Question

Evaluate $\int \frac{cosx-sinx}{1+sin2x}dx$

Answer 1

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

$\int \frac{\cos x-\sin x}{{\cos }^2{\sin }^{2}x+2\sin x\cos x}dx$ $[\therefore$ ${\cos }^{2}x+{\sin }^{2}x=$

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

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

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

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

$\Rightarrow \int \frac{1}{t^2}dt=\frac{t^{-1}}{-1}+c$

on agian putting value of $+$, we get

$(\sin x=\cos x{)}^{-1}+c$

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