Evaluate : ∫ tan x ( sec x + tan x ) d x

Given, ∫ tan x ( sec x + tan x ) d x Now multiply and divide by ( sec x - tan x ) = ∫ tan x ( sec x + tan x ) × sec x - tan x sec x - tan x d x = ∫ tan x sec x - tan 2 x sec 2 x - tan 2 x d x It can be written as = ∫ tan x sec x - tan 2 x d x ∵ sec 2 - tan 2 x = 1 So we get, = ∫ tan x sec x - sec 2 x - 1 d x By integrating w.r.t x ∴ ∫ tan x ( sec x + tan x ) d x = sec x - tan x + x + c .

Evaluate : ∫ tan x ( sec x + tan x ) d x

Given, ∫ tan x ( sec x + tan x ) d x Now multiply and divide by ( sec x - tan x ) = ∫ tan x ( sec x + tan x ) × sec x - tan x sec x - tan x d x = ∫ tan x sec x - tan 2 x sec 2 x - tan 2 x d x It can be written as = ∫ tan x sec x - tan 2 x d x ∵ sec 2 - tan 2 x = 1 So we get, = ∫ tan x sec x - sec 2 x - 1 d x By integrating w.r.t x ∴ ∫ tan x ( sec x + tan x ) d x = sec x - tan x + x + c .

Evaluate: ∫ cosec x ( cosec x - cot x ) d x

Given, ∫ cosec x ( cosec x - cot x ) d x Now multiply and divide by ( cosec x + cot x ) = ∫ cosec x ( cosec x - cot x ) × cosec x + cot x cosec x + cot x d x = ∫ cosec 2 x + cosec x cot x cosec 2 x - cot 2 x d x It can be written as = ∫ cosec 2 x + cosec x cot x d x ∵ cosec 2 x - cot 2 x = 1 By integrating w.r.t x ∴ ∫ cosec x ( cosec x - cot x ) d x = - cot x - cosec x + c .

Evaluate : ∫ cos x 1 + cos x d x

Given, ∫ cos x 1 + cos x d x Multiply and divide by ( 1 - cos x ) = ∫ cos x 1 + cos x × 1 - cos x 1 - cos x d x = ∫ cos x - ( cos x ) 2 1 - ( cos x ) 2 d x = ∫ cos x - ( cos x ) 2 ( sin x ) 2 d x ∵ sin 2 x + cos 2 x = 1 = ∫ cot x cosec x - ( cot x ) 2 d x = ∫ cot x cosec x - ( cosec x ) 2 + 1 d x ∵ cosec 2 x - cot 2 x = 1 ∴ ∫ cos x 1 + cos x d x = - cosec x + cot x + x + c

Evaluate : ∫ sin x ( 1 - sin x ) d x

Given, ∫ sin x ( 1 - sin x ) d x Now multiply and divide the integrand by ( 1 + sin x ) . = ∫ sin x + sin 2 x 1 - sin 2 x d x We can write it as, = ∫ sin x + sin 2 x cos 2 x d x = ∫ tan x sec x + tan 2 x d x It can be written as, = ∫ tan x sec x + sec 2 x - 1 d x ∵ sec 2 x - tan 2 x = 1 By integrating w.r.t x , ∴ ∫ sin x ( 1 - sin x ) d x = sec x + tan x - x + c .

Evaluate : ∫ 1 + cos 2 x d x

Given, ∫ 1 + cos 2 x d x = ∫ 2 ( cos x ) 2 d x ∵ cos 2 x = 2 cos 2 x - 1 = 2 ∫ cos x d x ∴ ∫ 1 + cos 2 x d x = 2 sin x + c .

Evaluate: ∫ 1 - cos 2 x d x

Given, ∫ 1 - cos 2 x d x It can be written as = ∫ 2 ( sin x ) 2 d x ∵ cos 2 x = 1 - 2 sin 2 x So we get = 2 ∫ sin x d x By integrating w.r.t x ∴ ∫ 1 - cos 2 x d x = - 2 cos x + c

Evaluate : ∫ 1 ( 1 + cos 2 x ) d x

Given, ∫ 1 ( 1 + cos 2 x ) d x = ∫ 1 2 ( cos x ) 2 d x ∵ cos 2 x = 2 cos 2 x - 1 = 1 2 ∫ ( sec x ) 2 d x ∵ sec x = 1 cos x ∴ ∫ 1 ( 1 + cos 2 x ) d x = 1 2 tan x + c .

Evaluate: ∫ 1 ( 1 - cos 2 x ) d x

Given, ∫ 1 ( 1 - cos 2 x ) d x It can be written as = ∫ 1 2 sin 2 x d x ∵ cos 2 x = 1 - 2 sin 2 x On further simplification = 1 2 ∫ cosec 2 x d x By integrating w.r.t x ∴ ∫ 1 ( 1 - cos 2 x ) d x = - 1 2 cot x + c

Evaluate : ∫ 1 + sin 2 x d x

Given, ∫ 1 + sin 2 x d x We can write it as = ∫ sin 2 x + cos 2 x + 2 sin x cos x d x ∵ sin 2 x = 2 sin x cos x & sin 2 x + cos 2 x = 1 = ∫ sin x + cos x 2 d x = ∫ ( sin x + cos x ) d x By integrating w.r.t x , ∴ ∫ 1 + sin 2 x d x = - cos x + sin x + c .