Evaluate the following integral : ∫ cos 4 x - 1 cot x - tan x d x .

Let I = ∫ cos 4 x - 1 cot x - tan x d x ⇒ I = ∫ cos 4 x - 1 cos x sin x - sin x cos x d x ⇒ I = ∫ cos 4 x - 1 cos 2 x - sin 2 x sin x cos x d x ⇒ I = - ∫ 2 sin 2 2 x cos 2 x sin x · cos x d x ⇒ I = ∫ cos 2 2 x - 1 sin 2 x cos 2 x · d x ∵ cos 2 x + sin 2 x = 1 , x ∈ R Now, substitute cos 2 x = t Differentiating with respect to x , then we get: - 2 sin 2 x · d x = d t Therefore, I = 1 2 ∫ - t 2 - 1 t · d t ⇒ I = 1 2 ∫ 1 - t 2 t · d t ⇒ I = 1 2 ∫ 1 t - t d t ⇒ I = 1 2 log t - t 2 2 + c ⇒ I = 1 2 log | cos 2 x | - 1 4 cos 2 2 x + c [Re-substituted t = cos 2 x ] Hence, ∫ cos 4 x - 1 cot x - tan x d x = 1 2 log | cos 2 x | - 1 4 cos 2 2 x + c .

Evaluate the following integral : ∫ cos 4 x + 1 cot x - tan x d x .

Let I = ∫ cos 4 x + 1 cot x - tan x d x ⇒ I = ∫ cos 4 x + 1 cos x sin x - sin x cos x d x ⇒ I = ∫ cos 4 x + 1 cos 2 x - sin 2 x sin x cos x d x ⇒ I = ∫ 2 cos 2 2 x cos 2 x sin x · cos x d x ⇒ I = ∫ cos 2 x · sin 2 x d x ⇒ I = 1 2 ∫ 2 cos 2 x · sin 2 x d x ⇒ I = 1 2 ∫ sin 4 x d x ⇒ I = - 1 8 cos 4 x + C Hence, ∫ cos 4 x + 1 cot x - tan x d x = - 1 8 cos 4 x + C .

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

Let I = ∫ cos 3 x sin 2 x + sin x d x ⇒ I = ∫ cos 2 x · cos x sin 2 x + sin x d x ⇒ I = ∫ 1 - sin 2 x · cos x sin 2 x + sin x d x Now, substitute sin x = t Differentiating with respect to x , then we get: cos x · d x = d t Therefore, I = ∫ 1 - t 2 · d t t 2 + t ⇒ I = ∫ ( 1 + t ) · ( 1 - t ) t ( 1 + t ) d t ⇒ I = ∫ ( 1 - t ) t d t ⇒ I = ∫ 1 t - 1 d t ⇒ I = ∫ 1 t d t - ∫ 1 · d t ⇒ I = log t - t + c ⇒ I = log | sin x | - sin x + c [Re-substituted t = sin x ] Hence, ∫ cos 3 x sin 2 x + sin x d x = log | sin x | - sin x + c .

Evaluate the following integral : ∫ cos 2 cot - 1 1 - x 1 + x d x .

Let I = ∫ cos 2 cot - 1 1 - x 1 + x d x Consider, y = cos 2 cot - 1 1 - x 1 + x Put x = cos 2 θ ⇒ 2 θ = cos - 1 x , then y = cos 2 cot - 1 1 - cos 2 θ 1 + cos 2 θ ⇒ y = cos 2 cot - 1 2 sin 2 θ 2 cos 2 θ ⇒ y = cos 2 cot - 1 sin θ cos θ ⇒ y = cos 2 cot - 1 tan θ ⇒ y = cos 2 cot - 1 cot π 2 - θ ⇒ y = cos 2 π 2 - θ [Using cot - 1 ( cot θ ) = θ , Provided that 0 < θ < π ] ⇒ y = cos π - 2 θ ⇒ y = - cos 2 θ ⇒ y = - cos cos - 1 x ⇒ y = - x [Using cos cos - 1 x = x , Provided that - 1 ≤ x ≤ 1 ] Therefore, I = ∫ - x d x ⇒ I = - x 2 2 + c Hence, ∫ cos 2 cot - 1 1 - x 1 + x d x = - x 2 2 + c .

Evaluate the following integral : ∫ d x sin 3 x cos ( x - a ) .

Let I = ∫ d x sin 3 x cos ( x - a ) ⇒ I = ∫ 1 sin 3 x ( cos x cos a + sin x sin a ) d x ⇒ I = ∫ 1 sin 3 x cos x cos a + sin 4 x sin a ) d x ⇒ I = ∫ 1 sin 2 x cot x cos a + sin a d x [Taking sin 4 x common from denominator] ⇒ I = ∫ cosec 2 x cot x cos a + sin a d x Let sin a + cot x cos a = t Differentiating with respect to x , then we get: - cosec 2 x cos a d x = d t ⇒ cosec 2 x d x = - 1 cos a d t Therefore, I = ∫ - 1 cos a d t t ⇒ I = - 1 cos a ∫ t - 1 2 d t ⇒ I = - 1 cos a t - 1 2 + 1 - 1 2 + 1 + c ⇒ I = - 1 cos a t 1 2 1 2 + c ⇒ I = - 2 cos a · sin a + cos a cot x + c Hence, ∫ d x sin 3 x cos ( x - a ) = - 2 cos a · sin a + cos a cot x + c .

Evaluate the following integral : ∫ cos 3 x + cos 5 x sin 2 x + sin 4 x d x .

Let I = ∫ cos 3 x + cos 5 x sin 2 x + sin 4 x d x ⇒ I = ∫ cos 3 x 1 + cos 2 x sin 2 x 1 + sin 2 x d x ⇒ I = ∫ cos 2 x 1 + cos 2 x sin 2 x 1 + sin 2 x cos x d x ⇒ I = ∫ 1 - sin 2 x 1 + 1 - sin 2 x sin 2 x 1 + sin 2 x cos x d x Let sin x = t Differentiating with respect to x , then we get: cos x d x = d t Therefore, I = ∫ 1 - t 2 2 - t 2 t 2 1 + t 2 d t ⇒ I = ∫ t 4 - 3 t 2 + 2 t 4 + t 2 d t ⇒ I = ∫ t 4 + t 2 - 4 t 2 + 2 t 4 + t 2 d t ⇒ I = ∫ t 4 + t 2 t 4 + t 2 - 2 2 t 2 - 1 t 4 + t 2 d t ⇒ I = ∫ 1 d t - 2 ∫ 2 t 2 - 1 t 2 t 2 + 1 d t ⇒ I = t - 2 ∫ 2 t 2 t 2 t 2 + 1 - 1 t 2 t 2 + 1 d t + c 1 , where c 1 = constant of integration ⇒ I = t - 2 ∫ 2 t 2 + 1 - t 2 + 1 - t 2 t 2 t 2 + 1 d t + c 1 ⇒ I = t - 4 ∫ 1 t 2 + 1 d t + 2 ∫ t 2 + 1 t 2 t 2 + 1 - t 2 t 2 t 2 + 1 d t + c 1 ⇒ I = t - 4 ∫ 1 t 2 + 1 d t + 2 ∫ 1 t 2 - 1 t 2 + 1 d t + c 1 ⇒ I = t - 4 tan - 1 t + 2 - 1 t - tan - 1 t + c 1 + c 2 , where c 2 = constant of integration ⇒ I = t - 6 tan - 1 t - 2 t + c , where c = c 1 + c 2 ⇒ I = sin x - 2 sin x - 6 tan - 1 ( sin x ) + c ⇒ I = sin x - 2 cosec x - 6 tan - 1 ( sin x ) + c Hence, ∫ cos 3 x + cos 5 x sin 2 x + sin 4 x d x = sin x - 2 cosec x - 6 tan - 1 ( sin x ) + c .

Evaluate the following integral : ∫ cos 5 x sin x d x .

Let I = ∫ cos 5 x sin x d x ⇒ I = ∫ cos 2 x cos 2 x cos x sin x d x ⇒ I = ∫ 1 - sin 2 x 1 - sin 2 x cos x sin x d x Let sin x = t Differentiating with respect to x , then we get: cos x d x = d t Therefore, I = ∫ 1 - t 2 1 - t 2 t d t ⇒ I = ∫ 1 - t 2 - t 2 + t 4 t d t ⇒ I = ∫ 1 - 2 t 2 + t 4 t d t ⇒ I = ∫ 1 t - 2 t + t 3 d t ⇒ I = ∫ 1 t d t - ∫ 2 t d t + ∫ t 3 d t ⇒ I = log t - t 2 + t 4 4 + c , where c = constant of integration ⇒ I = log sin x - sin x 2 + sin x 4 4 + c Hence, ∫ cos 5 x sin x d x = log sin x - sin x 2 + sin x 4 4 + c .

Evaluate the following integral : ∫ tan x tan 2 x tan 3 x d x .

Let I = ∫ tan x tan 2 x tan 3 x d x Now, tan ( 3 x ) = tan ( x + 2 x ) ⇒ tan ( 3 x ) = tan x + tan 2 x 1 - tan x · tan 2 x ⇒ tan ( 3 x ) - tan x tan ( 2 x ) tan ( 3 x ) = tan x + tan ( 2 x ) ∴ tan x tan ( 2 x ) tan ( 3 x ) = tan ( 3 x ) - tan ( x ) - tan ( 2 x ) Therefore, I = ∫ tan ( 3 x ) - tan ( x ) - tan ( 2 x ) d x ⇒ I = ∫ tan ( 3 x ) d x - ∫ tan ( x ) d x - ∫ tan ( 2 x ) d x ⇒ I = 1 3 log | sec ( 3 x ) | - log | sec x | - 1 2 log | sec ( 2 x ) | + c , where c = constant of integration Hence, ∫ tan x tan 2 x tan 3 x d x = 1 3 log | sec ( 3 x ) | - log | sec x | - 1 2 log | sec ( 2 x ) | + c .

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RUDIMENTS of MATHEMATICS For Class 12 of +2 Level>Calculus - B. Integral Calculus>Chapter 12 - Methods of Substitution>EXERCISE

Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE

Author:Parthasarathi Mukhopadhyay & Manabendra Nath Mukherjee

Parthasarathi Mukhopadhyay Mathematics Solutions for Exercise - Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE

Attempt the practice questions from Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. RUDIMENTS of MATHEMATICS For Class 12 of +2 Level solutions are prepared by Experienced Embibe Experts.