OP Malhotra Class 12 Maths Solutions Chapter 16 Definite Integrals Ex 16(c)

S Chand Class 12 ICSE Maths Solutions Chapter 16 Definite Integrals Ex 16(c)

Students can cross-reference their work with ISC Class 12 OP Malhotra Solutions Chapter 16 Definite Integrals Ex 16(c) to ensure accuracy.

Question 1.
(i) \(\int_0^6|x-2|\)dx = 10
(ii) \(\int_0^4|2-x| d x=4\)
(iii) \(\int_0^{\pi / 2} \sin ^2 x d x=\int_0^{\pi / 2} \cos ^2 x d x=\frac{\pi}{4}\)
(iv) \(\int_0^{\pi / 2} \frac{1}{1+\tan x} d x=\int_0^{\pi / 2} \frac{1}{1+\cos x} d x=\frac{\pi}{4}\)
(v) \(\int_0^{\pi / 2} \frac{f(\sin x) d x}{f(\sin x)+f(\cos x)}=\frac{\pi}{4}\)
(vi) \(\int_0^\pi x f(\sin x) d x=\frac{\pi}{2} \int_0^\pi f(\sin x) d x\)
(vii) \(\int_0^{\pi / 2} \log (\tan x+\cot x) d x=\pi \log 2\)
Solution:
(i) \(\int_0^6|x-2|\) dx
= \(\int_0^2|x-2| d x+\int_2^6|x-2|\)dx
when 0 < x < 2 ⇒ x – 2 < 0
⇒ |x – 2| = – (x – 2)
when 2 ≤ x ≤ 6 ⇒ x – 2 ≥ 0
⇒ |x – 2| = x – 2

(ii) Let I = \(\int_0^4|2-x| d x\)
= \(\int_0^2|2-x| d x+\int_2^4|2-x| d x\)
when 0 ≤ x < 2 ⇒ 2 – x > 0
⇒ |2 – x| = 2 – x
when 2 ≤ x ≤ 4 ⇒ 2 – x ≤ 0
⇒ |2 – x| = – (2 – x)
∴ I = \(\int_0^2(2-x) d x+\int_2^4-(2-x) d x\)
= \(\left.\left.\frac{(2-x)^2}{-2}\right]_0^2-\frac{(2-x)^2}{-2}\right]_2^4\)
= \(\frac{-1}{2}[0-4]+\frac{1}{2}[4-0]\)
= 2 + 2 = 4

(iii) Let I = \(\int_0^{\pi / 2} \sin ^2 x d x\) … (1)

Evaluate the following definite integrals:

Question 2.
(i) \(\int_1^4 f(x) d x\) where
f(x) = \(\left\{\begin{array}{l}
4 x+3, \text { if } 1 \leq x \leq 2 \\
3 x+5, \text { if } 2 \leq x \leq 4
\end{array}\right.\)
(ii) \(\int_{-1}^1 f(x) d x\) where
f(x) = \(\left\{\begin{array}{l}
1-2 x, \text { if } x \leq 0 \\
1+2 x, \text { if } x>0
\end{array}\right.\)
Solution:

Question 3.
(i) \(\int_0^{\pi / 2} \frac{d x}{1+\sqrt{\tan x}}\)
(ii) \(\int_0^{\pi / 2} \frac{1}{1+\sqrt{\cot x}} d x\)
(iii) \(\int_0^{\pi / 2} \frac{d x}{1+\tan x}\)
or \(\int_0^{\pi / 2} \frac{\cos x d x}{\sin x+\cos x}\)
(iv) \(\int_0^{\pi / 2} \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}} d x\)
(v) \(\int_0^{\pi / 2} \frac{\sqrt{(\cot x)}}{\sqrt{(\cot x)}+\sqrt{(\tan x)}} d x\)
or \(\int_0^{\pi / 2} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\)
(vi) \(\int_0^{\pi / 2} \frac{\sqrt{\sec x}}{\sqrt{\sec x}+\sqrt{cosec x}} d x\)
(vii) \(\int_0^{\pi / 2} \frac{\sin ^7 x}{\sin ^7 x+\cos ^7 x} d x\)
(viii) \(\int_0^{\pi / 2} \frac{\cos ^5 x}{\sin ^5 x+\cos ^5 x} d x\)
(ix) \(\int_0^a \frac{d x}{x+\sqrt{a^2-x^2}}\)
(x) \(\int_0^{\infty} \frac{x}{(1+x)\left(1+x^2\right)} d x\)
(xi) \(\int_0^{\pi / 2} \frac{\sqrt{\sin ^3 x}}{\sqrt{\sin ^3 x}+\sqrt{\cos ^3 x}} d x\)
(xii) \(\int_0^a \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x\)
(xiii) \(\int_0^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cos x} d x\)
(xiv) \(\int_0^{\pi / 2} \sin 2 x \log \tan x d x\)
(xv) \(\int_0^1 \log \left(\frac{1}{x}-1\right) d x\)
(xvi) \(\int_0^{\pi / 2}(2 \log \cos x-\log \sin 2 x) d x\)
(xvii) \(\int_{-\pi / 4}^{3 \pi / 2} \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}} d x\)
Solution:
(i) Let I = \(\int_0^{\pi / 2} \frac{d x}{1+\sqrt{\tan x}}\) … (1)

on adding eqn (1) & (2) ; we have
2I = \(\int_0^{\pi / 2} \frac{1+\sqrt{\tan x}}{1+\sqrt{\tan x}} d x=\int_0^{\pi / 2} d x=\frac{\pi}{2}\)
⇒ I = \(\frac { π }{ 4 }\)

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

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

(vii) Let I = \(\int_0^{\pi / 2} \frac{\sin ^7 x}{\sin ^7 x+\cos ^7 x} d x\) … (1)
we know that, \(\int_0^a f(x) d x=\int_0^a f(a-x) d x\)

(ix) Let I = \(\int_0^a \frac{d x}{x+\sqrt{a^2-x^2}}\)
put x = a sin θ ⇒ dx = a cosθ dθ
when x = 0 ⇒ θ = 0;
when x = a ⇒ θ = \(\frac { π }{ 2 }\)

(x) Let I = \(\int_0^{\infty} \frac{x}{(1+x)\left(1+x^2\right)} d x\)
put x = a sin θ ⇒ dx = sec² θ dθ
when x = 0 ⇒ θ = 0;
when x → ∞ ⇒ θ = \(\frac { π }{ 2 }\)

Question 4.
(i) \(\int_0^\pi \frac{x}{1+\sin x} d x\)
(ii) \(\int_0^\pi \frac{x \sin x}{1+\sin x} d x\)
(iii) \(\int_0^\pi \frac{x \sin x}{\sec x {cosec} x} d x\)
(iv) \(\int_0^\pi x \sin x \cos ^4 x d x\)
Solution:
(i) Let I = \(\int_0^\pi \frac{x}{1+\sin x} d x\) … (1)

Question 5.
(i) \(\int_0^2 x \sqrt{2-x} d x\)
(ii) \(\int_0^1 x(1-x)^{2 / 3} d x\)
(iii) \(\int_0^1 x(1-x)^5 d x\)
(iv) \(\int_0^1 \frac{x}{(1-x)^{3 / 4}} d x\)
Solution:

Question 6.
(i) \(\int_0^\pi \frac{d x}{\left(1+2 \sin ^2 x\right)}\)
(ii) \(\int_0^1 \frac{\log x}{\sqrt{1-x^2}} d x\)
(iii) \(\int_0^\pi \log (1+\cos \theta) d \theta\)
Solution:
(i) Let I = \(\int_0^\pi \frac{d x}{1+2 \sin ^2 x}\)
Divide Numerator and denominator by cos²x ; we have
I = \(\int_0^\pi \frac{\sec ^2 x d x}{\sec ^2 x+2 \tan ^2 x}\) = \(\int_0^\pi \frac{\sec ^2 x d x}{1+3 \tan ^2 x}\) = \(2 \int_0^{\pi / 2} \frac{\sec ^2 x d x}{1+3 \tan ^2 x}\)
[∵ \(\int_0^{2 a} f(x) d x=2 \int_0^a f(x)\)dx if f(2a – x) = f(x)]
put tan x = t ⇒ sec²x dx = dt
when x = 0 ⇒ t = 0 ;
when x = π/2 ⇒ t → ∞

(ii) Let I = \(\int_0^1 \frac{\log x}{\sqrt{1-x^2}} d x\)
put x = sin θ ⇒ dx = cosθ dθ
when x = 0 ⇒ θ = 0 ;
when x = 1 ⇒ θ = π/2

Question 7.
(i) \(\int_{-\frac{\pi}{2}}^{\pi / 2} \sin ^7 x d x\)
(ii) \(\int_{-1}^1 \sin ^{11} x d x\)
(iii) \(\int_{-\frac{\pi}{4}}^{\pi / 4} x^3 \sin ^4 x d x\)
(iv) \(\int_{-\pi}^\pi x^{10} \sin ^7 x d x\)
(v) \(\int_{-1}^1 \sin ^3 x \cos ^2 x d x\)
(vi) \(\int_{-8}^8\left(\sin ^{93} x+x^{295}\right) d x\)
Solution:
(i) Let f(x) = sin⁷x
∴ f(- x) = sin⁷(-x)
= [sin(-x)]⁷ = [- sin x]⁷
= – sin⁷x = – f (x)
∴ f(x) be an odd function.
Thus \(\int_{-\pi / 2}^{\pi / 2} f(x) d x\) = 0
⇒ \(\int_{-\pi / 2}^{\pi / 2} \sin ^7 x d x\) = 0

(ii) Let f(x) = sin¹¹x
∴ f(- x) = sin¹¹(-x)
= [sin(-x)]¹¹ = [- sin x]¹¹
= – sin¹¹x
= – f(x)
∴ f(x) be an odd function.
Thus, \(\int_{-1}^1 f(x) d x\) = 0
⇒ \(\int_{-1}^1 \sin ^{11} x d x\) = 0

(iii) Let f(x) = x³sin⁴x
∴ f(-x) = (-x)³sin⁴(-x)
= – x³[sin(-x)]⁴
= – x³(- sin x)⁴
= – x³ sin⁴x
= – f (x)
∴ f(x) be an odd function.
Thus, \(\int_{-\pi / 4}^{\pi / 4} f(x) d x\) = 0
⇒ \(\int_{-\pi / 4}^{\pi / 4} x^3 \sin ^4 x d x\) = 0

(iv) Let f(x) = x¹⁰sin⁷x
∴ f(-x) = (-x)¹⁰sin⁷(-x)
= x¹⁰[sin(-x)]⁷
= x¹⁰(-sin x)⁷
= – x¹⁰sin⁴x
∴ f (x) be an odd function.
Thus, \(\int_{-\pi}^\pi f(x) d x\) = 0
⇒ \(\int_{-\pi}^\pi x^{10} \sin ^7 x d x\) = 0

(v) Let f(x) = sin³x cos²x
∴ f(-x) = sin³(-x) cos²(-x)
= [sin(-x)]³ [cos(-x)]²
= [- sin x]³ [cos x]²
= – sin³x cos²x = – f (x)
∴ f(x) be an odd function.
Thus, \(\int_{-1}^1 f(x) d x\) = 0
⇒ \(\int_{-1}^1 \sin ^3 x \cos ^2 x d x\) = 0

(vi) Let f(x) = sin⁹³x + x²⁹⁵
∴ f(-x) = sin⁹³(-x) + (-x)²⁹⁵
= [sin(-x)]⁹³ – x²⁹⁵
= – sin⁹³x – x²⁹⁵
= – [sin⁹³x + x²⁹⁵] = – f (x)
∴ f(x) be an odd function.
Thus, \(\int_{-8}^8 f(x) d x\) = 0
⇒ \(\int_{-8}^8\left(\sin ^{93} x+x^{295}\right) d x\) = 0

Question 8.
Prove that :
\(\int_0^2 x^2 \sqrt{(2-x)} d x=\frac{128 \sqrt{2}}{105}\)
Solution:

Question 9.
\(\int_0^\pi \theta \sin ^2 \theta \cos ^2 \theta d \theta=\frac{\pi^2}{16}\).
Solution: