Question
$\int \frac{\cos 2 x}{(\sin x+\cos x)^2} d x$
✓
Answer
$\int \frac{\cos 2 x}{(\sin x+\cos x)^2} d x=\int \frac{\cos ^2 x-\sin ^2 x}{(\cos x+\sin x)^2} d x$
$=\int \frac{(\cos x+\sin x)(\cos x-\sin x)}{(\cos x+\sin x)^2} d x$
$=\int \frac{\cos x-\sin x}{\cos x+\sin x} d x$
$=\log |\cos x+\sin x|+ c \quad \ldots \ldots \ldots\left[\because \frac{ f ^{\prime}(x)}{ f (x)} d x=\log | f (x)|+ c \right]$
$=\int \frac{(\cos x+\sin x)(\cos x-\sin x)}{(\cos x+\sin x)^2} d x$
$=\int \frac{\cos x-\sin x}{\cos x+\sin x} d x$
$=\log |\cos x+\sin x|+ c \quad \ldots \ldots \ldots\left[\because \frac{ f ^{\prime}(x)}{ f (x)} d x=\log | f (x)|+ c \right]$
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