$\int_{0}^{\frac{\pi}{2}} \log \left[\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}\right] d x=$

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Aicharya · Accepted Answer

$\int_{0}^{\frac{\pi}{2}} \log \sqrt{\frac{2 \sin ^{2} x}{2 \cos ^{2} x}} d x$ $=\int_{0}^{\pi / 2} \log \left(\frac{\sin x}{\cos x}ight) d x \Rightarrow \int_{0}^{\pi / 2} \log (	an x) d x$...(1) $=\int_{0}^{\pi / 2}\left[	an \left(\frac{\pi}{2}-xight)ight] d x=\int_{0}^{\pi / 2} \log \cot x d x$...(2) Eq. (1) $+(2)$ gives $\begin{aligned} 2 \mathrm{I} &=\int_{0}^{\pi / 2}[\log (	an \mathrm{x})+\log (\cot \mathrm{x})] \mathrm{dx}=\int_{0}^{\pi / 2} \log [	an \mathrm{x} \cot \mathrm{x}] \mathrm{d} \mathrm{x} \ &=\int_{0}^{\pi / 2} \log 1 \mathrm{dx}=0 \Rightarrow \mathrm{I}=0 \end{aligned}$

$\int_{0}^{\frac{\pi}{2}} \log \left[\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}\right] d x=$

Solution