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

Let I = ∫ sin 3 x cos x d x Put sin x = t ⇒ cos x d x = d t ⇒ I = ∫ sin 3 x cos x d x = ∫ t 3 d t On Integrating: I = t 4 4 + C Again, substituting the value t = sin x , we get I = sin 4 x 4 + C

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

Let I = ∫ sin 3 x cos x d x Put sin x = t ⇒ cos x d x = d t ⇒ I = ∫ sin 3 x cos x d x = ∫ t 3 d t On Integrating: I = t 4 4 + C Again, substituting the value t = sin x , we get I = sin 4 x 4 + C

Evaluate the following integral: ∫ sec 4 x d x

Let I = ∫ sec 4 x d x = ∫ sec 2 x · sec 2 x d x ⇒ I = ∫ sec 2 x 1 + tan 2 x d x Put tan x = t ⇒ sec 2 x d x = d t ⇒ I = ∫ 1 + t 2 d t On Integrating, = t + t 3 3 + c Again substituting the value of t = tan x , we get I = tan x + tan 3 x 3 + C Where C is constant of integration

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

Let I = ∫ cos 3 x sin 4 x d x ⇒ I = ∫ cos x sin 4 x cos 2 x d x = ∫ cos x sin 4 x 1 - sin 2 x d x Put sin x = t ⇒ cos x d x = d t ⇒ I = ∫ t 4 1 - t 2 d t ⇒ I = ∫ t 4 d t - ∫ t 6 d t On Integrating: I = t 5 5 - t 7 7 + c Again substituting the value of t = sin x , we get I = sin 5 x 5 - sin 7 x 7 + C

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

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

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

Let I = ∫ cos 3 x sin 2 3 x d x = ∫ cos x cos 2 x sin 2 3 x d x ⇒ I = ∫ cos x 1 - sin 2 x sin 2 3 x d x Put sin x = t ⇒ cos x d x = d t ⇒ I = ∫ t 2 3 1 - t 2 d t = ∫ t 2 3 d t - ∫ t 8 3 d t = t 5 3 5 3 - t 11 3 11 3 + C Again substituting the value of t = sin x we get I = 3 sin 5 / 3 x 5 - 3 sin 11 / 3 x 11 + C

Evaluate the following integral: ∫ cos 3 5 x sin 3 x d x

Let I = ∫ cos 3 5 x sin 3 x d x = ∫ sin x sin 2 x cos 3 5 x d x = ∫ sin x 1 - cos 2 x cos 3 5 x d x Put cos x = t ⇒ - sin x d x = d t ⇒ I = ∫ t 3 5 t 2 - 1 d t = ∫ t 13 5 d t - ∫ t 3 5 d t On Integrating: ⇒ t 18 5 18 5 - t 8 5 8 5 + C Again substituting the value of t = cos x We get I = 5 cos 18 5 x 18 - 5 cos 8 5 x 8 + C

Evaluate the following integral: ∫ cosec 4 2 x d x

Let I = ∫ cosec 4 2 x d x = ∫ cosec 2 2 x cosec 2 2 x d x = ∫ cosec 2 2 x 1 + cot 2 2 x d x Let cot 2 x = t - 2 cosec 2 2 x d x = d t ⇒ I = - 1 2 ∫ 1 + t 2 d t = - 1 2 ∫ d t - 1 2 ∫ t 2 d t On Integrating, I = - 1 2 t - t 3 6 + C Again substituting the value of t = cot 2 x we get I = - cot 2 x 2 - cot 3 2 x 6 + C

Evaluate the following integral: ∫ cos 2 x cos x d x

Let I = ∫ cos 2 x cos x d x = ∫ 2 cos 2 x - 1 cos x d x = ∫ 2 cos x d x - ∫ sec x d x On Integrating: I = 2 sin x - log sec x + tan x + C

Evaluate the following integrals: ∫ cos x cos ( x + α ) d x

Let I = ∫ cos x cos ( x + α ) d x Let x + α = t ⇒ d x = d t ⇒ I = ∫ cos ( t - α ) cos t d t = ∫ cos t cos α + sin t sin α cos t d t = ∫ cos α d t + ∫ tan t sin α d t Here α is a constant On Integrating, = t cos α + sin α log sec t + c = ( x + α ) cos α - sin α log cos ( x + α ) + c Hence, I = x cos α - sin α log cos ( x + α ) + C