[2023年藤田医科大学・数学問題2]ルートの中に逆数の定積分

\begin{align*} \int_{1}^{3}\sqrt{\frac{4}{x}-1}dx & =\int_{\sqrt{3}}^{\frac{1}{\sqrt{3}}}y\frac{-8y}{\left(y^{2}+1\right)^{2}}dy\cmt{\sqrt{\frac{4}{x}-1}=y,dx=\frac{-8y}{\left(1+y^{2}\right)^{2}}dy}\\ & =-8\int_{\sqrt{3}}^{\frac{1}{\sqrt{3}}}\frac{y^{2}}{\left(1+y^{2}\right)^{2}}dy\\ & =-8\int_{\frac{\pi}{3}}^{\frac{\pi}{6}}\frac{\tan^{2}\theta}{\left(1+\tan^{2}\theta\right)^{2}}\cdot\frac{1}{\cos^{2}\theta}d\theta\cmt{y=\tan\theta,dy=\frac{1}{\cos^{2}\theta}d\theta}\\ & =8\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\tan^{2}\theta}{\cos^{-4}\theta}\cdot\frac{1}{\cos^{2}\theta}d\theta\\ & =8\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\sin^{2}\theta d\theta\\ & =8\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1-\cos2\theta}{2}d\theta\\ & =8\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\left(1-\cos2\theta\right)d\theta\\ & =4\left[\theta-\frac{1}{2}\sin2\theta\right]_{\frac{\pi}{6}}^{\frac{\pi}{3}}\\ & =4\left(\left(\frac{\pi}{3}-\frac{1}{2}\cdot\frac{\sqrt{3}}{2}\right)-\left(\frac{\pi}{6}-\frac{1}{2}\cdot\frac{\sqrt{3}}{2}\right)\right)\\ & =4\left(\frac{\pi}{3}-\frac{\pi}{6}\right)\\ & =\frac{2}{3}\pi \end{align*}