RBSE Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.3

Rajasthan Board RBSE Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.3 Textbook Exercise Questions and Answers.

RBSE Class 12 Maths Solutions Chapter 7 Integrals Ex 7.3

Question 1.
sin² (2x + 5)
Answer:

Question 2.
sin 3x cos 4x
Answer:

Question 3.
cos 2x cos 4x cos 6x
Answer:
Let I = ∫cos 2x cos 4x cos 6x dx

Question 4.
sin³ (2x + 1)
Answer:

Question 5.
sin³ x cos³ x
Answer:
Let I = ∫ sin³ x cos³ x dx
= ∫sin x sin² x cos³ x dx
= ∫sin x (1 - cos² x) cos³ x dx
Putting cos x = t
⇒ - sin x dx = dt
⇒ sin x dx = - dt
∴ ∫sin x(1 - cos² x) cos³ x dx
= - ∫(1 - t²) t³ dt
= - ∫ (t³ - t⁵) dt
= ∫t⁵ dt - ∫t³ dt
= \(\frac{t^{6}}{6}-\frac{t^{4}}{4}\) + C
= \(\frac{1}{6}\) cos⁶ x - \(\frac{1}{4}\) cos⁴ x + C

Question 6.
sin x sin 2x sin 3x
Answer:
Let I = ∫sin x sin 2x sin 3x
= \(\frac{1}{2}\) ∫2 sin x sin 2x sin 3x dx
[Multiplying numerator and denominator by 2]
= \(\frac{1}{2}\) ∫ (2 sin x sin 2x) sin 3x dx
= \(\frac{1}{2}\) ∫ [cos(x - 2x) - cos(x + 2x)] sin 3x dx
[∵ 2 sin A sin B = cos(A - B) - cos (A + B)]
= \(\frac{1}{2}\) ∫ [cos (- x) - cos 3x] sin 3x dx
= \(\frac{1}{2}\) ∫ (cos x - cos 3x) sin 3x dx [∵ cos (- θ) = cos θ)]
= \(\frac{1}{2}\) ∫cos x sin x dx - \(\frac{1}{2}\) ∫ cos 3x sin 3x dx
= \(\frac{1}{4}\) ∫ 2 sin 3x cos x dx - \(\frac{1}{4}\) ∫ 2 cos 3x sin 3x dx
[Multiplying numerator and denominator by 2]

Question 7.
sin 4x sin 8x
Answer:

Question 8.
\(\frac{1-\cos x}{1+\cos x}\)
Answer:
Let

Question 9.
\(\frac{\cos x}{1+\cos x}\)
Answer:

Question 10.
sin⁴ x
Answer:

Question 11.
cos⁴ 2x
Answer:

Question 12.
\(\frac{\sin ^{2} x}{1+\cos x}\)
Answer:
∫\(\frac{\sin ^{2} x}{1+\cos x} dx = ∫\frac{1-\cos ^{2} x}{1+\cos x}\) dx
= ∫ \(\frac{(1-\cos x)(1+\cos x)}{(1+\cos x)}\) dx
= ∫ (1 - cos x) dx = ∫ dx - ∫ cos x dx
= x - sin x + C

Question 13.
\(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\)
Answer:

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

Question 14.
\(\frac{\cos x-\sin x}{1+\sin 2 x}\)
Answer:

Question 15.
tan³ 2x sec 2x
Answer:
Let I = ∫tan³ 2x
= ∫tan² 2x.tan 2x sec 2x dx
= ∫ (sec² 2x - 1) tan 2x sec 2x dx
Putting sec 2x = t
⇒ 2 sec 2x tan 2x dx = dt
⇒ sec 2x tan 2x dx = \(\frac{1}{2}\) dt

Question 16.
tan⁴ x
Answer:
Let I = ∫ tan⁴ x dx = ∫ tan² x.tan² x dx
= ∫ tan² x (sec² x - 1) dx
= ∫ tan² x sec² x dx - ∫ tan² x dx
= ∫ tan² x sec² x dx - ∫ (sec² x - 1) dx
= ∫ tan² x sec² x dx - ∫ (sec² x - 1) dx
= ∫ tan² x sec² x dx - ∫ sec² x dx + ∫ dx .......... (i)
Putting tan x = t
⇒ sec² x dx = dt
∴ ∫ tan² x sec² x dx = ∫ t² dt
= \(\frac{t^{3}}{3}=\frac{\tan ^{3} x}{3}\)
From (i),
∴ ∫ tan² x sec² x dx - ∫ sec² x dx + ∫ dx
= \(\frac{\tan ^{3} x}{3}\) - tan x + x + C
= \(\frac{1}{3}\)tan³ x - tan x + x + C

Question 17.
\(\frac{\sin ^{3} x+\cos ^{3} x}{\sin ^{2} x \cos ^{2} x}\)
Answer:

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

Question 18.
\(\frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x}\)
Answer:

Question 19.
\(\frac{1}{\sin x \cos ^{3} x}\)
Answer:

Question 20.
\(\frac{\cos 2 x}{(\cos x+\sin x)^{2}}\)
Answer:

Question 21.
sin^-1 (cos x)
Answer:

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

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
Answer:

= ∫ sec² x dx - ∫ cosec² x dx
= tan x + cot x + C
Hence, (A) is the correct answer.

Question 24.
∫ \(\frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)}\) dx equals:
(A) - cot (ex^x) + C
(B) tan (xe^x) + C
(C) tan (e^x) + C
(D) cot (e^x) + C
Answer:
∫ \(\frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)}\) dx
Putting e^x x = t
⇒ (xe^x + e^x) dx = dt
⇒ e^x (1 + x) dx = dt
∴ ∫\(\frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)}\) dx = ∫\(\frac{d t}{\cos ^{2} t}\)
= ∫ sec² t dt
= tan t + C
= tan (e^xx) + C
Hence, (B) is correct answer.