Question Number 103512 by bemath last updated on 15/Jul/20

∫ ((x dx)/((cot x+tan x)^2 )) = (a) (x/(16))−((x sin 4x)/(32))−((cos 4x)/(128))+c (b) (x/(16))+((x sin 4x)/(32))−((cos 4x)/(128))+c (c) (x/(16))+((xsin 4x)/(64))+((cos 4x)/(128))+c (d)(x/(16))+((xcos 4x)/(32))+((sin 4x)/(128))+c

Commented bybobhans last updated on 15/Jul/20

nothing answer too

Commented bybemath last updated on 15/Jul/20

yes. i think the question wrong

Answered by Dwaipayan Shikari last updated on 15/Jul/20

∫(x/((((cosx)/(sinx))+((sinx)/(cosx)))^2 ))dx=(1/4)∫xsin^2 2x=∫(x/8)(1−cos4x) ∫(x/8)−(1/8)∫xcos4x (x^2 /(16))−(1/(32))xsin4x+(1/(32))∫sin4x (x^2 /(16))−(1/(32))xsin4x−(1/(128))cos4x+Constant

Answered by bramlex last updated on 15/Jul/20

cot x+tan x = ((cos x)/(sin x))+((sin x)/(cos x)) = (2/(sin 2x)) ; I = ∫ ((x dx)/(((4/(sin ^2 2x))))) = (1/4)∫ x((1/2)−(1/2)cos 4x) dx = (1/8)∫ (x−xcos 4x) dx = (1/8)((x^2 /2)−((xsin 4x)/4)−((cos 4x)/(16)))+c = (x^2 /(16))−((xsin 4x)/(32))−((cos 4x)/(128))+c ■