$\int\left(\frac{\cos x-\sin x}{1+\sin 2x}\right)dx$ is equal to

$\frac{-1}{\cos x+\sin x}+C$, where C is constant of integration

The correct answer is Option (2) → $\frac{-1}{\cos x+\sin x}+C$, where C is constant of integration

Given integral

$\int \frac{\cos x-\sin x}{1+\sin 2x}\,dx$

$1+\sin2x=1+2\sin x\cos x=(\sin x+\cos x)^2$

Hence

$\int \frac{\cos x-\sin x}{(\sin x+\cos x)^2}\,dx$

Let $t=\sin x+\cos x$

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

Integral becomes

$\int \frac{dt}{t^2}$

$=-\frac{1}{t}+C$

$=-\frac{1}{\sin x+\cos x}+C$

The value of the integral is $-\frac{1}{\cos x+\sin x}+C$.