[2024年東京医科歯科大学数学第3問]3角関数のルートを分母にもつ定積分

\begin{align*} \int_{0}^{\frac{\pi}{2}}\frac{\sin x}{1+\sqrt{\sin2x}}dx & =\frac{1}{2}\left(\int_{0}^{\frac{\pi}{2}}\frac{\sin x}{1+\sqrt{\sin2x}}dx+\int_{0}^{\frac{\pi}{2}}\frac{\sin\left(\frac{\pi}{2}-x\right)}{1+\sqrt{\sin\left(2\left(\frac{\pi}{2}-x\right)\right)}}dx\right)\cmt{\because\text{キングプロパティー}}\\ & =\frac{1}{2}\left(\int_{0}^{\frac{\pi}{2}}\frac{\sin x}{1+\sqrt{\sin2x}}dx+\int_{0}^{\frac{\pi}{2}}\frac{\sin\left(\frac{\pi}{2}-x\right)}{1+\sqrt{\sin\left(\pi-2x\right)}}dx\right)\\ & =\frac{1}{2}\left(\int_{0}^{\frac{\pi}{2}}\frac{\sin x}{1+\sqrt{\sin2x}}dx+\int_{0}^{\frac{\pi}{2}}\frac{\cos x}{1+\sqrt{\sin2x}}dx\right)\\ & =\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{\sin x+\cos x}{1+\sqrt{\sin2x}}dx\\ & =\frac{1}{2}\int_{-1}^{1}\frac{1}{1+\sqrt{1-t^{2}}}dt\cmt{t=-\cos x+\sin x,dt=\left(\sin x+\cos x\right)dx,t^{2}=1-\sin2x}\\ & =\int_{0}^{1}\frac{1}{1+\sqrt{1-t^{2}}}dt\\ & =\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\sqrt{1-\sin^{2}\theta}}\cos\theta d\theta\cmt{t=\sin\theta}\\ & =\int_{0}^{\frac{\pi}{2}}\frac{\cos\theta}{1+\cos\theta}d\theta\\ & =\int_{0}^{\frac{\pi}{2}}\left(1-\frac{1}{1+\cos\theta}\right)d\theta\\ & =\int_{0}^{\frac{\pi}{2}}\left(1-\frac{1}{2\cos^{2}\frac{\theta}{2}}\right)d\theta\\ & =\left[\theta-\tan\frac{\theta}{2}\right]_{0}^{\frac{\pi}{2}}\\ & =\frac{\pi}{2}-\tan\frac{\pi}{4}\\ & =\frac{\pi}{2}-1 \end{align*}