Question

Find the integral of the function $\frac{\cos 2x}{(\cos x+\sin x{)}^2}$

Answer 1

$\frac{\cos 2x}{(\cos x+\sin x{)}^2}=\frac{\cos 2x}{{\cos }^{2}x+{\sin }^{2}x+2\sin x\cos x}=\frac{\cos 2x}{1+\sin 2x}$
$\therefore \int \frac{\cos 2x}{(\cos x+\sin x{)}^2}dx=\int \frac{\cos 2x}{(1+\sin 2x)}dx$
Let $1+\sin 2x=t\Rightarrow 2\cos 2xdx=dt$
$\therefore \int \frac{\cos 2x}{(\cos x+\sin x{)}^2}dx=\frac{1}{2}\int \frac{1}{t}dt$
$=\frac{1}{2}\log |t|+C$
$=\frac{1}{2}\log |1+\sin 2x|+C$
$=\frac{1}{2}\log |(\sin x+\cos x{)}^2|+C$
$=\log |\sin x+\cos x|+C$