[2025年京都大学・数学第1問]簡単な定積分

\begin{align*} \int_{0}^{\frac{\pi}{2}}\sqrt{\frac{1-\cos x}{1+\cos x}}dx & =\int_{0}^{\frac{\pi}{2}}\sqrt{\frac{2\sin^{2}\frac{x}{2}}{2\cos^{2}\frac{x}{2}}}dx\cmt{\because\text{半角の公式}}\\ & =\int_{0}^{\frac{\pi}{2}}\sqrt{\tan^{2}\frac{x}{2}}dx\\ & =\int_{0}^{\frac{\pi}{2}}\left|\tan\frac{x}{2}\right|dx\\ & =\int_{0}^{\frac{\pi}{2}}\tan\frac{x}{2}dx\cmt{\because0\leq x\leq\frac{\pi}{2}\rightarrow0\leq\tan\frac{x}{2}}\\ & =2\left[-\log\left|\cos\frac{x}{2}\right|\right]_{0}^{\frac{\pi}{2}}\\ & =-2\left(\log\frac{1}{\sqrt{2}}-\log1\right)\\ & =-2\left(\log2^{-\frac{1}{2}}-\log e^{0}\right)\\ & =-2\left(-\frac{1}{2}\log2\right)\\ & =\log2 \end{align*}