Question

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

Answer 1

Solve

$\frac{cosx-sinx}{1+sin2x}$

$=\frac{cosx-sinx}{{\cos }^{2}x+{\sin }^{2}x+2\sin x\cos x}$

$=\frac{cosx-sinx}{{(cosx+sinx)}^2}$

On integrating, we get

$\int \frac{cosx-sinx}{{(cosx+sinx)}^2}dx$

Let $cosx+sinx=t$

Then,}

$\frac{d}{dx}(cosx+sinx)=\frac{d}{dx}t$

$\Rightarrow -\sin x+\cos x=\frac{dt}{dx}$

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

Now,

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

$=\int t^{-2}dt$

$=\frac{t^{-2+1}}{-2+1}$

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

Hence, this is the answer.