dt
When x = 0 ⇒ t = 0 ;
When x = 1 ⇒ t = 1

Access to comprehensive Understanding ISC Mathematics Class 12 Solutions Chapter 8 Integrals Ex 8.16 encourages independent learning.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.16

Very Short answer type questions (1 to 5) :

Evaluate the following (1 to 19) integrals :

Question 1.
(i) \(\int_2^3\) x² dx (NCERT)
(ii) \(\int_{-1}^1\) (x + 1) dx (NCERT)
(iii) \(\int_2^3 \frac{1}{x}\) dx
Solution:
(i) \(\int_2^3\) x² dx
= \(\frac{1}{3}\) (3³ – 2³)
= \(\frac{1}{3}\) (27 – 8)
= \(\frac{19}{3}\)

(ii) \(\int_{-1}^1\) (x + 1) dx
= \(\frac{1}{2}\) [(1 + 1)² – (- 1 + 1)²]
= \(\frac{1}{2}\) [4 – 0]
= 2

(iii) \(\int_2^3 \frac{1}{x}\) dx = log |x|\(]_2^3\)
= log 3 – log 2
= log \(\frac{3}{2}\)

Question 2.
(i) \(\int_{-4}^{-1} \frac{1}{x}\) dx
(ii) \(\int_0^1 \frac{1}{2 x-3}\) dx
(iii) \(\int_0^1 \sqrt{5 x+4}\) dx
Solution:
(i) \(\int_{-4}^{-1} \frac{1}{x}\) dx
= log \(|x|]_{-4}^{-1}\)
= log |- 1| – log |- 4|
= 0 – log 4
= – log 4

(ii) \(\left.\int_0^1 \frac{1}{2 x-3} d x=\frac{\log |2 x-3|}{2}\right]_0^1\)
= \(\frac{1}{2}\) [log |2 – 3| – log |0 – 3|]
= \(\frac{1}{2}\) [log |- 1| – log |- 3|]
= \(\frac{1}{2}\) [0 – log 3]
= – \(\frac{1}{2}\) log 3

(iii) \(\int_0^1 \sqrt{5 x+4}\) dx
= \(\left.\frac{(5 x+4)^{\frac{1}{2}+1}}{5\left(\frac{1}{2}+1\right)}\right]_0^1\)
= \(\left.\frac{2}{15}(5 x+4)^{3 / 2}\right]_0^1\)
= \(\frac{2}{15}\) [9^3/2 – 4^3/2]
= \(\frac{2}{15}\) [27 – 8]
= \(\frac{38}{15}\)

Question 3.
(i) \(\int_2^3\) 3^x dx (NCERT)
(ii) \(\int_0^{\pi / 4}\) tan x dx (NCERT)
(iii) \(\int_{\pi / 4}^{\pi / 2}\) cot x dx
Solution:
(i) Let I = \(\int_2^3\) 3^x dx
= \(\left.\frac{3^x}{\log 3}\right]_2^3\)
= \(\frac{1}{\log 3}\) [3³ – 3²]
= \(\frac{18}{\log 3}\)

(ii) \(\int_0^{\pi / 4}\) tan x dx
= \(\int_0^{\pi / 4} \frac{\sin x}{\cos x}\) dx
= – log |cos x|\(]_0^{\pi / 4}\)
= – log \(\frac{1}{\sqrt{2}}\)
= – log 2^{– 1/2}
= \(\frac{1}{2}\) log 2

(iii) \(\int_{\pi / 4}^{\pi / 2}\) cot x dx
= \(\int_{\pi / 4}^{\pi / 2} \frac{\cos x}{\sin x}\) dx
= log |sin x| \(]_{\pi / 4}^{\pi / 2}\)
= log sin \(\frac{\pi}{2}\) – log sin \(\frac{\pi}{4}\)
= 0 – log 2^{– 1/2}
= \(\frac{1}{2}\) log 2.

Question 4.
(i) \(\int_0^{\pi / 4}\) sin 2x dx (NCERT)
(ii) \(\int_0^{\pi / 2}\) cos 2x dx (NCERT)
(iii) \(\int_0^\pi\) (sin² \(\frac{x}{2}\) – cos² \(\frac{x}{2}\)) dx (NCERT)
Solution:
(i) \(\int_0^{\pi / 4}\) sin 2x dx
= – \(\left.\frac{\cos 2 x}{2}\right]_0^{\pi / 4}\)
= – \(\frac{1}{2}\) [cos \(\frac{\pi}{2}\) – cos 0]
= – \(\frac{1}{2}\) (0 – 1)
= \(\frac{1}{2}\)

(ii) \(\int_0^{\pi / 2}\) cos 2x dx
= \(\left.\frac{\sin 2 x}{2}\right]_0^{\pi / 2}\)
= \(\frac{1}{2}\) [sin π – sin 0] = 0

(iii) Let I = \(\int_0^\pi\) (sin² \(\frac{x}{2}\) – cos² \(\frac{x}{2}\)) dx
= \(\int_0^\pi\) – cos (2 × \(\frac{x}{2}\)) dx
[∵ cos 2θ = cos² θ – sin² θ]
= – \(\int_0^\pi\) cos x dx
= – [sin x]₀ ^π
= – [sin π – sin 0]
= – (0 – 0) = 0

Question 5.
(i) \(\int_0^{1 / \sqrt{2}} \frac{d x}{\sqrt{1-x^2}}\)
(ii) \(\int_0^3 \frac{d x}{9+x^2}\)
(iii) \(\int_1^{\sqrt{3}} \frac{d x}{1+x^2}\)
Solution:
(i) \(\left.\int_0^{1 / \sqrt{2}} \frac{d x}{\sqrt{1-x^2}}=\sin ^{-1} x\right]_0^{1 / \sqrt{2}}\)
= sin^-1 \(\left(\frac{1}{\sqrt{2}}\right)\) – sin^-1 0
= \(\frac{\pi}{4}\) – 0
= \(\frac{\pi}{4}\)

(ii) \(\int_0^3 \frac{d x}{9+x^2}\)

(iii) \(\left.\int_1^{\sqrt{2}} \frac{d x}{1+x^2}=\tan ^{-1} x\right]_1^{\sqrt{3}}\)
= tan^-1 √3 – tan^-1 1
= \(\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}\)
[∵ tan \(\frac{\pi}{4}\) = 1
and tan \(\frac{\pi}{3}\) = √3]

Question 6.
(i) \(\int_2^3 \frac{d x}{x^2-1}\) (NCERT)
(ii) \(\int_{-\pi / 2}^{\pi / 2} \frac{1}{1+\cos x}\) dx
Solution:
(i) \(\int_2^3 \frac{d x}{x^2-1}\)
= \(\left.\frac{1}{2 \times 1} \log \left|\frac{x-1}{x+1}\right|\right]_2^3\)
= \(\frac{1}{2}\left[\log \left(\frac{1}{2}\right)-\log \left(\frac{1}{3}\right)\right]\)
= \(\frac{1}{2} \log \frac{3}{2}\)

(ii) \(\int_{-\pi / 2}^{\pi / 2} \frac{1}{1+\cos x}\)

Question 6 (old).
(ii) \(\int_0^\pi \frac{1}{1+\sin x} d x\)
Solution:
Let I = \(\int_0^\pi \frac{1}{1+\sin x} d x\)

= [(tan π – sec π) – (tan 0 – sec 0)]
= [0 – (- 1)] – (0 – 1)
= 1 + 1 = 2.

Question 7.
(i) \(\int_1^2\) (4x³ – 5x² + 6x + 9) dx (NCERT)
(ii) \(\int_0^8\left(\sqrt{8 x}-\frac{x^2}{8}\right)\) dx
Solution:
(i) \(\int_1^2\) (4x³ – 5x² + 6x + 9) dx

(ii) \(\int_0^8\left(\sqrt{8 x}-\frac{x^2}{8}\right)\) dx

Question 8.
(i) \(\int_{\pi / 6}^{\pi / 4}\) cosec x dx (NCERT)
(ii) \(\int_0^{\pi / 4}\) (2 sec² x + x³ + 2) dx (NCERT)
Solution:
(i) \(\int_{\pi / 6}^{\pi / 4}\) cosec x dx
= log |cosec x – cot x|\(]_{\pi / 6}^{\pi / 4}\)
= log |cosec \(\frac{\pi}{4}\) – cot \(\frac{\pi}{4}\)| – log |cosec \(\frac{\pi}{6}\) – cot \(\frac{\pi}{6}\)|
= log |√2 – 1| – log |2 – √3|
= log |√2 – 1| – log |2 – √3|

(ii) \(\int_0^{\pi / 4}\) (2 sec² x + x³ + 2) dx

Question 9.
(i) \(\int_0^1 \frac{d x}{\sqrt{1+x}-\sqrt{x}}\) (NCERT)
(ii) \(\int_0^1 \frac{1-x}{1+x}\) dx
Solution:
(i) Let I = \(\int_0^1 \frac{d x}{\sqrt{1+x}-\sqrt{x}}\)

(ii) Let I = \(\int_0^1 \frac{1-x}{1+x}\) dx
= – \(\int_0^1 \frac{x-1}{x+1}\) dx
= – \(\int_0^1 \frac{x+1-2}{x+1}\) dx
= – \(\int_0^1\left[1-\frac{2}{x+1}\right]\) dx
= – [x – 2 log |x + 1|\(]_0^1\)
= – [(1 – 2 log |1 + 1|) – (0 – 2 log |0 + 1|)]
= – 1 + 2 log 2 – (0 – 2 log 1)
= – 1 + 2 log 2 – 0
= – 1 + 2 log 2

Question 10.
(i) \(\int_0^{\pi / 2}\) sin³ x dx
(ii) \(\int_0^{\pi / 4}\) (tan x + cot x)^-1 dx (ISC 2003)
Solution:
(i) I = \(\int_0^{\pi / 2}\) sin³ x dx
= \(\int_0^{\pi / 2}\) sin² x sin x dx
= \(\int_0^{\pi / 2}\) (1 – cos² x) sin x dx
put cos x = t
⇒ – sin x dx = dt
When x = 0
⇒ t = cos 0 = 1
and when x = π/2
⇒ t = cos π/2 = 0
∴ I = \(\int_1^0\) (1 – t²) (- dt)
= – \(\left[t-\frac{t^3}{3}\right]_1^0\)
= \(\left[\frac{t^3}{3}-t\right]_1^0\)
= (0 – 0) – (\(\frac{1}{3}\) – 1)
= \(\frac{2}{3}\)

(ii) Let I = \(\int_0^{\pi / 4}\) (tan x + cot x)^-1 dx

Question 11.
(i) \(\int_0^{\pi / 2} \sqrt{1-\cos 2 x}\) dx
(ii) \(\int_0^\pi \sqrt{1+\sin x}\) dx
Solution:
(i) \(\int_0^{\pi / 2} \sqrt{1-\cos 2 x}\) dx

= – √2 (cos \(\frac{\pi}{2}\) – cos 0)
= – √2 (0 – 1)
= √2

(ii) I = \(\int_0^\pi \sqrt{1+\sin x}\) dx

Question 12.
(i) \(\int_0^{\pi / 2} \frac{\sin x+\cos x}{\sqrt{1+\sin 2 x}}\) dx
(ii) \(\int_0^{\pi / 4} \frac{\cos 3 x}{\cos x}\) dx
Solution:
(i) Let I = \(\int_0^{\pi / 2} \frac{\sin x+\cos x}{\sqrt{1+\sin 2 x}}\) dx

(ii) Let I = \(\int_0^{\pi / 4} \frac{\cos 3 x}{\cos x}\) dx

Question 13.
(i) \(\int_0^{\pi / 4}\) sin 2x sin 3x dx
(ii) \(\int_0^{\pi / 2}\) (a² cos² x + b² sin² x) dx
Solution:
(i) Let I = \(\int_0^{\pi / 4}\) sin 2x sin 3x dx
= \(\frac{1}{2}\) \(\int_0^{\pi / 4}\) (2 sin 3x sin 2x) dx
= \(\frac{1}{2}\) \(\int_0^{\pi / 4}\) (cos x – cos 5x) dx
= \(\frac{1}{2}\left[\sin x-\frac{\sin 5 x}{5}\right]_0^{\pi / 4}\)
= \(\frac{1}{2}\left[\sin \frac{\pi}{4}-\frac{1}{5} \sin \frac{5 \pi}{4}-0+0\right]\)
= \(\frac{1}{2}\left(\frac{1}{\sqrt{2}}+\frac{1}{5 \sqrt{2}}\right)=\frac{3}{5 \sqrt{2}}\)

(ii) Let I = \(\int_0^{\pi / 2}\) (a² cos² x + b² sin² x) dx

Question 14.
(i) \(\int_{\pi / 3}^{\pi / 4}\) (tan x + cot x)² dx
(ii) \(\int_1^2 \frac{d x}{(x+1)(x+2)}\)
Solution:
(i) Let I = \(\int_{\pi / 3}^{\pi / 4}\) (tan x + cot x)² dx

(ii) \(\int_1^2 \frac{d x}{(x+1)(x+2)}\)

Question 15.
(i) \(\int_1^2 \frac{x+3}{x(x+2)}\) dx
(ii) \(\int_1^3 \frac{d x}{x^2(x+1)}\) (NCERT)
Solution:
(i) Let I = \(\int_1^2 \frac{x+3}{x(x+2)}\) dx
= \(\frac{1}{2} \int_1^2 \frac{2 x+6}{x^2+2 x}\) dx

(ii) Let \(\frac{1}{x^2(x+1)}=\frac{\mathrm{A}}{x}+\frac{\mathrm{B}}{x^2}+\frac{\mathrm{C}}{x+1}\) ……………….(1)
Multiply eq. (1) by x² (x + 1) ; we have
1 = Ax (x + 1) + B ( x + 1) + Cx² ………………..(2)
putting x = 0 in eqn. (2) ; we have 1 = B
puiting x = – 1 ineqn. (2); we have 1 = C
coeff. of x² ;
0 = A + C = A – 1
∴ from (1); we have

Question 15 (old).
(i) \(\int_1^2 \frac{d x}{(x+1)\left(x^2-7 x+12\right)}\)
(ii) \(\int_1^2 \frac{5 x^2}{x^2+4 x+3}\) dx
Solution:
(i) Let \(\frac{1}{(x+1)\left(x^2-7 x+12\right)}=\frac{1}{(x+1)(x-3)(x-4)}\)
= \(\frac{A}{x+1}+\frac{B}{x-3}+\frac{C}{x-4}\) ………………..(1)
Multiplying both sides of eqn. (1) by (x + 1) (x – 3) (x – 4) ; we have
1 = A (x – 3) (x – 4) + B (x + 1) (x – 4) + C (x + 1) (x – 3) …………….(2)
putting x = – 1, 3, 4 successively in eqn. (2) ; we have
1 = A (- 4) (- 5)
⇒ A = \(\frac{1}{20}\)
1 = B (4) (- 1)
⇒ B = – \(\frac{1}{4}\)
and 1 = C (5) . 1
⇒ C = \(\frac{1}{5}\)
∴ from (1) ; we have

(ii) \(\int_1^2 \frac{5 x^2}{x^2+4 x+3}\) dx
= \(\int_1^2\left[5-\frac{20 x+15}{x^2+4 x+3}\right]\) dx

Question 16.
(i) \(\int_0^{\pi / 2} \frac{5 \sin x+3 \cos x}{\sin x+\cos x}\) dx
(ii) \(\int_0^{\pi / 2} \frac{3 \sin x+4 \cos x}{\sin x+\cos x}\) dx
Solution:
(i) Numerator = l (denominator) + m \(\frac{d}{d x}\) (deno.)
∴ 5 sin x + 3 cos x = l (sin x + cos x) + m \(\frac{d}{d x}\) (sin x + cos x)
⇒ 5 sin x + 3 cos x = l (sin x + cos x) + m (cos x – sin x)
Coeff. of sin x ;
5 = l – m;
Coeff. of cos x ;
3 = l + m
On solving these eqn’s; we have
l = 4, m = – 1

(ii) Let I = \(\int_0^{\pi / 2} \frac{3 \sin x+4 \cos x}{\sin x+\cos x}\) dx
= \(\int_0^{\pi / 2}\left[3+\frac{\cos x}{\sin x+\cos x}\right]\) dx
= \(3 \int_0^{\pi / 2} d x+\int_0^{\pi / 2} \frac{\cos x d x}{\sin x+\cos x}\) dx
= 3 \(\frac{\pi}{2}\) + I₁ ………………(*)
Let cos x = l (sin x + cos x) + m (cos x – sin x) ………..(1)
Coeff. of cos x ;
1 = l + m ………………(2)
Coeff. of sin x ;
0 = l – m ………………..(3)
On solving (2) and (3) ; we have
l = \(\frac{1}{2}\) = m
On dividing eqn. (1) throughout by sin x + cos x ; we have

Question 16 (old).
(ii) \(\int_0^\pi \frac{\sin x}{\sin x+\cos x}\) dx
Solution:
Numerator = l (deno.) + m \(\frac{d}{d x}\) (deno.)
sin x = l (sin x + cos x) + m \(\frac{d}{d x}\) (sin x + cos x)
sin x = l (sin x + cos x) + m (cos x – sin x)
Coeff. of sin x;
1 = l – m;
Coeff. of cos x;
0 = l + m
On solving these eqn’s,
l = \(\frac{1}{2}\) ; m = – \(\frac{1}{2}\)
∴ \(\int_0^\pi \frac{\sin x d x}{\sin x+\cos x}=\int_0^\pi \frac{l(\sin x+\cos x)}{\sin x+\cos x} d x+m \int_0^\pi \frac{(\cos x-\sin x) d x}{\sin x+\cos x}\)
= l \(\int_0^\pi\) dx + m log |sin x + cos x|\(]_0^\pi\)
= \(\frac{1}{2}\) × π + (- \(\frac{1}{2}\)) [log 1 – log 1]
= \(\frac{\pi}{2}\)

Question 17.
(i) \(\int_0^1\) x e^x dx (NCERT)
(ii) \(\int_0^1\) (x esup>x + sin \(\frac{\pi}{4}\) x) dx (NCERT)
Solution:
(i) \(\int_0^1\) x e^x dx
= \(\left.x e^x\right]_0^1-\int_0^1 e^x d x\)
= \(\left[x e^x-e^x\right]_0^1\)
= (1 . e¹ – e¹) – (0 – e⁰)
= 1

(ii) \(\int_0^1\) (x e^x + sin \(\frac{\pi}{4}\) x) dx

Question 18.
(i) \(\int_0^1\left(x e^{2 x}+\sin \frac{\pi x}{2}\right)\) dx
(ii) \(\int_0^{\pi / 2}\) x² cos 2x dx
Solution:
(i) Let I = \(\int_0^1\left(x e^{2 x}+\sin \frac{\pi x}{2}\right)\) dx

(ii) \(\int_0^{\pi / 2}\) x² cos 2x dx

Question 19.
(i) \(\int_1^2 e^x\left(\frac{1}{x}-\frac{1}{x^2}\right)\) dx
(ii) \(\int_{\pi / 2}^\pi e^x\left(\frac{1-\sin x}{1-\cos x}\right)\) dx (NCERT)
Solution:
(i) Let I = \(\int_1^2 e^x\left(\frac{1}{x}-\frac{1}{x^2}\right)\) dx

(ii) Let I = \(\int_{\pi / 2}^\pi e^x\left(\frac{1-\sin x}{1-\cos x}\right)\) dx

Question 20.
(i) If \(\int_0^a\) 3x² dx = 8, find the value of a.
(ii) If \(\int_1^a\) (3x² + 2x + 1) dx = 11, find the value(s) of a.
Solution:
(i) Given \(\int_0^a\) 3x² dx = 8
⇒ 3 \(\left[\frac{x^3}{3}\right]_0^a\) = 8
⇒ a³ – 0 = 8 = 2³
⇒ a³ – 2³ = 0
⇒ (a – 2) (a² + 2a + 4) = 0
⇒ a = 2, \(\frac{-2 \pm \sqrt{4-16}}{2}\)
i.e., a = 2, \(\frac{-2 \pm \sqrt{4-16}}{2}\) ; i.e., a = 2, – 1 ± √3 i
Thus, only real value of a be 2.

(ii) \(\int_1^a\) (3x² + 2x + 1) dx = 11
⇒ \(\left.\frac{3 x^3}{3}+\frac{2 x^2}{2}+x\right]_1^a\) = 11
⇒ (a³ + a² + a) – (1 + 1 + 1) = 11
⇒ a³ + a² + a – 14 = 0
⇒ (a – 2) (a² + 3a + 7) = 0
either a – 2 = 0 or a² + 3a + 17 = 0
it does not gives real values of a
⇒ a = 2
Hence a = 2.

Question 21.
(i) If \(\int_0^a\) √x dx = 4a \(\int_0^{\pi / 4}\) sin 2x dx, find the value of a.
(ii) If \(\int_a^b\) x³ dx = 0 and \(\int_a^b\) x² dx = \(\frac{2}{3}\), find the values of a and b.
Solution:
(i) Given, \(\int_0^a\) √x dx

(ii) Given \(\int_a^b\) x³ dx = 0
⇒ \(\left.\frac{x^4}{4}\right]_a^b\) = 0
⇒ \(\frac{1}{4}\) (b⁴ – a⁴) = 0
⇒ b⁴ – a⁴ = 0
⇒ (b – a) (b + a) (b² + a²) = 0 ………………(1)
and \(\int_a^b\) x² dx = \(\frac{2}{3}\)
⇒ \(\left.\frac{x^3}{3}\right]_a^b=\frac{2}{3}\)
⇒ \(\frac{b^3-a^3}{3}=\frac{2}{3}\)
⇒ (b – a) (b² + ab + a²) = 2 …………..(2)
When b – a = 0 ⇒ b = a, does not satisfies given integrals.
from (1) ; when b + a = 0 ⇒ b = – a
∴ from (2); we have
– 2a (a² – a² + a²) = + 2
⇒ a³ = – 1
⇒ a = – 1 [other two values of a are complex numbers]
and b = – a + 1
and other eqn’s does not gives real values of a and b.
Thus, a = – 1 and b = + 1.

Students appreciate clear and concise ML Aggarwal Maths for Class 12 Solutions Chapter 8 Integrals Ex 8.15 that guide them through exercises.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.15

Very Short answer type questions (1 to 9) :

Evaluate the following (1 to 21) integrals :

Question 1.
(i) \(\int_1^3\) (2x – 1) dx
(ii) \(\int_3^5\) (2 – x) dx
Solution:
(I) Comparing \(\int_1^3\) (2x – 1) dx with \(\int_a^b\) f(x) dx
Here, f(x) = 2x – 1; a = 1; b = 3
∴ nh = b – a = 3 – 1 = 2
∴ f(a) = f(1) = 2 . 1 – 1 = 1
f(a + h) = f(1 + h) = 2(1 + h) – 1 = 1 + 2h
f(a + 2h) = f(1 + 2h) = 2(1 + 2h) – 1 = 1 + 4h
………………………………………………
………………………………………………

(ii) We know that \(\int_a^b\) f(x) dx = \(\ {Lt}_{h \rightarrow 0} h[f(a)+f(a+h)+f(a+2 h) \ldots .+f(a+\overline{n-1} h)]\)
On comparing \(\int_3^5\) (2 – x) dx with \(\int_a^b\) f(x)
Here a = 3 ; b = 5 ;
f(x) = 2 – x
and nh = b – a = 5 – 3 = 2

Question 1 (old).
(i) \(\int_0^5\left(x+\frac{1}{2}\right)\) dx (ISC 2009)
(ii) \(\int_a^b\) x dx (NCERT)
Solution:
(i) Comparing \(\int_0^5\left(x+\frac{1}{2}\right)\) dx with \(\int_a^b\) f(x) dx
Here f(x) = x + \(\frac{1}{2}\) ;
a = 0 ; b = 5
∴ nh = b – a = 5
∴ f(a) = f(0) = 0 + \(\frac{1}{2}\) = \(\frac{1}{2}\)
f(a + h) = f (0 + h) = f(h) = h + \(\frac{1}{2}\)
f(a + 2h) = f(2h) = 2h + \(\frac{1}{2}\)
…………………………………………
…………………………………………

(ii) We know that

Question 2.
(i) \(\int_0^2\) (x² + 3) dx (NCERT Exemplar)
(ii) \(\int_2^3\) x² dx (NCERT)
Solution:
(i) Here f(x) = x² + 3 ;
a = 0, b = 2 and nh = 2 – 0 = 2
Now f(0) = 0² + 3 ;
f(0 + h) = h² + 3 ;
f(0 + 2h) = 2²h² + 3 ………………….

(ii) On comparing \(\int_2^3\) x² dx with \(\int_a^b\) f(x) dx
Here a = 2, b = 3, nh = 3 – 2 = 1,
f(x) = x²
f(a) = f(2) = 2² ;
f(2 + h) = (2 + h)² ;
f(2 + 2h) = (2 + 2h)²

Question 3.
(i) \(\int_0^3\) (2x² – 5) dx
(ii) \(\int_0^3\) (2x² + 3x + 5) dx
Solution:
(i) Comparing \(\int_0^3\) (2x² – 5) dx with \(\int_a^b\) f(x) dx
Here f(x) = 2x² – 5 ;
a = 0 ;
b = 3 ;
nh = b – a = 3
∴ f(a) = f(0) = 2 × 0² – 5 = – 5
f(a + h) = f(h) = 2h² – 5
f(a + 2h) = f(2h) = 2 (2h)² – 5
…………………………………….
…………………………………….

(ii) We know that
\(\int_a^b\) f(x) dx = \(\ {Lt}_{h \rightarrow 0} h[f(a)+f(a+h)+f(a+2 h) \ldots .+f(a+\overline{n-1} h)]\)
On comparing \(\int_0^3\) (2x² + 3x + 5) with \(\int_a^b\) f(x) dx
We have a = 0 ; b = 3;
f(x) = 2x² + 3x + 5
and nh = b – a = 3 – 0 = 3

Question 4.
(i) \(\int_1^2\) (3x² – 1) dx
(ii) \(\int_1^3\) (2x² + 5x) dx
Solution:
(i) Here f(x) = 3x² – 1 ;
a = 1 ; b = 2
and nh = b – a = 2 – 1 = 1
Now f(1) = 3 . 1² – 1 = 2 ;
f(1 + h) = 3 (1 + h)² – 1 = 2 + 6h + 3h² ;
f(1 + 2h) = 3 (1 + 2h)² – 1
= 2 + 12h + 12h²
and so on,
\(f(1+\overline{n-1} h)\) = 3 \((1+\overline{n-1} h)^2\) – 1
= 2 + 6 (n – 1) h + 3 (n – 1)² h²
∴ By definition

(ii) Here f(x) = 2x² + 5x ;
a = 1 ; b = 3 and nh = 3 – 1 = 2
f(1) = 7 ;
f(1 + h) = 2 (1 + h)² + 5 (1 + h) = 7 + 9h + 2h² ;
f(1 + 2h) = 2 (1 + 2h)² + 5 (1 + 2h) = 7 + 18h + 18h²
………………………………………………………………………………..
\(\) = 7 + 9 (n – 1) h + 2 (n – 1)² h²
∴ By def. we have

Question 5.
(i) \(\int_a^b\) x² dx
(ii) \(\int_1^3\) (3x² + 2x + 1) dx
Solution:
(i) Comparing \(\int_a^b\) x² dx with \(\int_a^b\) f(x) dx
Here f(x) = x² ; nh = b – a
f(a) = a²
f(a + h) = (a + h)²
f(a + 2h) = (a + 2h)²
……………………………..
……………………………..
\(f(a+\overline{n-1} h)=(a+\overline{n-1} h)^2\)

= a² (b – a) + a (b – a)² + \(\frac{2}{6}\) (b – a)³
= \(\frac{1}{3}\) [3a² (b – a) + 3a (b – a)² + (b – a)³]
= \(\frac{1}{3}\) (b – a) [3a² + 3ab – 3a² + b² + a² – 2ab]
= \(\frac{1}{3}\) (b – a) (b² + ab + a²)
= \(\frac{b^3-a^3}{3}\)

(ii) Comparing \(\int_1^3\) (3x² + 2x + 1) dx with \(\int_a^b\) f(x) dx
f(x) = 3x² + 2x + 1 ;
a = 1 ; b = 3
and nh = b – a = 3 – 1 = 2
∴ f(a) = f(1) = 3 . 1² + 2 . 1 + 1 = 6
f(a + h) = f(1 + h)
= 3 (1 + h)² + 2 (1 + h) + 1
= 3h² + 8h + 6
f(a + 2h) = f(1 + 2h)
= 3 (1 + 2h)² + 2 (1 + 2h) + 2 (1 + 2h) + 1
= 12h² + 16h + 6
…………………………………………..
……………………………………………

Question 6.
(i) \(\int_0^2\) e^x dx (NCERT)
(ii) \(\int_{-1}^5\) e^x dx
Solution:
(i) We know that
\(\int_a^b\) f(x) dx = \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) h [f(a) + f(a + h) + f(a + 2h) + …………….. + \(f(a+\overline{n-1} h)\)]
Here a = 0 ;
b = 2;
f(x) = e^x ;
nh = b – a = 2 – 0 = 2
Thus, \(\int_0^2\) e^x dx = \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) h [f(0) + f(h) + f(2h) + ………… + f(n – 1) h]

(ii) Comparing \(\int_{-1}^5\) e^x dx with \(\int_a^b\) f(x) dx
Here, f(x) = e^x ;
a = – 1 ; b = 5 ;
nh = b – a = 5 + 1 = 6
∴ f(a) = f(- 1) = e^-1
f(a + h) = f(- 1 + h)
= e^{-1 + h}
f(a + 2h) = f(- 1 + 2h)
= e^{-1 + 2h}
…………………………..
…………………………..

Question 6 (old).
\(\int_1^4\) (3x² + 2x) dx
Solution:
Here f(x) = 3x² + 2x ;
a = 1, b = 4
and nh = 4 – 1 = 3
Now f(1) = 3 . 1² + 2 . 1 = 5
f(1 + h) = 3 (1 + h)² + 2 (1 + h)
= 5 + 8h + 3h² ;
f(1 + 2h) = 3 (1 + 2h)² + 2 (1 + 2h)
= 5 + 16h + 12h² ;
………………………………
………………………………
\(f(1+\overline{n-1} h)\) = 5 + 8 (n – 1) h + 3 (n – 1)² h²
∴ By definition, we have

= 5 × 3 + 4 × 3 × 3 + \(\frac{1}{2}\) × 3 (3) (6)
= 15 + 36 + 27 = 78

Question 7.
\(\int_1^3\) (x² + 3x + e^x) dx
Solution:
Comparing \(\int_1^3\) (x² + 3x + e^x) dx with \(\int_a^b\) f(x) dx f(x) dx, we have
f(x) = x² + 3x + e^x; a = 1; b = 3;
nh = b – a = 3 – 1 = 2
f(a) = f(1) = 1² + 3.1 + e¹
f(a + h) = f (1+ h)
= (1+ h)² +3 (1+ h) + e^{1 + h}
f(a + 2h) = f(1 + 2h)
= (1 + 2h)² + 3(1 + 2h) + e^{1 + 2h}
…………………………
…………………………

Interactive ML Aggarwal Class 12 ISC Solutions Chapter 8 Integrals Ex 8.14 engage students in active learning and exploration.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.14

Evaluate the following (1 to 5) integrals :

Question 1.
(i) ∫ \(\sqrt{x^2+4 x+6}\) dx (NCERT)
(ii) ∫ \(\sqrt{5-2 x+x^2}\) dx (NCERT Exemplar)
Solution:
(i) Let I = ∫ \(\sqrt{x^2+4 x+6}\) dx

(ii) Let I = ∫ \(\sqrt{5-2 x+x^2}\) dx

Question 1 (old).
(i) ∫ \(\sqrt{x^2+8 x+4}\) dx
(ii) ∫ \(\sqrt{x^2+2 x+5}\) dx (NCERT)
Solution:
(i) I = ∫ \(\sqrt{x^2+8 x+4}\) dx

(ii) Let I = ∫ \(\sqrt{x^2+2 x+5}\) dx

Question 2.
(i) ∫ \(\sqrt{3-2 x-x^2}\) dx (NCERT)
(ii) ∫ \(\sqrt{1+3 x-x^2}\) dx (NCERT)
Solution:
(i) Let I = ∫ \(\sqrt{3-2 x-x^2}\) dx

(ii) Let I = ∫ \(\sqrt{1+3 x-x^2}\) dx

Question 3.
(i) ∫ (x + 3) \(\sqrt{3-4 x-x^2}\) dx
(ii) ∫ (3x + 1) \(\sqrt{4-3 x-2 x^2}\) dx
Solution:
(i) ∫ (x + 3) \(\sqrt{3-4 x-x^2}\) dx

(ii) Let I = ∫ (3x + 1) \(\sqrt{4-3 x-2 x^2}\) dx
= – ∫ – 3 \(\left(x+\frac{1}{3}\right) \sqrt{4-3 x-3 x^2}\) dx

Question 4.
(i) ∫ \(\sqrt{1-4 x-x^2}\) dx (NCERT)
(ii) ∫ \(\sqrt{2 a x-x^2}\) dx (NCERT Exemplar)
Solution:
(i) Let I = ∫ \(\sqrt{1-4 x-x^2}\) dx

(ii) Let I = ∫ \(\sqrt{2 a x-x^2}\) dx

Question 5.
∫ (2x + 3) \(\sqrt{4 x^2+5 x+6}\) dx
Solution:
Let I = ∫ (2x + 3) \(\sqrt{4 x^2+5 x+6}\) dx
Let 2x +3 = A \(\frac{d}{d x}\) (4x² + 5x + 6) + B
= A (8x + 5) + B
∴ 2 = 8A
⇒ A = \(\frac{1}{4}\)
and 5A + B = 3
⇒ B = 3 – \(\frac{5}{4}\) = \(\frac{7}{4}\)

Utilizing ML Aggarwal Class 12 Solutions ISC Chapter 8 Integrals Ex 8.13 as a study aid can enhance exam preparation.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.13

Very Short answer type questions (1 to 3) :

Evaluate the following (1 to 8) integrals :

Question 1.
(i) ∫ \(\sqrt{4-x^2}\) dx (NCERT)
(ii) ∫ \(\sqrt{x^2-9}\) dx
Solution:
(i) ∫ \(\sqrt{4-x^2}\) dx

(ii) ∫ \(\sqrt{x^2-9}\) dx

Question 2.
(i) ∫ \(\sqrt{1+x^2}\) dx (NCERT)
(ii) ∫ \(\sqrt{1-4 x^2}\) dx (NCERT)
Solution:
(i) ∫ \(\sqrt{1+x^2}\) dx

(ii) ∫ \(\sqrt{1-4 x^2}\) dx

Question 3.
(i) ∫ \(\sqrt{4 x^2-9}\) dx
(ii) ∫ \(\sqrt{4-9 x^2}\) dx
Solution:
(i) ∫ \(\sqrt{4 x^2-9}\) dx

(ii) ∫ \(\sqrt{4-9 x^2}\) dx

Question 4.
(i) ∫ \(\frac{x^2}{\sqrt{x^2+6}}\) dx
(ii) ∫ \(\frac{1}{x-\sqrt{x^2-1}}\) dx
Solution:
(i) ∫ \(\frac{x^2}{\sqrt{x^2+6}}\) dx

(ii) ∫ \(\frac{d x}{x-\sqrt{x^2-1}}\)

Question 5.
(i) ∫ \(\frac{x^2+3}{\sqrt{x^2-25}}\) dx
(ii) ∫ (x – 3) \(\sqrt{\frac{x+2}{x-2}}\) dx
Solution:
(i) ∫ \(\frac{x^2+3}{\sqrt{x^2-25}}\) dx

(ii) Let I = ∫ (x – 3) \(\sqrt{\frac{x+2}{x-2}}\) dx

Question 6.
(i) ∫ x \(\sqrt{x^4-1}\) dx
(ii) ∫ \(\frac{\sqrt{9-(\log x)^2}}{x}\) dx
Solution:
(i) Let I = ∫ x \(\sqrt{x^4-1}\) dx ;
put x² = t
⇒ x dx = \(\frac{d t}{2}\)

(ii) Let I = ∫ \(\frac{\sqrt{9-(\log x)^2}}{x}\) dx

put log x = t
⇒ \(\frac{1}{x}\) dx = dt

Question 7.
(i) ∫ sin x cos x \(\sqrt{\sin ^4 x+4}\) dx
(ii) ∫ \(\frac{\log x}{x} \sqrt{(\log x)^4-1}\) dx
Solution:
(i) Let I = ∫ sin x cos x \(\sqrt{\sin ^4 x+4}\) dx
put sin² x = t
⇒ 2 sin x cos x dx = dt

(ii) Let I = ∫ \(\frac{\log x}{x} \sqrt{(\log x)^4-1}\) dx
put (log x)² = t
⇒ 2 log x . \(\frac{1}{x}\) dx = dt
= ∫ \(\sqrt{t^2-1^2} \frac{d t}{2}\)
= \(\frac{1}{2}\left[\frac{t \sqrt{t^2-1}}{2}-\frac{1}{2} \log \left|t+\sqrt{t^2-1}\right|\right]\)
= \(\frac{1}{4}\left[(\log x)^2 \sqrt{(\log x)^4-1}-\log \left|(\log x)^2+\sqrt{(\log x)^4-1}\right|\right]\) + C

Question 8.
(i) ∫ x cos^-1 x dx (NCERT)
(ii) ∫ cos^-1 √x dx
Solution:
(i) ∫ x cos^-1 x dx

(ii) Let I = ∫ cos^-1 √x dx
put √x = t
⇒ x = t²
⇒ dx = 2t dt

The availability of step-by-step ML Aggarwal Class 12 Solutions Chapter 8 Integrals Ex 8.12 can make challenging problems more manageable.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.12

Very Short answer type questions (1 to 5) :

Evaluate the following (1 to 12) integrals

Question 1.
(i) ∫ e^x (sin x + cos x) dx (NCERT)
(ii) ∫ e^x (tan x + sec² x) dx
Solution:
(i) ∫ e^x (sin x + cos x) dx
= ∫ e^x sin x dx + ∫ e^x cos x dx
= sin x e^x – ∫ cos x e^x dx + ∫ e^x cos x dx + C
= sin x . e^x + C

(ii) Let I = ∫ e^x (tan x + sec² x) dx
= ∫ e^x tan x dx + ∫ e^x sec² x dx
= tan x . e^x – ∫ sec² x . e^x dx + ∫ e^x . sec² x dx + C
= tan x . e^x + C

Question 2.
(i) ∫ e^x (log x + \(\frac{1}{x}\)) dx
(ii) ∫ e^x (cot x – cosec² x) dx
Solution:
(i) Let I = ∫ e^x (log x + \(\frac{1}{x}\)) dx
= ∫ e^x log x dx + ∫ \(\frac{e^x}{x}\) dx
= log x e^x – ∫ \(\frac{1}{x}\) e^x dx + ∫ \(\frac{e^x}{x}\) dx + c
= e^x log x + c

(ii) Let I = ∫ e^x (cot x – cosec² x) dx
= ∫ e^x [cot x + (- cosec² x)] dx
= ∫ e^x cot x dx – ∫ e^x cosec² x dx
= e^x cot x + ∫ cosec² x e^x dx – ∫ e^x cosec² x dx
= e^x cot x + c

Question 3.
(i) ∫ e^x (x² + 2x) dx
(ii) ∫ (x + 1) e^x dx
Solution:
(i) Let I = ∫ e^x (x² + 2x) dx
= ∫ e^x x² dx + ∫ e^x . 2x dx
= x² e^x – 2x e^x dx + 2x . e^x dx + C
= x² e^x + C’

(ii) Let I = ∫ (x + 1) e^x dx
= ∫ x e^x dx + ∫ e^x dx
= x e^x – ∫ 1 . e^x dx + ∫ e^x + C
= x e^x + C

Question 4.
(i) ∫ e^x \(\left(\frac{1}{x}-\frac{1}{x^2}\right)\) dx
(ii) ∫ (1 + log x) dx
Solution:
(i) Let I = ∫ e^x \(\left(\frac{1}{x}-\frac{1}{x^2}\right)\) dx

(ii) Let I = ∫ (1 + log x) dx
= ∫ 1 . dx + ∫ log x . 1 dx
= x + log x . x – ∫ \(\frac{1}{x}\) . x dx
= x + x log x – x + C
= x log x + C

Question 5.
(i) ∫ e^x (sin^-1 x + \(\frac{1}{\sqrt{1-x^2}}\)) dx
(ii) ∫ (x cos x + sin x) dx
Solution:
(i) Let I = ∫ e^x (sin^-1 x + \(\frac{1}{\sqrt{1-x^2}}\)) dx
= ∫ e^x sin^-1 x dx + ∫ \(\frac{e^x}{\sqrt{1-x^2}}\) dx
= e^x sin^-1 x – ∫ \(\frac{1}{\sqrt{1-x^2}}\) e^x dx + ∫ \(\frac{e^x d x}{\sqrt{1-x^2}}\) + c
= e^x sin^-1 x + c

(ii) Let I = ∫ (x cos x + sin x) dx + ∫ sin x dx
= x sin x – ∫ sin x dx + ∫ sin x dx + C
= x sin x + C

Question 6.
(i) Given ∫ e^x (tan x + 1) sec x dx = e^x f(x) + C. Write f(x) satisfying the above.
(ii) If ∫ \(\left(\frac{x-1}{x^2}\right)\) e^x dx = f(x) e^x + C, then write the value of f(x).
Solution:
(i) Let I = ∫ e^x (tan x + 1) sec x dx
= ∫ e^x tan x sec x dx + ∫ e^x sec x dx
= e^x secx – ∫ e^x sec x dx + ∫ e^x sec x dx + C
= e^x sec x + C ……………..(1)
Also I = e^x f(x) + c ……………….(2)
From (1) and (2) ; we have
f(x) = sec x

(ii) Let I = ∫ \(\left(\frac{x-1}{x^2}\right)\) e^x dx

Also given I = f(x) e^x + C …………(2)
From (1) and (2) ; we have
f(x) = \(\frac{1}{x}\)

Question 7.
(i) ∫ e^x (tan x + log sec x) dx
(ii) ∫ \(\frac{2 x-1}{4 x^2}\) e^2x dx
Solution:
(i) Let I = ∫ e^x (tan x + log sec x) dx
= ∫ e^x [tan x + (- log cos x)] dx
= – [ (log cos x) e^x – ∫ \(\frac{1}{cos x}\) (- sin x) e^x dx] + ∫ e^x tan x dx
= – e^x log cos x + c
= e^x log (cos x)^-1 + c
= e^x log sec x + c

(ii) Let I = ∫ \(\frac{2 x-1}{4 x^2}\) e^2x dx

Question 8.
(i) ∫ \(\frac{\sin x \cos x-1}{\sin ^2 x}\) e^x dx
(ii) ∫ e^x (cot x + log |sin x|) dx
Solution:
(i) Let I = ∫ \(\frac{\sin x \cos x-1}{\sin ^2 x}\) e^x dx
= ∫ e^x [cot x – cosec² x] dx
= ∫ e^x cot x – ∫ e^x cosec² x dx
= cot x e^x – ∫ (- cosec² x) e^x dx – ∫ e^x cosec² x dx
= e^x cot x + ∫ e^x cosec² x – ∫ e^x cosec² x dx + c
= e^x cot x + c

(ii) Let I = ∫ e^x (cot x + log |sin x|) dx
= ∫ e^x log sin x dx + ∫ e^x cot x dx
= (log sin x) e^x – ∫ \(\frac{1}{sin x}\) cos x e^x dx + ∫ e^x cot x dx
= e^x log sin x – ∫ cot x e^x dx + ∫ e^x cot x dx + C
= e^x log sin x + c

Question 9.
(i) ∫ \(\frac{x e^x}{(x+1)^2}\) dx (NCERT)
(ii) ∫ \(\frac{x-1}{(x+1)^3}\) e^x dx
Solution:
(i) Let I = ∫ \(\frac{x e^x}{(x+1)^2}\) dx

(ii) Let I = ∫ e^x \(\frac{x-1}{(x+1)^3}\) dx

Question 10.
(i) ∫ \(\frac{x-3}{(x-1)^3}\) e^x dx (NCERT)
(ii) ∫ \(\frac{x e^{2 x}}{(1+2 x)^2}\) dx
Solution:
(i) Let I = ∫ \(\frac{x-3}{(x-1)^3}\) e^x dx

(ii) Let I = ∫ \(\frac{x e^{2 x}}{(1+2 x)^2}\) dx

Question 11.
(i) ∫ \(\frac{1+\sin x}{1+\cos x}\) e^x dx
(ii) ∫ \(\frac{2+\sin 2 x}{1+\cos 2 x}\) e^x dx
Solution:
(i) Let I = ∫ \(\frac{1+\sin x}{1+\cos x}\) e^x dx

(ii) Let I = ∫ \(\frac{2+\sin 2 x}{1+\cos 2 x}\) e^x dx
= ∫ \(\left[\frac{2+2 \sin x \cos x}{2 \cos ^2 x}\right]\) e^x dx
= ∫ sec² x . e^x dx + ∫ e^x . tan x dx
= e^x tan x – ∫ e^x . tan x dx + ∫ e^x tan x dx + C
= e^x tan x + C

Question 12.
(i) ∫ \(\frac{\sin 4 x-4}{1-\cos 4 x}\) e^x dx
(ii) ∫ (cos x + 3 sin x) e^3x dx
Solution:
(i) Let I = ∫ e^x \(\frac{\sin 4 x-4}{1-\cos 4 x}\) dx
= ∫ e^x \(\left(\frac{\sin 4 x-4}{2 \sin ^2 2 x}\right)\) dx
= ∫ e^x \(\left[\frac{2 \sin 2 x \cos 2 x-4}{2 \sin ^2 d x}\right]\) dx
[Form ∫ e^x [f(x) + f'(x)] dx]
= ∫ e^x [cot 2x – 2 cosec² 2x] dx
∴ I = ∫ e^x cot 2x dx – 2 ∫ e^x cosec² 2x dx
= cot 2x e^x – ∫ – cosec² 2x . 2 e^x dx – 2 ∫ e^x cosec² 2x dx
= e^x . cot 2x + C

(ii) Let I = ∫ (cos x + 3 sin x) e^3x dx
= ∫ e^3x cos x dx + ∫ sin x e^3x dx
= e^3x sin x – ∫ 3 e^3x sin x dx + 3 ∫ sin x e^3x dx + C
= e^3x sin x + C

Question 12 (old).
(i) ∫ (1 + log x) dx
Solution:
(i) Let I = ∫ (1 + log x) dx
= ∫ log x . dx + ∫ dx
= x log x – ∫ \(\frac{1}{x}\) . x dx + ∫ dx + C
= x log x + C

Question 13.
∫ (sin (log x) + cos (log x)) dx
Solution:
Let I = ∫ (sin (log x) + cos (log x)) dx
put log x = t
⇒ x = e^t
⇒ dx = e^t dt
∴ I = ∫ [sin t + cos t] e^t dt
= ∫ e^t sin t dt + ∫ e^t cos t dt
= sin t e^t – ∫ cos t e^t dt + ∫ e^t cos t dt
= e^t sin t + C
= x sin (log x) + C

Students often turn to Class 12 ISC Maths Solutions Chapter 8 Integrals Ex 8.11 to clarify doubts and improve problem-solving skills.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.11

Very Short answer type questions (1 to 3) :

Evaluate the following (1 to 22) integrals

Question 1.
(i) ∫ x e^x dx
(ii) ∫ x sin x dx (NCERT)
Solution:
(i) ∫ x e^x dx = x . e^x – ∫ 1 . e^x dx
[Integrating by parts]
= x e^x – e^x + C
= (x – 1) e^x + C

(ii) ∫ x sin x dx = – x cos x – ∫ 1 . (- cos x) dx
= – x cos x + sin x + C

Question 2.
(i) ∫ x sec² x dx (NCERT)
(ii) ∫ x sin 3x dx (NCERT)
Solution:
(i) ∫ x sec² x dx
= x tan x – ∫ 1 . tan x dx
= x tan x – ∫ \(\frac{\sin x}{\cos x}\) dx
= x log x + log |cos x| + C

(ii) ∫ x sin 3x dx
= x \(\left(-\frac{\cos 3 x}{3}\right)\) + ∫ 1 . \(\frac{\cos 3 x}{3}\) dx
= – x \(\frac{\cos 3 x}{3}\) + \(\frac{\sin 3 x}{9}\) + C

Question 3.
(i) ∫ x e^2x dx
(ii) ∫ x cos 2x dx
Solution:
(i) ∫ x e^2x dx = \(\frac{x e^{2 x}}{2}-\int 1 \cdot \frac{e^{2 x}}{2} d x\)
= \(\frac{x e^{2 x}}{2}-\frac{e^{2 x}}{4}\)
= \(\frac{1}{4}\) (2x – 1) e^2x + C

(ii) ∫ x cos 2x dx = \(x \frac{\sin 2 x}{2}-\int 1 \cdot \frac{\sin 2 x}{2} d x\)
= \(\frac{x \sin 2 x}{2}+\frac{\cos 2 x}{4}\) + C

Question 4.
(i) ∫ x sec² x tan x dx
(ii) ∫ (e^{log x} + sin x) cos x dx
Solution:
(i) ∫ x sec² x tan x dx
= x . \(\frac{\tan ^2 x}{2}\) – ∫ 1 . \(\frac{\tan ^2 x}{2}\) dx
[∵ ∫ sec² x tan x dx = ∫ tan x . sec² x dx
= \(\frac{\tan ^2 x}{2}\) ;
Since ∫ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\), n ≠ – 1]
= \(\frac{x \tan ^2 x}{2}\) – \(\frac{1}{2}\) ∫ (sec² x – 1) dx
= \(\frac{x \tan ^2 x}{2}\) – \(\frac{1}{2}\) tan x + \(\frac{x}{2}\) + C
= \(\frac{1}{2}\) [x (1 + tan² x) – tan x] + C
= \(\frac{1}{2}\) [x sec² x – tan x] + C

(ii) Let I = ∫ (e^{log x} + sin x) cos x dx
= ∫ (x + sin x) cos x dx
= ∫ x cos x dx + ∫ \(\frac{1}{2}\) sin 2x dx
= x sin x – ∫ 1 . sin x dx – \(\frac{\cos 2 x}{4}\) + c
= x sin x + cos x – \(\frac{\cos 2 x}{4}\) + c

Question 5.
(i) ∫ x log x dx (NCERT)
(ii) ∫ x log 2x (NCERT)
Solution:
(i) Let I = ∫ x log x dx
using integrating by parts
∴ I = log x . \(\frac{x^2}{2}\) – ∫ \(\frac{1}{x} \cdot \frac{x^2}{2}\) dx
= \(\frac{x^2}{2}\) log x – \(\frac{x^2}{4}\) + C
= \(\frac{x^2}{4}\) (2 log x – 1) + C

(ii) ∫ x log 2x dx = log 2x . \(\frac{x^2}{2}\) – ∫ \(\frac{2}{2 x} \frac{x^2}{2}\) dx
[Integrating by parts]
= \(\frac{x^2}{2}\) log 2x – \(\frac{x^2}{4}\) + C
= \(\frac{x^2}{4}\) [2 log 2x – 1] + C

Question 6.
(i) ∫ x⁴ log x dx
(ii) ∫ log x dx
Solution:
(i) Let I = ∫ x⁴ log x dx
= log x . \(\frac{x^5}{5}\) – ∫ \(\frac{1}{x} \cdot \frac{x^5}{5}\) dx + C
= \(\frac{x^5}{5} \log x-\frac{1}{5} \cdot \frac{x^5}{5}\) + C
= \(\frac{x^5}{25}\) (5 log x – 1) + C

(ii) ∫ log x dx = ∫ log x . 1 dx
= x log x – ∫ \(\frac{1}{x}\) . x dx
= x log x – x + C
= x (log x – 1) + C

Question 6 (old).
(i) ∫ x² log x dx (NCERT)
Solution:
(i) ∫ x²
= \(\log x \cdot \frac{x^3}{3}-\int \frac{1}{x} \cdot \frac{x^3}{3} d x\)
= \(\frac{x^3}{3} \log x-\frac{x^3}{9}\) + C
= \(\frac{x^3}{9}\) (3 log x – 1) + C

Question 7.
(i) ∫ (x² + 1) log x dx (NCERT)
(ii) ∫ x² log (1 + x) dx
Solution:
(i) ∫ (x² + 1) log x dx
= log x . (\(\frac{x^3}{3}\) + x) – ∫ \(\frac{1}{x}\left(\frac{x^3}{3}+x\right)\)
[using integrating by parts]
∴ I = (\(\frac{x^3}{3}\) + x) log x – \(\frac{1}{3} \frac{x^3}{3}\) – x + C
= \(\frac{x^3}{9}\) (3 log x – 1) + x (log x – 1) + C

(ii) ∫ x² log (1 + x) dx

Question 8.
(i) ∫ tan^-1 x dx (NCERT)
(ii) ∫ cos^-1 \(\left(\frac{1}{x}\right)\) dx
Solution:
(i) ∫ tan^-1 x dx
= ∫ tan^-1 x . 1 dx
= x tan^-1 x – ∫ \(\frac{x}{1+x^2}\) dx
= x tan^-1 x – \(\frac{1}{2}\) log (1 + x²) + C
[∵ ∫ \(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)| + C]

(ii) ∫ cos^-1 \(\frac{1}{x}\) dx
= ∫ sec^-1 x dx
= ∫ sec^-1 x . 1 dx
= x sec^-1 x – ∫ \(\frac{1}{x \sqrt{x^2-1}}\) x dx
= x sec^-1 x – ∫ \(\frac{d x}{\sqrt{x^2-1}}\)
= x sec^-1 x – log |x + \(\sqrt{x^2-1}\) + C

Question 9.
(i) ∫ x³ tan^-1 x dx
(ii) ∫ x² sin^-1 x dx
Solution:
(i) Let I = ∫ x³ tan^-1 x dx

(ii) Let I = ∫ x² sin^-1 x dx

Question 10.
(i) ∫ \(\frac{\sin ^{-1} x}{\sqrt{1-x}}\) dx
(ii) ∫ x² e^x dx
Solution:
(i) Let I = ∫ \(\frac{\sin ^{-1} x}{\sqrt{1-x}}\) dx

(ii) ∫ x² e^x dx
= x² e^x – ∫ 2x . e^x dx [Integrating by parts]
= x² e^x – 2 [x e^x – ∫ 1 . e^x dx] [Integrating by parts]
= x² e^x – 2 [x e^x – e^x] + C
= e^x (x² – 2x + 2) + C

Question 11.
(i) ∫ x² e^{3 x} dx
(ii) ∫ x² cos x dx
Solution:
(i) ∫ x² e^{3 x} dx

(ii) ∫ x² cos x dx
= x² sin x – ∫ 2x sin x dx
= x² sin x – 2 [- x cos x + ∫ 1 . cos x dx]
= x² – 2 [- x cos x + ∫ 1 . cos x dx]
= x² – 2 [- x cos x + sin x] + c
= x² sin x + 2x cos x – 2 sin x + c

Question 12.
(i) ∫ (log x)² dx
(ii) ∫ x³ sin (x²) dx
Solution:
(i) ∫ (log x)² dx
= ∫ (log x)² . 1 dx
= (log x)² x – ∫ 2 log x . \(\frac{1}{x}\) . x dx
= x (log x)² – 2 ∫ log x . 1 dx
= x (log x)² – 2 [x log x – x] + C
= x (log x)² – 2x log x + 2x + C

(ii) Let I = ∫ x³ sin (x²) dx
put x² = t
⇒ 2x dx = dt
= ∫ t sin t \(\frac{dt}{2}\)
= \(\frac{1}{2}\) [- t cos t + ∫ 1 . cos t dt] + C
= \(\frac{1}{2}\) [- t cos t + sin t] + C
= \(\frac{1}{2}\) [- x² cos² x² + sin x²] + C

Question 13.
(i) ∫ 2x³ ex² dx
(ii) ∫ cos √x dx
Solution:
(i) Let I = ∫ 2x³ ex² dx
put x² = t
⇒ 2x dx = dt
= ∫ t . e^t dt
= t e^t – ∫ 1 . e^t dt + C
= t e^t – e^t + C
= (t – 1) e^t + C
= (x² – 1) e^x2 + C

(ii) Let I = ∫ cos √x + C
put √x = t
⇒ x = t²
⇒ dx = 2t dt
= ∫ cos t (2t dt)
= 2 ∫ t cos t dt
= 2 [t sin t – ∫ 1 . sin t dt] + c
= 2 [t sin t + cos t] + c
= 2 [√x sin √x + cos √x] + c

Question 14.
(i) ∫ tan^-1 √x dx
(ii) ∫ x³ tan^-1 (x²) dx
Solution:
(i) Let I = ∫ tan^-1 √x dx
put √x = t
⇒ x = t²
⇒ dx = 2t dt
= ∫ tan^-1 t (2t dt)
= 2 ∫ tan^-1 t t dt
= 2 [tan^-1 t . \(\frac{t^2}{2}\) – ∫ \(\frac{1}{1+t^2} \cdot \frac{t^2}{2}\)] + c
= 2 \(\left[\frac{t^2}{2} \tan ^{-1} t-\frac{1}{2} \int \frac{1+t^2-1}{1+t^2} d t\right]\)
= t² tan^-1 t – ∫ (1 – \(\frac{1}{1+t^2}\)) dt
= t² tan^-1 t – t + tan^-1 t + c
= (1 + t²) tan^-1 t – t + c
= (1 + x) tan^-1 √x – √x + c

(ii) Let I = ∫ x³ tan^-1 (x²) dx
put x² = t
⇒ 2x dx = dt
∴ I = ∫ t . tan^-1 t \(\frac{dt}{2}\)

Question 15.
(i) ∫ sin³ √x dx
(ii) ∫ \(\frac{x \cos ^{-1} x}{\sqrt{1-x^2}}\) dx (NCERT)
Solution:
(i) Let I = ∫ sin³ √x dx
put √x = t
⇒ x = t²
⇒ dx = 2t dt
= ∫ sin³ t (2t dt)
= 2 ∫ \(\frac{t}{4}\) [3 sin t – sin 3t] dt
[∵ sin 3t = 3 sin t – 4 sin³ t]
⇒ sin³ t = \(\frac{1}{4}\) [3 sin t – sin 3t]

(ii) Let I = ∫ \(\frac{x \cos ^{-1} x}{\sqrt{1-x^2}}\) dx
put cos^-1 x = t
x = cos t
dx = – sin t dt
= ∫ \(\frac{\cos t \cdot t(-\sin t d t)}{\sqrt{1-\cos ^2 t}}\)
= – ∫ t cos tdt
= – [t sin t – ∫ 1 . sin t dt] + c
= – [t sin t + cos t] + c
= – [\(\sqrt{1-x^2}\) cos^-1 x + x] + c

Question 16.
(i) ∫ sin^-1 (3x – 4x³) dx
(ii) ∫ tan^-1 \(\left(\frac{3 x-x^3}{1-3 x^2}\right)\) dx
Solution:
(i) Let I = ∫ sin^-1 (3x – 4x³) dx
put x = sin θ
⇒ dx = cos θ dθ
= ∫ sin^-1 (3 sin θ – 4 sin³ θ) cos θ dθ
= ∫ sin^-1 (sin 3θ) cos θ dθ
= 3 ∫ θ . cos θ dθ
= 3 [θ sin θ – ∫ 1 . sin θ] + c
= 3 [θ sin θ + cos θ] + c
= 3 [sin^-1 x . x + \(\sqrt{1-x^2}\)] + c
[∵ cos θ = \(\sqrt{1-\sin ^2 \theta}\)
= \(\sqrt{1-x^2}\)]
= 3 [x sin^-1 x + \(\sqrt{1-x^2}\)] + c

(ii) Let I = ∫ tan^-1 \(\left(\frac{3 x-x^3}{1-3 x^2}\right)\) dx
put x = tan θ
⇒ dx = sec² θ dθ
= ∫ tan^-1 \(\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right)\) sec² θ dθ
= ∫ tan^-1 (tan 3θ) sec² θ dθ
= 3 ∫ θ sec² θ dθ

Question 17.
(i) ∫ \(\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}\) dx
(ii) ∫ \(\frac{x^2 \tan ^{-1} x}{1+x^2}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}\) dx
put x = sin θ
⇒ dx = cos θ dθ
∴ θ = sin^-1 x
⇒ I = ∫ \(\frac{\theta \cos \theta d \theta}{\left(1-\sin ^2 \theta\right)^{3 / 2}}\)
= \(\frac{\theta \cos \theta d \theta}{\cos ^3 \theta}\)
= ∫ θ sec² θ dθ
= θ tan θ – ∫ 1 . tan θ dθ
= θ tan θ – ∫ \(\frac{\sin \theta}{\cos \theta}\) dθ
= θ tan θ + log |cos θ| + C

(ii) Let I = ∫ \(\frac{x^2 \tan ^{-1} x}{1+x^2}\) dx
put tan^-1 x = θ
⇒ x = tan θ
⇒ dx = sec² θ dθ
= ∫ \(\frac{\tan ^2 \theta \cdot \theta}{1+\tan ^2 \theta}\) . sec² θ dθ
= ∫ θ tan² θ dθ
= ∫ θ (sec² θ – 1) dθ

= ∫ θ sec² θ dθ – ∫ θ dθ
= θ tan θ – ∫ 1 . tan θ dθ – \(\frac{\theta^2}{2}\) + c
= θ tan θ + log |cos θ| – \(\frac{\theta^2}{2}\) + c
= x tan^-1 x + log \(\left|\frac{1}{\sqrt{1+x^2}}\right|\) – \(\frac{1}{2}\) (tan^-1 x)² + c
= x tan^-1 x – \(\frac{1}{2}\) log |1 + x²| – \(\frac{1}{2}\) (tan^-1 x)² + c

Question 18.
(i) ∫ e^x sin x dx
(ii) ∫ e^2x sin x dx
Solution:
(i) Let I = ∫ e^x sin x dx
= sin x e ^x – ∫ cos x e^x dx
= e^x sin x – [cos x . e^x – ∫ – sin x e^x dx]
= (sin x – cos x) e^x – 1
⇒ 2I = (sin x – cos x) e^x – 1
⇒ I = \(\frac{e^x}{2}\) (sin x – cos x) + C

[Students can also take first function as sin x]
= – e^2x cos x + ∫ 2 e^2x cos x dx
∴ I = – e^2x cos x + 2 [e^2x sin x – ∫ 2 e^2x sin x dx]
⇒ I = e^2x (- cos x + 2 sin x) – 4 I
⇒ I = \(\frac{e^{2 x}}{5}\) [2 sin x – cos x] + C

Question 19.
(i) ∫ e^ax sin bx dx
(ii) ∫ e^ax cos (bx + c) dx
Solution:
(i) Let I = ∫ e^ax sin bx dx

(ii) Let I = ∫ e^ax cos (bx + c) dx
= cos (bx + c) \(\frac{e^{a x}}{a}\) + ∫ sin (bx + c) . b . \(\frac{e^{a x}}{a}\) dx

Question 20.
(i) ∫ x² e^x3 cos x³ dx
(ii) ∫ e^x sin² x dx
Solution:
(i) Let I = ∫ x² e^x3 cos x³ dx
put x³ = t
⇒ 3x² dx = dt
= ∫ e^t cos t \(\frac{d t}{3}\)
= \(\frac{1}{3}\) ∫ e^t cos t dt
= \(\frac{1}{3}\) I₁ ………………(1)
where I₁ = ∫ e^t cos t dt
= e^t sin t – ∫ e^t sin t dt
= e^t sin t – [- e^t cos t – ∫ e^t (- cos t) dt]
∴ I₁ = e^t sin t + e^t cos t – I₁
⇒ I₁ = \(\frac{e^t}{2}\) [sin t + cos t]
∴ from (1) ;
∴ I = \(\frac{e^t}{6}\) [sin t + cos t]
I = \(\frac{e^{x^3}}{6}\) [sin x³ + cos x³] + c

(ii) Let I₁ = ∫ e^x sin² x dx

Question 21.
(i) ∫ e^{sin-1 x} dx
(ii) ∫ \(\frac{e^{m \tan ^{-1} x}}{\left(1+x^2\right)^{3 / 2}}\) dx
(iii) ∫ cosec³ x dx
Solution:
(i) put sin^-1 x = t
⇒ x = sin t
⇒ dx = cos t dt
∴ I = ∫ e^{sin-1 x} dx
= ∫ e^t cos t dt
= e^t sin t – ∫ e^t sin t dt
∴ I = e^t sin t – [e^t (- cos t) + ∫ e^t cos t dt]
⇒ I = e^t (sin t + cos t) – I
⇒ I = \(\frac{e^t}{2}\) (sin t + cos t) + C
⇒ I = \(\frac{e^{\sin ^{-1} x}}{2}\left[x+\sqrt{1-x^2}\right]\) + C

(ii) Let I = ∫ \(\frac{e^m \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}}\) dx
put tan^-1 x = t
⇒ x = tan t
⇒ dx = sec² t dt

(iii) Let I = ∫ cosec³ x dx
= ∫ cosec x cosec² x dx
= cosec x (- cot x) – ∫ – cot x cosec x (- cot x) dx
= – cot x cosec x – ∫ cosec x (cosec² x – 1) dx
= – cot x cosec x – ∫ cosec³ x dx + ∫ cosec x dx
⇒ I = – cot x cosec x – I + ∫ cosec x dx
⇒ 2I = – cot x cosec x + log |tan \(\frac{x}{2}\)| + c
⇒ I = \(\frac{-\cot x \ {cosec} x}{2}+\frac{1}{2} \log \left|\tan \frac{x}{2}\right|\) + c

Question 22.
(i) ∫ cos (log x) dx
(ii) ∫ \(\frac{\sin ^{-1} x}{x^2}\) dx
Solution:
(i) Let I = ∫ cos (log x) . 1 dx
= cos (log x) . x – ∫ – sin (log x) . \(\frac{1}{x}\) . x dx
= x cos (log x) + ∫ sin (log x) . 1 dx
= x cos (log x) + sin (log x) . x – ∫ cos (log x) . \(\frac{1}{x}\) . x dx
∴ I = x [cos (log x) + sin (log x)] – I
⇒ 2I = x [cos (log x) + sin (log x)]
⇒ I = \(\frac{x}{2}\) [cos (log x) + sin (log x)] + c

(ii) Let I = ∫ \(\frac{\sin ^{-1} x}{x^2}\) dx
put sin^-1 x = t
x = sin t
⇒ dx = cos t dt
= ∫ \(\frac{t}{\sin ^2 t}\) cos t dt
= ∫ t (cot t cosec t dt)
= t (- cosec t) – ∫ 1 . (- cosec t) dt + c

= – t cosec t + ∫ cosec t dt + c
= – t cosec t – log |cosec t + cot t| + c
= \(-\frac{1}{x} \sin ^{-1} x-\log \left|\frac{1}{x}+\frac{\sqrt{1-x^2}}{x}\right|\) + c
= \(\frac{-\sin ^{-1} x}{x}-\log \left|\frac{1+\sqrt{1-x^2}}{x}\right|\) + c