Question

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

Answer 1

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

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

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

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

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

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

Differentiating w.r.t. $x$

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

On substituting

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

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

$=\left(\frac{t^{-2+1}}{-2+1}\right)+C$

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

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

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

$[\because t=\sin x+\cos x]$

Where $C$ is constant of integration.