PSEB 12th Class Maths Solutions Chapter 7 Integrals Ex 7.3

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 7 Integrals Ex 7.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.3

Direction (1 – 22): Find the integrals of the functions:

Question 1.
sin² (2x + 5)
Solution.
= ∫ sin² (2x + 5) dx = ∫ \(\frac{1-\cos (4 x+10)}{2}\) dx
= ∫ 1 dx + \(\frac{1}{2}\) ∫ cos (4x + 10) dx
= \(\frac{1}{2}\) x – \(\frac{1}{2}\) (\(\left(\frac{\sin (4 x+10)}{4}\right)\)) + C
= \(\frac{x}{2}\) – \(\frac{1}{8}\) sin(4x + 10) + C

Question 2.
sin 3x cos 4x
Solution.
∫ sin 3x cos 4x dx = \(\frac{1}{2}\) ∫ {sin(3x + 4x) + sin(3x – 4x)} dx
[∵ 2 sin A cos B = sin (A + B) + sin(A – B)]
= \(\frac{1}{2}\) ∫ {sin 7x + sin (- x)}dx

= \(\frac{1}{2}\) ∫ {sin 7x – sin x} dx

= \(\frac{1}{2}\) ∫ sin 7x dx – \(\frac{1}{2}\) ∫ sin x dx

= \(\frac{1}{2}\) \(\left(\frac{-\cos 7 x}{7}\right)\) – \(\frac{1}{2}\) (- cos x) + C

= \(\frac{-\cos 7 x}{14}+\frac{\cos x}{2}\) + C

Question 3.
cos 2x cos 4x cos 6x
Solution.
∫ cos 2x (cos 4x cos 6x) dx = ∫ cos 2x [\(\frac{1}{2}\) {cos (4x +6x) + cos (4x – 6x)}] dx
[∵ 2 cos A cos B = cos (A + B) + cos (A – B)]
= \(\frac{1}{2}\) ∫ {cos 2x cos 10x + cos 2x cos (- 2x)}dx 2J

= \(\frac{1}{2}\) ∫ {cos 2x cos 10x + cos² 2x}dx

= \(\frac{1}{2}\) ∫ (cos 12x + cos 8x + 1 + cos 4x)dx

= \(\frac{1}{4}\left[\frac{\sin 12 x}{12}+\frac{\sin 8 x}{8}+x+\frac{\sin 4 x}{4}\right]\) + C

Question 4.
sin³ (2x + 1)
Solution.
Let I = ∫ sin³ (2x + 1) dx
⇒ = ∫ sin² (2x + 1) . sin (2x + 1) dx
= ∫ (1 – cos² (2x + 1)) sin(2x +1) dx
Let cos (2x + 1) = t
⇒ – 2 sin (2x + 1) dx = dt
⇒ sin (2x + 1) dx = \(\frac{-d t}{2}\)
∴ I = – \(\frac{1}{2}\) ∫ (1 – t²) dt
= \(\frac{-1}{2}\left\{t-\frac{t^{3}}{3}\right\}\) + C

Question 5.
sin³ x cos³ x
Solution.
Let I = ∫ sin³ x cos³ x dx
= ∫ cos³ x . sin² x . sin x dx
= ∫ cos³ x (1 – cos² x) sin x dx
Let cos x = t
⇒ – sin x dx = dt
⇒ dx = \(\frac{d t}{-\sin x}\)
⇒ I = – ∫ t³ (1 – t²) dt
= – ∫ (t³ – t⁵) dt
= – \(\left\{\frac{t^{4}}{4}-\frac{t^{6}}{6}\right\}\) + C

= – \(\left\{\frac{\cos ^{4} x}{4}-\frac{\cos ^{6} x}{6}\right\}\) + C

= \(\frac{\cos ^{6} x}{6}-\frac{\cos ^{4} x}{4}\) + C

Question 6.
sin x sin 2x sin 3x
Solution.
∫ sin x sin 2x sin 3x dx = ∫ [sin x . \(\frac{1}{2}\) {cos(2x – 3x) – cos(2x + 3x)}] dx
[∵ 2 sin A sin B = cos(A – B) – cos(A + B)]
= \(\frac{1}{2}\) ∫ (sin x cos (- x) – sin x cos 5x) dx
= \(\frac{1}{2}\) ∫ (sin x cos x – sin x cos 5x) dx

Question 7.
sin 4x sin 8x
Solution.
∫ sin 4x sin 8x dx = ∫ {\(\frac{1}{2}\) cos(4x – 8x) – cos(4x + 8x)} dx
= \(\frac{1}{2}\) ∫ (cos (- 4x) – cos 12x) dx
= \(\frac{1}{2}\) ∫(cos 4x – cos 12x) dx
= \(\frac{1}{2}\) \(\) + C
= \(\) [sin 4x – \([latex]\)[/latex] sin 12x] + C

Question 8.
\(\frac{1-\cos x}{1+\cos x}\)
Solution.
∫ \(\frac{1-\cos x}{1+\cos x}\) dx = ∫ \(\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\) dx

[∵ 1 – cos x = 2 sin² \(\frac{x}{2}\) and
1 + cos x = 2 cos² \(\frac{x}{2}\)]

= ∫ tan² \(\frac{x}{2}\) dx

= (sec² \(\frac{x}{2}\) – 1) dx

= ∫ sec² \(\frac{x}{2}\) dx – ∫ 1 dx

= \(\left[\frac{\tan \frac{x}{2}}{\frac{1}{2}}-x\right]\) + C

= 2 tan \(\frac{x}{2}\) – x + C

Question 9.
\(\frac{\cos x}{1+\cos x}\)
Solution.
∫ \(\frac{\cos x}{1+\cos x}\) dx = ∫ \(\frac{\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\) dx
[∵ cos x = cos² \(\frac{x}{2}\) – sin² \(\frac{x}{2}\) and
cos x = 2 cos² \(\frac{x}{x}\) – 1]

= \(\frac{1}{2}\) ∫ (1 – tan² \(\frac{x}{2}\)) dx

= \(\frac{1}{2}\) ∫ (1 – sec² \(\frac{x}{2}\) + 1 ) dx

= \(\frac{1}{2}\) ∫ (2 – sec² \(\frac{x}{2}\)) dx

= \(\frac{1}{2}\) ∫ 2 dx – \(\frac{1}{2}\) ∫ sec² \(\frac{x}{2}\) dx

= \(\frac{1}{2}\) [2x – \(\frac{\tan \frac{x}{2}}{\frac{1}{2}}\)] + C

= x – tan \(\frac{x}{2}\) + C

Question 10.
sin⁴ x
Solution.
∫ sin⁴ x dx = ∫ sin² x sin² x dx

Question 11.
cos⁴ 2x
Solution.
∫ cos⁴ 2x dx = ∫ (cos² 2x)² dx

Question 12.
\(\frac{\sin ^{2} x}{1+\cos x}\)
Solution.
∫ \(\frac{\sin ^{2} x}{1+\cos x}\) dx = ∫ \(\frac{\left(1-\cos ^{2} x\right)}{1+\cos x}\) dx

= ∫ \(\frac{(1-\cos x)(1+\cos x)}{1+\cos x}\) dx

[∵ a² – ² = (a – b) (a + b)]

= ∫ (1 – cos x) dx = x – sin x + C

Question 13.
\(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\)
Solution.
∫ \(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\) dx = ∫ \(\frac{\left(2 \cos ^{2} x-1\right)-\left(2 \cos ^{2} \alpha-1\right)}{\cos x-\cos \alpha}\) dx

= ∫ \(\frac{2\left(\cos ^{2} x-\cos ^{2} \alpha\right)}{\cos x-\cos \alpha}\) dx

= ∫ \(\frac{2(\cos x-\cos \alpha)(\cos x+\cos \alpha)}{\cos x-\cos \alpha}\) dx

= 2 ∫(cos x + cos α) dx
= 2 sin x + 2 cos α x + C
= 2 (sin x + x cos α) + C

Question 14.
\(\frac{\cos x-\sin x}{1+\sin 2 x}\)
Solution.
∫ \(\frac{\cos x-\sin x}{1+\sin 2 x}\) dx = ∫ \(\frac{\left(2 \cos ^{2} x-1\right)-\left(2 \cos ^{2} \alpha-1\right)}{\cos x-\cos \alpha}\) dx

Question 15.
tan³ 2x sec 2x
Solution.
∫ tan³ 2x sec 2x dx = ∫ tan² 2x tan 2x sec 2x dx
= ∫ (sec² 2x – 1) tan 2x sec 2x dx
= ∫ (sec² 2x tan 2x sec 2x – tan 2x sec 2x) dx
= ∫ sec² 2x tan 2x sec 2x dx – ∫ tan 2xsec 2x dx
= ∫ sec² 2x tan 2x sec 2x dx – \(\frac{\sec 2 x}{2}\) + C
Let sec 2x = t
∴ 2 sec 2x tan 2x dx = dt
∴ ∫ tan³ 2x sec 2x dx = \(\frac{1}{2}\) ∫ t² dt – \(\frac{\sec 2 x}{2}\) + C
= \(\frac{t^{3}}{6}-\frac{\sec 2 x}{2}\) + C

= \(\frac{(\sec 2 x)^{3}}{6}-\frac{\sec 2 x}{2}\) + C

Question 16.
tan4 x
Sol.
tan⁴ x dx = ∫ tan² x . tan² x dx
= ∫ [(sec² x – 1) tan² x] dx
= ∫ [sec² x tan² x – tan² x] dx
= ∫ sec² x tan² x dx – ∫ (sec² x – 1)dx
= ∫ sec² x tan² x dx – ∫ sec² x dx + ∫ 1 dx
= ∫ sec² x tan² x dx – tan x + x + C ……………….(i)
= ∫ sec² x tan² x dx
Let tan x = t
⇒ sec² x dx = dt
∫ sec² x tan² x dx = ∫ t² dt
= \(\frac{\tan ^{3} x}{3}\)
From Eq. (i), we get
∫ tan⁴ x dx = \(\frac{1}{3}\) tan³ x – tan x + x + C

Question 17.
\(\frac{\sin ^{3} x+\cos ^{3} x}{\sin ^{2} x \cos ^{2} x}\)
Solution.
∫ \(\frac{\sin ^{3} x+\cos ^{3} x}{\sin ^{2} x \cos ^{2} x}\) dx = \(\int \frac{\sin ^{3} x}{\sin ^{2} x \cos ^{2} x} d x+\int \frac{\cos ^{3} x}{\sin ^{2} x \cos ^{2} x} d x\)

= \(\int \frac{\sin x}{\cos ^{2} x} d x+\int \frac{\cos x}{\sin ^{2} x} d x\)

= ∫ tan x sec x dx + ∫ cot x cosec x dx

= sec x – cosec x + C

Question 18.
\(\frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x}\)
Solution.
∫ \(\frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x}\) dx = ∫ \(\frac{\cos 2 x+(1-\cos 2 x)}{\cos ^{2} x}\) dx
[∵ cos 2x = 1 – 2 sin² x]

∫ \(\frac{1}{\cos ^{2} x}\) dx
= ∫ sec² x dx = tan x + C

Question 19.
\(\frac{1}{\sin x \cos ^{3} x}\)
Solution.
∫ \(\frac{1}{\sin x \cos ^{3} x}\) dx

= ∫ tan x sec² x dx + ∫ \(\frac{\sec ^{2} x}{\tan x}\) dx
Let tan x = t
⇒ sec² x dx = dt
= ∫ t dt + ∫ \(\frac{1}{t}\) dt
= \(\frac{t^{2}}{2}\) + log |t| + C
= \(\frac{1}{2}\) tan² x + log |tan x| + C

Question 20.
\(\frac{\cos 2 x}{(\cos x+\sin x)^{2}}\)
Solution.
∫ \(\frac{\cos 2 x}{(\cos x+\sin x)^{2}}\) dx = ∫ \(\frac{\cos 2 x}{\cos ^{2} x+\sin ^{2} x+2 \sin x \cos x}\) dx
= ∫ \(\frac{\cos 2 x}{(1+\sin 2 x)}\) dx
Let 1 + sin 2x = t
⇒ 2 cos 2x dx = dt
∴ ∫ \(\frac{\cos 2 x}{(\cos x+\sin x)^{2}}\) dx = \(\frac{1}{2}\) ∫ \(\frac{1}{t}\) dt

= \(\frac{1}{2}\) log |t| + C

= \(\frac{1}{2}\) log|1 + sin 2x| + C

= \(\frac{1}{2}\) log |(sin x + cos x)²| + C

= log |sin x + cos x| + C

Question 21.
sin^-1 x (cos x)
Solution.
∫ sin^-1 x (cos x) dx = ∫ sin^-1 x [sin (\(\frac{\pi}{2}\) – x)] dx
= ∫ (\(\frac{\pi}{2}\) – x) = \(\frac{\pi}{2}\) ∫ dx – ∫ x dx
= \(\frac{\pi x}{2}-\frac{x^{2}}{2}\) + C

Question 22.
\(\frac{1}{\cos (x-a) \cos (x-b)}\)
Solution.

Question 23.
∫ \(\frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x}\) dx is equal to
(A) tan x + cot x + C
(B) tan x + cosec x + C
(C) tan x + cot x + C
(D) tan x + sec x + C
Solution.
∫ \(\frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x}\) dx = ∫ \(\left(\frac{\sin ^{2} x}{\sin ^{2} x \cos ^{2} x}-\frac{\cos ^{2} x}{\sin ^{2} x \cos ^{2} x}\right)\) dx
= ∫ (sec² x – cosec² x) dx
= tan x + cot x + C
Hence, the correct answer is (A).

Question 25.
∫ \(\frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)}\) dx equals
(A) – cot (e x^x) + C
(B) tan (x e^x) + C
(C) tan (e^x) + C
(D) cot (e^x) + C
Solution.
∫ \(\frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)}\) dx
Let x e^x = t
⇒ (e^x . x + e^x . 1) dx = dt
⇒ e^x (x + 1) dx = dt
∴ \(\int \frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)} d x=\int \frac{d t}{\cos ^{2} t}\) = ∫ sec² t dt
= tan t + C
= tan (e^x x) + C
Hence, the correct answer is (B).