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

Let I = ∫ sin 3 x cos x d x ⇒ I = ∫ sin 2 x sin x cos x d x ⇒ I = ∫ 1 - cos 2 x sin x cos x d x Let cos x = t Differentiate it, ⇒ - sin x d x = d t Now, I = ∫ - 1 - t 2 t 1 2 d t = - ∫ 1 t 1 2 - t 2 t 1 2 d t = - ∫ t - 1 2 - t 2 - 1 2 d t = ∫ t 3 2 - t - 1 2 d t = t 3 2 + 1 3 2 + 1 - t - 1 2 + 1 - 1 2 + 1 + C = t 5 2 5 2 - t 1 2 1 2 + C = 2 5 t 5 2 - 2 t 1 2 + C ⇒ I = 2 5 cos x 5 2 - 2 cos x 1 2 + C , where C is the constant of integration.

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

Let I = ∫ sin 3 x cos x d x ⇒ I = ∫ sin 2 x sin x cos x d x ⇒ I = ∫ 1 - cos 2 x sin x cos x d x Let cos x = t Differentiate it, ⇒ - sin x d x = d t Now, I = ∫ - 1 - t 2 t 1 2 d t = - ∫ 1 t 1 2 - t 2 t 1 2 d t = - ∫ t - 1 2 - t 2 - 1 2 d t = ∫ t 3 2 - t - 1 2 d t = t 3 2 + 1 3 2 + 1 - t - 1 2 + 1 - 1 2 + 1 + C = t 5 2 5 2 - t 1 2 1 2 + C = 2 5 t 5 2 - 2 t 1 2 + C ⇒ I = 2 5 cos x 5 2 - 2 cos x 1 2 + C , where C is the constant of integration.

Evaluate the following Integral: ∫ 1 tan - 1 x 1 + x 2 d x

Let I = ∫ 1 tan - 1 x 1 + x 2 d x Let tan - 1 x = t Differentiate it, ⇒ d tan - 1 x = d t ⇒ 1 1 + x 2 d x = d t Now, I = ∫ d t t = ∫ t - 1 2 d t = t - 1 2 + 1 - 1 2 + 1 + C = t 1 2 1 2 + C = 2 t + C ⇒ I = 2 tan - 1 x + C , where C is the constant of integration.

Evaluate the following Integral: ∫ tan x sin x cos x d x

Let I = ∫ tan x sin x cos x d x Multiply cos x with numerator and denominator ⇒ I = ∫ tan x × cos x sin x cos x × cos x d x = ∫ tan x sin x cos x × cos 2 x d x ⇒ I = ∫ tan x × sec 2 x tan x d x Let tan x = u Differentiate it, ⇒ d tan x = d u ⇒ sec 2 x d x = d u Now, I = ∫ u 1 2 u d u = ∫ u 1 2 - 1 d u = ∫ u - 1 2 d u = u - 1 2 + 1 - 1 2 + 1 + C = u 1 2 1 2 + C = 2 u + C ⇒ I = 2 tan x + C , where C is the constant of integration.

Evaluate the following Integral: ∫ 1 x log x 2 d x

Let I = ∫ 1 x log x 2 d x Let log x = t Differentiate it, ⇒ d log x = d t ⇒ 1 x d x = d t Now, I = ∫ t 2 d t ⇒ I = t 3 3 + C ⇒ I = log x 3 3 + C Hence, ∫ 1 x log x 2 d x = log x 3 3 + C , where C is the constant of integration.

Evaluate the following Integral: ∫ sin 5 x cos x d x

Let I = ∫ sin 5 x cos x d x Let sin x = t Differentiate it, ⇒ d sin x = d t ⇒ cos x d x = d t Now, I = ∫ t 5 d t = t 5 + 1 5 + 1 + C = t 6 6 + C ⇒ I = sin x 6 6 + C , where C is the constant of integration.

Evaluate the following Integral: ∫ tan 3 2 x sec 2 x d x

Let I = ∫ tan 3 2 x sec 2 x d x Let tan x = u Differentiate it, ⇒ d tan x = d u ⇒ sec 2 x d x = d u Now, I = ∫ u 3 2 d u ⇒ I = u 3 2 + 1 3 2 + 1 + C = u 5 2 5 2 + C = 2 5 u 5 2 + C ⇒ I = 2 5 tan x 5 2 + C

Evaluate the following Integral: ∫ x 3 x 2 + 1 3 d x

Let I = ∫ x 3 x 2 + 1 3 d x ⇒ I = ∫ x 2 x x 2 + 1 3 d x Let 1 + x 2 = t Differentiate it, ⇒ d 1 + x 2 = t ⇒ 2 x d x = d t ⇒ x d x = d t 2 Now, I = 1 2 ∫ t - 1 t 3 d t = 1 2 ∫ t t 3 - 1 t 3 d t = 1 2 ∫ t - 2 - t - 3 d t = 1 2 t - 2 + 1 - 2 + 1 - t - 3 + 1 - 3 + 1 + C = 1 2 t - 1 - 1 - t - 2 - 2 + C = - 1 2 t + 1 4 t 2 + C ⇒ I = - 1 2 1 + x 2 + 1 4 1 + x 2 2 + C

Evaluate the following Integral: ∫ 4 x + 2 x 2 + x + 1 d x

Let I = ∫ 4 x + 2 x 2 + x + 1 d x Let x 2 + x + 1 = t Differentiate it, ⇒ d x 2 + x + 1 = d t ⇒ 2 x + 1 d x = d t Now, I = ∫ 2 2 x + 1 x 2 + x + 1 d x = ∫ 2 t d t = 2 ∫ t 1 2 d t = 2 t 1 2 + 1 1 2 + 1 + C = 2 t 3 2 3 2 + C = 4 3 t 3 2 + C ⇒ I = 4 3 x 2 + x + 1 3 2 + C , where C is the constant of integration.

Evaluate the following Integral: ∫ 4 x + 3 2 x 2 + 3 x + 1 d x

Let I = ∫ 4 x + 3 2 x 2 + 3 x + 1 d x Let 2 x 2 + 3 x + 1 = t Differentiate it, ⇒ d 2 x 2 + 3 x + 1 = d t ⇒ 4 x + 3 d x = d t Now, I = ∫ d t t = ∫ t - 1 2 d t = t - 1 2 + 1 - 1 2 + 1 + C = 2 t + C ⇒ I = 2 2 x 2 + 3 x + 1 + C , where C is the constant of integration.