Question
Evaluate: $\int_0^{\frac{\pi}{4}} \frac{\cos 2 x}{1+\cos 2 x+\sin 2 x} d x$
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Answer
$\text { Let } I =\int_0^{\frac{\pi}{4}} \frac{\cos 2 x}{1+\cos 2 x+\sin 2 x} d x$
$=\int_0^{\frac{\pi}{4}} \frac{\cos ^2 x-\sin ^2 x}{2 \cos ^2 x+2 \sin x \cos x} d x$
$=\int_0^{\frac{\pi}{4}} \frac{(\cos x+\sin x)(\cos x-\sin x)}{2 \cos (\cos x+\sin x)} d x$
$=\frac{1}{2} \int_0^{\frac{\pi}{4}} \frac{\cos x-\sin x}{\cos x} d x$
$=\frac{1}{2} \int_0^{\frac{\pi}{4}}(1-\tan x) d x$
$=\frac{1}{2} \int_0^{\frac{\pi}{4}} d x-\frac{1}{2} \int_0^{\frac{\pi}{4}} \tan x d x$
$=\frac{1}{2}[x]_0^{\frac{\pi}{4}}-\frac{1}{2}[\log |\sec x|]_0^{\frac{\pi}{4}}$
$=\frac{1}{2}\left(\frac{\pi}{4}-0\right)-\frac{1}{2}\left[\log \left|\sec \frac{\pi}{4}\right|-\log |\sec 0|\right]$
$=\frac{\pi}{8}-\frac{1}{2}(\log \sqrt{2}-\log 1)$
$=\frac{\pi}{8}-\frac{1}{2}(\log \sqrt{2}-0)$
$\therefore I =\frac{1}{2}\left(\frac{\pi}{4}-\log \sqrt{2}\right)$
$=\int_0^{\frac{\pi}{4}} \frac{\cos ^2 x-\sin ^2 x}{2 \cos ^2 x+2 \sin x \cos x} d x$
$=\int_0^{\frac{\pi}{4}} \frac{(\cos x+\sin x)(\cos x-\sin x)}{2 \cos (\cos x+\sin x)} d x$
$=\frac{1}{2} \int_0^{\frac{\pi}{4}} \frac{\cos x-\sin x}{\cos x} d x$
$=\frac{1}{2} \int_0^{\frac{\pi}{4}}(1-\tan x) d x$
$=\frac{1}{2} \int_0^{\frac{\pi}{4}} d x-\frac{1}{2} \int_0^{\frac{\pi}{4}} \tan x d x$
$=\frac{1}{2}[x]_0^{\frac{\pi}{4}}-\frac{1}{2}[\log |\sec x|]_0^{\frac{\pi}{4}}$
$=\frac{1}{2}\left(\frac{\pi}{4}-0\right)-\frac{1}{2}\left[\log \left|\sec \frac{\pi}{4}\right|-\log |\sec 0|\right]$
$=\frac{\pi}{8}-\frac{1}{2}(\log \sqrt{2}-\log 1)$
$=\frac{\pi}{8}-\frac{1}{2}(\log \sqrt{2}-0)$
$\therefore I =\frac{1}{2}\left(\frac{\pi}{4}-\log \sqrt{2}\right)$
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