Question
$\int \frac{\cos 2 x}{(\sin x+\cos x)^{2}} d x$ is equal to

Answer

Given Integral is: $\int \frac{\cos 2 x}{(\sin x+\cos x)^{2}} d x$ 
Let $I=\int \frac{\cos 2 x}{(\sin x+\cos x)^{2}} d x$ 
= $\int \frac{\cos ^{2} x-\sin ^{2} x}{(\sin x+\cos x)^{2}} d x$ 
= $\int \frac{(\cos x-\sin x)(\cos x+\sin x)}{(\sin x+\cos x)^{2}} d x$ 
= $\int \frac{(\cos x-\sin x)}{(\sin x+\cos x)} d x$ 
Put sin x + cos x = t $\Rightarrow$ (cos x - sin x)dx = dt
$\Rightarrow \int \frac{(\cos x-\sin x)}{(\sin x+\cos x)} d x=\int \frac{d t}{t}$ 
= log |t| + C
= log |sin x + cos x| + C

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