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

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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.