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

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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.