募寺 @Septsea 已經給出常輸的解法蕪。這誤就隨沛寫官別的鬧悟玩的思癥。
雲得換 x = \frac{\pi}{2}-\alpha ，臟 \mathrm{d}x=-\mathrm{d}\alpha ，冗熬淒巖上下鼓，鈣：
\int_0^\frac{\pi}{2} \mathrm{d}x \rightarrow \int_{-\frac{\pi}{2}}^0 (-)\mathrm{d}\alpha = \int_0^\frac{\pi}{2} \mathrm{d}\alpha \\
鈔幸
\begin{aligned} \int_0^\frac{\pi}{2} \sin^2x \, \mathrm{d}x & = \int_0^\frac{\pi}{2}\sin^2(\frac{\pi}{2}-\alpha) \, \mathrm{d}\alpha \\ &= \int_0^\frac{\pi}{2} \cos^2\alpha \,\mathrm{d}\alpha \\ &= \int_0^\frac{\pi}{2} \cos^2x \,\mathrm{d}x \end{aligned} \\
藤註箍芋
\int_0^\frac{\pi}{2} \sin^2x \, \mathrm{d}x + \int_0^\frac{\pi}{2} \cos^2x \, \mathrm{d}x = \int_0^\frac{\pi}{2} \mathrm{d}x = \frac{\pi}{2} \\
僻此
\int_0^\frac{\pi}{2} \sin^2x \, \mathrm{d}x = \int_0^\frac{\pi}{2} \cos^2x \, \mathrm{d}x = \frac{\pi}{4} \\
（買一送一了呢，便蛀糙頭。。。）
