$∫\frac{cosx-sinx}{1+sin2x}dx$ is equal to :

$\frac{-1}{sinx+cosx}+c,$ where C is a constant.

The correct answer is Option (2) → $\frac{-1}{sinx+cosx}+c,$ where C is a constant.

$∫\frac{\cos x-\sin x}{\cos^2x+\sin^2x+2\sin x\cos x}dx$   

$1=\sin^2x+\cos^2x$

$\sin 2x=2\sin x\cos x$

$=∫\frac{(\cos x-\sin x)dx}{(\cos x-\sin x)^2}$

let $y=\cos x+\sin x$

$dy=(\cos x-\sin x)dx$

$⇒∫\frac{dy}{y^2}=\frac{-1}{y}+c$

$=\frac{-1}{sinx+cosx}+c$