一、题目
$$
\int_{0}^{\pi} x f(\sin x) \mathrm{d} x = ?
$$
难度评级：
二、解析 
$$
\int_{0}^{\pi} x f(\sin x) \mathrm{d} x \Rightarrow
$$
令 $u$ $=$ $\pi$ $-$ $x$, $x$ $=$ $\pi$ $-$ $u$, 则：
$$
– \int_{\pi}^{0} (\pi – u) f[\sin (\pi – u)] \mathrm{d} u \Rightarrow
$$
$$
\int_{0}^{\pi} (\pi – u) f[\sin (\pi – u)] \mathrm{d} u \Rightarrow
$$
Next 
由于 $\sin(\pi – u)$ $=$ $\sin u$ $\Rightarrow$
$$
\int_{0}^{\pi} (\pi – u) f(\sin u) \mathrm{d} u \Rightarrow
$$
$$
\pi \int_{0}^{\pi} f(\sin u) \mathrm{d} u – u \int_{0}^{\pi} f(\sin u) \mathrm{d} u \Rightarrow
$$
Next
再令 $u$ $=$ $x$ $\Rightarrow$
用什么字母表示变量都只是一种形式，因此，可以直接将式子中所有的变量 $x$ 替换为 $u$.
$$
\pi \int_{0}^{\pi} f(\sin x) \mathrm{d} x – x \int_{0}^{\pi} f(\sin x) \mathrm{d} x \Rightarrow
$$
$$
\int_{0}^{\pi} x f(\sin x) \mathrm{d} x = \pi \int_{0}^{\pi} f(\sin x) \mathrm{d} x – x \int_{0}^{\pi} f(\sin x) \mathrm{d} x \Rightarrow
$$
$$
2 \int_{0}^{\pi} x f(\sin x) \mathrm{d} x = \pi \int_{0}^{\pi} f(\sin x) \mathrm{d} x \Rightarrow
$$
$$
\int_{0}^{\pi} x f(\sin x) \mathrm{d} x = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \mathrm{d} x.
$$
