Question

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

Answer 1

Consider the given integral.

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

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

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

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

Let $\cos x+\sin x=t$. Now,

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

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

$dx=\frac{1}{\cos x-\sin x}dt$

Therefore,

$I=\int \frac{(\cos x-\sin x)}{t}\times \frac{dt}{(\cos x-\sin x)}$

$I=\int \frac{1}{t}dt$

$I=\log \left|t\right|+C$

$I=\log |\cos x+\sin x|+C$

Hence, this is the required value of the integral.