Show that


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

A. $\frac{-1}{\sin x+\cos x}+\mathrm{C}$

B. $\log |\sin x+\cos x|+\mathrm{C}$

C. $\log |\sin x-\cos x|+\mathrm{C}$

D. $\frac{1}{(\sin x+\cos x)^{2}}$


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

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

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

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

Let $\cos x+\sin x=t \Rightarrow(\cos x-\sin x) d x=d t$

$\therefore I=\int \frac{d t}{t}$

$=\log |t|+\mathrm{C}$

$=\log |\cos x+\sin x|+\mathrm{C}$

Hence, the correct answer is B.

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