Question

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

Answer 1

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

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

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

On integrating, we get

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

Let $\cos x+\sin x=t$ --(1)

Then,

$\frac{d}{dx}(\cos x+\sin x)=\frac{d}{dx}t$

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

$\Rightarrow (\cos x-\sin x)dx=dt$---(2)

Now,

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

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

$=\int t^{-2}dt$ [Since, from (1) and (2)]

$=\frac{t^{-2+1}}{-2+1}+C$ [Since, $\int x^n=\frac{x^{n+1}}{n+1},n\ne -1$]

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

$=-\frac{1}{\cos x+\sin x}+C$ [Where, $C$ is any real constant]

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