# CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Ex 9(c)

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 9 Integration Ex 9(c) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Exercise 9(c)

Integrate the following:
Question 1.
(i) ∫sin 4x cos 3x dx
Solution:

(ii) ∫cos 5x cos 2x dx
Solution:

(iii) ∫sin x cos 4x dx
Solution:

(iv) ∫sin 6x sin 3x dx
Solution:

(v) ∫cos 4x cos 5x sin 2x dx
Solution:

(vi) ∫sin$$\frac{3 x}{4}$$ cos$$\frac{x}{2}$$ dx
Solution:

(vii) ∫cos 2x cos$$\frac{x}{2}$$ dx
Solution:

(viii) ∫sin$$\frac{x}{2}$$ sin$$\frac{x}{3}$$ cos$$\frac{x}{4}$$ dx
Solution:

Question 2.
(i) ∫cos2 x dx
Solution:

(ii) ∫sin3 x dx
Solution:
(ii) ∫sin3 x dx = ∫sin2 x . sin x dx
= ∫(1 – cos2 x) . sin x dx
[Put cos x = t
Then sin x dx = -dt]
= ∫(1 – t2) . (-dt) = $$\frac{1}{3}$$t3 – t + C
= $$\frac{1}{3}$$cos3 x – cos x + C

(iii) ∫cos4 x dx
Solution:

(iv) ∫sin5 x dx
Solution:
∫sin5 x dx = ∫sin4 x sin x dx
= ∫(1 – cos2 x)2. sin x dx
[Put cos x = t
Then sin x dx = -dt]
= -∫(1 – t2)2 dt
= -∫(1 – 2t + t4) dt
= -(t – $$\frac{2}{3}$$t3 + $$\frac{1}{5}$$t5) + C
= $$\frac{2}{3}$$cos3 x – cos x – $$\frac{1}{5}$$cos5 x + C

(v) ∫cos7 x dx
Solution:

(vi) ∫sin6 x dx
Solution:

(vii) ∫cos5 x sin3 x dx
Solution:
∫cos5 x sin3 x dx
= ∫cos5 x (1 – cos2 x) sin x dx
[Put cos x = t Then sin x dx = -dt]
= ∫t5 (1 – t2 ). (-dt) = ∫(t7 – t5) dt
=  $$\frac{1}{8}$$t8 – $$\frac{1}{8}$$t6 + C
= $$\frac{1}{8}$$cos8 x – $$\frac{1}{8}$$cos6 x + C

(viii) ∫sin20 x cos20 x dx
Solution:
∫sin20 x cos20 x dx
= ∫sin20 x (1 – sin2 x) cos x dx
[Put sin x = t
Then cos x dx = -dt]
= ∫t20 (1 – t20). (-dt) = ∫(t20 – t22) dt
=  $$\frac{1}{21}$$t21 – $$\frac{1}{23}$$t23 + C
= $$\frac{1}{21}$$sin21 x – $$\frac{1}{23}$$sin23 x + C

(ix) ∫$$\frac{\sin ^3 x}{\cos ^6 x}$$ dx
Solution:

(x) ∫cot3 x cosec16 x dx
Solution:

(xi) ∫sec30 x tan x dx
Solution:
∫sec30 x tan x dx
= ∫sec29 x . sec x . tan x dx
[Put sec x = t
Then sec x. tan x dx = dt]
= ∫t29 dt
= $$\frac{1}{30}$$t30 + C
= $$\frac{1}{30}$$sec30 x + C

(xii) ∫sin3 x sec14 x dx
Solution:

Question 3.
(i) ∫sin4 x cos4 x dx
Solution:

(ii) ∫sin 3x cos2 x dx
Solution:

(iii) ∫cos 2x sin3 x dx
Solution:

(iv) ∫sin4 x cos2 x dx
Solution:

Question 4.
(i) ∫tan5 θ sec4 θ dθ
Solution:
∫tan5 θ sec4 θ dθ
= ∫tan5 θ (1 + tan2 θ) sec2 θ dθ
[Put tan θ = t
Then sec2 θ dθ = dt]
= ∫t5 (1 + t2) dt = ∫(t5 + t7) dt
= $$\frac{1}{6}$$t6 + $$\frac{1}{8}$$t8 + C
= $$\frac{1}{6}$$tan6 θ + $$\frac{1}{8}$$tan8 θ + C

(ii) ∫cot4 θ cosec4 θ dθ
Solution:
∫cot4 θ cosec4 θ dθ
= ∫cot4 θ (1 + cot2 θ) cosec2 θ dθ
[Put cot θ = t
Then cosec2 θ dθ = -dt]
= ∫t4 (1 + t2) (-dt) = ∫(t4 + t6) dt
= -{$$\frac{1}{5}$$t5 + $$\frac{1}{7}$$t7} + C
= -{$$\frac{1}{5}$$cot5 θ+ $$\frac{1}{7}$$cot7 θ} + C

(iii) ∫sec11 θ tan θ dθ
Solution:
∫sec11 θ tan θ dθ
= ∫sec10 θ sec θ tan θ dθ
[Put sec θ = t
Then sec θ tan θ dθ = -dt]
= ∫t10 dt = $$\frac{1}{11}$$t11 + C
= $$\frac{1}{11}$$sec11 θ + C

(iv) ∫cot θ cosec7 θ dθ
Solution:
∫cot θ cosec7 θ dθ
= ∫cosec6 θ . (cosec θ – cot θ) dθ
[Put sec θ = t
Then cosec θ . cot θ dθ = dt]
= ∫t6 (dt) = –$$\frac{1}{17}$$t7 + C
= –$$\frac{1}{17}$$cosec7 θ + C

(v) ∫tan3 θ dθ
Solution:
∫tan3 θ d0 = ∫tan θ – (tan2 θ) dθ
= ∫tan θ (sec2 θ – 1) dθ
= ∫tan θ sec2 θ dθ – ∫tan θ dθ
= ∫tan θ d(tan θ) – ∫tan θ dθ
= $$\frac{1}{2}$$tan2 θ + ln |cos θ| + C

(vi) ∫cot4 θ dθ
Solution:
∫cot4 θ dθ = ∫cot2 θ cot2 θ dθ
= ∫cot2 θ (cosec2 θ – 1) dθ
= ∫cot2 θ . cosec2 θ dθ – ∫cot2 θ dθ
= ∫cot2 θ . cosec2 θ dθ – ∫(cosec2 θ – 1) dθ
= $$\frac{1}{3}$$cot3 θ + cot θ + θ + C

(vii) ∫tan5 θ dθ
Solution:
∫tan5 θ dθ = ∫tan3 θ . (sec2 θ – 1) dθ
= ∫tan3 θ . sec2 θ dθ – ∫tan3 θ dθ
= ∫tan3 θ . sec2 θ dθ – ∫tan θ (sec2 θ – θ) dθ
= ∫tan3 θ d(tan θ) – ∫tan θ d(tan θ) + ∫tan θ dθ
= $$\frac{1}{4}$$tan4 θ – $$\frac{1}{2}$$tan2 θ + ln |sec θ| + C

(viii) ∫cot6 θ dθ
Solution:
∫cot6 θ dθ
= ∫cot4 θ (cosec2 θ – 1) dθ
= ∫cot4 θ . cosec2 θ dθ – ∫cot4 θ dθ
= ∫cot4 θ cosec2 θ dθ – ∫cot2 θ (cosec2 θ – 1) dθ
= ∫cot4 θ . cosec2 θ dθ – ∫cot2 θ cosec2 θ dθ + ∫(cosec2 θ – 1) dθ
= -cot4 θ d(cot θ) + ∫cot2 θ d(cot θ) + ∫cosec2 θ dθ – ∫dθ
= –$$\frac{1}{5}$$cot5 θ + $$\frac{1}{3}$$cot3 θ – cot θ – θ + C

Question 5.
(i) $$\frac{\sin a x-\sin b x}{\cos a x-\cos b x}$$ dx
Solution:

(ii) $$\frac{\cos p x+\cos q x}{\sin p x+\sin q x}$$ dx
Solution:
$$\frac{\cos p x+\cos q x}{\sin p x+\sin q x}$$ dx

(iii) $$\frac{\sin 4 x-\sin 2 x}{\cos x}$$ dx
Solution:

(iv) $$\frac{\sin 2 x}{a \cos ^2 x+b \sin ^2 x+c}$$ dx
Solution: