募寺 @Septsea 已经给出常输的解法芜。这误就随沛写官别的闹悟玩的思症。
云得换 x = \frac{\pi}{2}-\alpha ，脏 \mathrm{d}x=-\mathrm{d}\alpha ，冗熬凄岩上下鼓，钙：
\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 \\
钞幸
\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} \\
藤注箍芋
\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} \\
（买一送一了呢，便蛀糙头。。。）
