Evaluate the following integral: ∫ cos 9 x sin x d x

Let I = ∫ cos 9 x sin x d x = ∫ cos 8 x · cos x sin x d x = ∫ cos 2 x 4 · cos x sin x d x = ∫ 1 - sin 2 x 4 · cos x sin x d x Let sin x = t ⇒ cos x d x = d t ⇒ I = ∫ 1 - t 2 4 t d t = ∫ 1 - 2 t 2 + t 4 2 t d t = ∫ 1 + 4 t 4 + t 8 - 4 t 2 - 4 t 6 + 2 t 4 t d t = ∫ 1 + 6 t 4 + t 8 - 4 t 2 - 4 t 6 t d t = ∫ 1 t + 6 t 3 + t 7 - 4 t - 4 t 5 d t = log t + 6 t 4 4 + t 8 8 - 4 t 2 2 - 4 t 6 6 + C = log t + 3 2 t 4 + t 8 8 - 2 t 2 - 2 3 t 6 + C Again substituting the value of t , we get I = log sin x + 3 2 sin x 4 + sin x 8 8 - 2 sin x 2 - 2 3 sin x 6 + C

Evaluate the following integral: ∫ d x ( 3 cos x + 4 sin x )

Let I = ∫ d x 3 cos x + 4 sin x Multiply and divide the denominator by 5 I = 1 5 ∫ d x 3 5 cos x + 4 5 sin x Put 4 5 = cos y , 3 5 = sin y ∴ 3 2 + 4 2 = 5 2 , tan y = 3 4 I = 1 5 ∫ d x ( sin y cos x + cos y sin x ) = 1 5 ∫ d x sin ( x + y ) = 1 5 ∫ cosec ( x + y ) d x Integrating it, we get = 1 5 log cosec ( x + y ) - cot ( x + y ) + C I = 1 5 log cosec ( x + y ) - cot ( x + y ) + C = 1 5 log 1 - cos ( x + y ) sin ( x + y ) + C = 1 5 log tan ( x + y 2 ) + C ∵ cos x + y = 1 - 2 sin 2 x + y 2 and sin x + y = 2 sin x + y 2 cos x + y 2 Hence, I = 1 5 log tan x 2 + 1 2 tan - 1 3 4 + C

Evaluate the following integral: ∫ d x ( a cos x + b sin x ) 2 , a > 0 and b > 0

Let I = ∫ d x ( a cos x + b sin x ) 2 Put a = cos y , b = sin y Then, a 2 + b 2 = 1 , tan y = b a ⇒ I = ∫ d x ( cos y cos x + sin y sin x ) 2 = ∫ d x cos 2 ( x - y ) = ∫ sec 2 ( x - y ) d x Integrating it, we get I = tan ( x - y ) + C I = tan x - tan - 1 b a + C Hence, I = 1 a 2 + b 2 tan x - tan - 1 b a + C ∵ a 2 + b 2 = 1

Evaluate the following integral: ∫ d x ( cos x - sin x )

Let I = ∫ d x cos x - sin x Multiply and divide the denominator by 2 I = 1 2 ∫ d x 1 2 cos x - 1 2 sin x = 1 2 ∫ d x sin π 4 cos x - cos π 4 sin x = 1 2 ∫ d x sin ( π 4 - x ) = 1 2 ∫ cosec ( π 4 - x ) d x After integrating it, We get = - 1 2 log cosec ( π 4 - x ) - cot ( π 4 - x ) + C = - 1 2 log 1 sin π 4 - x - cos π 4 - x sin π 4 - x + C = - 1 2 log 1 - cos π 4 - x sin π 4 - x = - 1 2 log 1 - 1 - 2 sin 2 π 8 - x 2 2 sin π 8 - x 2 cos π 8 - x 2 Hence, I = - 1 2 log tan ( π 8 - x 2 ) + C

Evaluate the following integral: ∫ ( 2 tan x - 3 cot x ) 2 d x

Let I = ∫ ( 2 tan x - 3 cot x ) 2 d x = ∫ 4 tan 2 x + 9 cot 2 x - 12 tan x cot x d x = ∫ 4 sec 2 x - 1 + 9 cosec 2 x - 1 - 12 d x = ∫ 4 sec 2 x d x + ∫ 9 cosec 2 x d x - ∫ 25 d x Hence, I = 4 tan x - 9 cot x - 25 x + C

Evaluate the following integral: ∫ sin x sin 2 x sin 3 x d x

Let I = ∫ sin x sin 2 x sin 3 x d x I = 1 2 ∫ ( cos 2 x - cos 4 x ) sin 2 x d x ∵ 2 sin A sin B = cos A - B - cos A + B = 1 2 ∫ sin 2 x cos 2 x d x - 1 2 ∫ sin 2 x cos 4 x d x = 1 4 ∫ sin 4 x d x - 1 4 ∫ ( sin 6 x - sin 2 x ) d x ∵ sin A + B + sin A - B = 2 sin A cos B Hence, I = - cos 4 x 16 + cos 6 x 24 - cos 2 x 8 + C

Evaluate the following integral: ∫ 1 - cot x 1 + cot x d x

Let I = ∫ 1 - cot x 1 + cot x d x ⇒ I = ∫ 1 - cos x sin x 1 + cos x sin x d x I = ∫ sin x - cos x sin x + cos x d x = - ∫ cos x - sin x sin x + cos x d x Put sin x + cos x = t ⇒ ( cos x - sin x ) d x = d t ⇒ I = - ∫ d t t = - log t + C ∴ I = - log sin x + cos x + C

Evaluate the following integral: ∫ d x ( 2 sin x + cos x + 3 )

Let I = ∫ d x cos x + 2 sin x + 3 = ∫ d x 1 - tan 2 x 2 1 + tan 2 x 2 + 2 2 tan x 2 1 + tan 2 x 2 + 3 ⇒ I = ∫ sec 2 x 2 d x 3 + 1 + 2 tan 2 x 2 + 4 tan x 2 Put tan x 2 = t ∴ 1 2 sec 2 x 2 d x = d t ⇒ I = ∫ 2 d t 4 + 4 t + 2 t 2 = ∫ d t 2 + 2 t + t 2 = ∫ d t ( t + 1 ) 2 + 2 - 1 = ∫ d t ( t + 1 ) 2 + 1 = ∫ d t ( 1 ) 2 + ( t + 1 ) 2 = tan - 1 ( t + 1 ) + C Again substituting the value of t we get I = tan - 1 tan x 2 + 1 + C

E M B I B E

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Integral Calculus>Chapter 13 - Methods of Integration>EXERCISE

R S Aggarwal Solutions for Exercise 3: EXERCISE

Author:R S Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 3: EXERCISE

Attempt the free practice questions from Exercise 3: EXERCISE with hints and solutions to strengthen your understanding. SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12 solutions are prepared by Experienced Embibe Experts.