Question

Find the integrals of the function $\frac{\cos x-\sin x}{1+\sin 2x}$

Answer 1

$\frac{\cos x-\sin x}{1+\sin 2x}=\frac{\cos x-\sin x}{({\sin }^{2}x+{\cos }^{2}x)+2\sin x\cos x}$
$=\frac{\cos x-\sin x}{(\sin x+\cos x{)}^2}$
Put $\sin x+\cos x=t$
$\therefore (\cos x-\sin x)dx=dt$
$\Rightarrow \int \frac{\cos x-\sin x}{1+\sin 2x}dx=\int \frac{\cos x-\sin x}{(\sin x+\cos x{)}^2}dx$
$=\int \frac{dt}{t^2}=\int t^{-2}dt$
$=-t^{-1}+C=-\frac{1}{t}+C=\frac{-1}{\sin x+\cos x}+C$