Question Number 114102 by bemath last updated on 17/Sep/20

∫ (dx/(tan x−sin x))

Answered by bobhans last updated on 17/Sep/20

∫ (dx/(tan x−sin x)) = −(1/2)cosec ^2 x −(1/2)cot x cosec x               + (1/2)ln ∣cot x+cosec x∣ + c

Answered by 1549442205PVT last updated on 17/Sep/20

Put tan(x/2)=t⇒dt=(1/2)(1+t^2 )dx  ∫ (dx/(tan x−sin x))=∫((2dt)/((((2t)/(1−t^2 ))−((2t)/(1+t^2 )))(1+t^2 )))  =∫(((1−t^2 ))/(2t^3 ))dt=∫((1/(2t^3 ))−(1/(2t)))dt=−(1/(4t^2 ))−(1/2)ln∣t∣  =−(1/(4tan^2 (x/2)))−(1/2)ln∣tan(x/2)∣+C