Question

Integrate:

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

Answer 1

Consider the given integral.

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

Let $t={\tan }^{-1}(\sin x+\cos x)$

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

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

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

Therefore,

$I=\int tdt$

$I=\frac{t^2}{2}+C$

On putting the value of $t$, we get

$I=\frac{{\left({\tan }^{-1}(\sin x+\cos x)\right)}^2}{2}+C$

Hence, this is the answer.