Evaluate the following integral: ∫ 1 e x + 1 d x

Let I = ∫ 1 e x + 1 d x = ∫ 1 e x 1 + e - x d x . . . . . 1 Put 1 + e - x = u ⇒ d x = - e - x d u So, I = - ∫ d u u = - log u + C = - log 1 + e - x + C

Evaluate the following integral: ∫ 1 e x + 1 d x

Let I = ∫ 1 e x + 1 d x = ∫ 1 e x 1 + e - x d x . . . . . 1 Put 1 + e - x = u ⇒ d x = - e - x d u So, I = - ∫ d u u = - log u + C = - log 1 + e - x + C

Evaluate the following integral: ∫ cot x log sin x d x

Let I = ∫ cot x log sin x d x Put log sin x = t ⇒ 1 sin x × cos x = d t ⇒ cot x d x = d t So, I = ∫ d t t = log t + C = log log sin x + C

Evaluate the following integral: ∫ e 2 x e 2 x - 2 d x

Let ∫ e 2 x e 2 x - 2 d x Put e 2 x - 2 = t ⇒ 2 e 2 x d x = d t So, I = 1 2 ∫ d t t = 1 2 log t + C = 1 2 log e 2 x - 2 + C

Evaluate the following integral: ∫ 2 cos x - 3 sin x 6 cos x + 4 sin x d x

Let I = ∫ 2 cos x - 3 sin x 6 cos x + 4 sin x d x = ∫ 2 cos x - 3 sin x 2 3 cos x + 2 sin x d x Let 3 cos x + 2 sin x = t ⇒ - 3 sin x + 2 cos x = d t So, I = 1 2 ∫ d t t = 1 2 log t + C = 1 2 log 3 cos x + 2 sin x + C

Evaluate the following integral: ∫ cos 2 x + x + 1 x 2 + sin 2 x + 2 x d x

Let I = ∫ cos 2 x + x + 1 x 2 + sin 2 x + 2 x d x Put x 2 + sin 2 x + 2 x = t ⇒ 2 x + 2 cos 2 x + 2 d x = d t So, I = 1 2 ∫ d t t = 1 2 log t + C = 1 2 log x 2 + sin 2 x + 2 x + C

Evaluate the following integral: ∫ 1 cos x + a cos x + b d x

Let I = ∫ 1 cos x + a cos x + b d x = 1 sin a - b ∫ sin a - b cos x + a cos x + b d x = 1 sin a - b ∫ sin x + a - x + b cos x + a cos x + b d x = 1 sin a - b ∫ sin x + a cos x + b - cos x + a sin x + b cos x + a cos x + b ∵ sin x - y = sin x cos y - cos x sin y = 1 sin a - b ∫ tan x + a d x - ∫ tan x + b d x = 1 sin a - b - log cos x + a + log cos x + b + C = 1 sin a - b log cos x + b cos x + a + C

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

Let I = ∫ - sin x + 2 cos x 2 sin x + cos x d x Put 2 sin x + cos x = t ⇒ 2 cos x - sin x d x = d t So, I = ∫ d t t = log t + C = log 2 sin x + cos x + C

Evaluate the following integral: ∫ cos 4 x - cos 2 x sin 4 x - sin 2 x d x

Let I = ∫ cos 4 x - cos 2 x sin 4 x - sin 2 x d x = ∫ - 2 sin 4 x + 2 x 2 sin 4 x - 2 x 2 2 cos 4 x + 2 x 2 sin 4 x - 2 x 2 d x ∵ sin α - sin β = 2 cos α + β 2 sin α - β 2 , cos α - cos β = - 2 sin α + β 2 sin α - β 2 = - ∫ sin 3 x cos 3 x d x = - ∫ tan 3 x d x = 1 3 log cos 3 x + C

Evaluate the following integral: ∫ sec x log sec x + tan x d x

Let I = ∫ sec x log sec x + tan x Put log sec x + tan x = t ⇒ 1 sec x + tan x sec 2 x + sec x tan x d x = d t ⇒ sec x sec x + tan x sec x + tan x d x = d t ⇒ sec x d x = d t So, I = ∫ d t t = log t + C = log log ( sec x + tan x + C