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: