题目中没有给出的等式可以通过“嵌套”的方式构造出来

一、题目

设 $f(x)$ $=$ $x^{2} \arcsin x-\int_{0}^{\pi} f(\sin x) \mathrm{~d} x$, 则 $\int_{0}^{\pi} f(\sin x) \mathrm{~d} x=?$

难度评级：

二、解析

若令：

$$
\int_{0}^{\pi} f(\sin x) \mathrm{~d} x=A
$$

则由 $f(x)$ $=$ $x^{2} \arcsin x-\int_{0}^{\pi} f(\sin x) \mathrm{~d} x$ 可得：

$$
\textcolor{orangered}{
f(\sin x) = \sin ^{2} x \arcsin (\sin x)-A
}
$$

因此，利用“嵌套”的方式，可得：

$$
\begin{aligned}
& A \\ \\
& = \int_{0}^{\pi} f(\sin x) \mathrm{~d} x \\ \\
& = \int_{0}^{\pi} \left[ \sin ^{2} x \arcsin (\sin x)-A \right] \mathrm{~d} x \\ \\
& = \int_{\textcolor{orangered}{0}}^{\textcolor{orangered}{\pi}} \sin ^{2} x \arcsin (\sin x) \mathrm{~d} x – \pi A \\ \\
& = \int_{\textcolor{orangered}{0}}^{\textcolor{orangered}{\frac{\pi}{2}}} \textcolor{orangered}{x} \sin ^{2} x \mathrm{~d} x + \int_{\textcolor{orangered}{\frac{\pi}{2}}}^{\textcolor{orangered}{\pi}} x \sin ^{2} \textcolor{orangered}{\arcsin (\sin x)} \mathrm{~d} x – \pi A
\end{aligned}
$$

注意：

$\int_{0}^{\pi} \sin ^{2} x \arcsin (\sin x) \mathrm{~d} x$ $\neq$ $\int_{0}^{\pi} x \sin ^{2} x \mathrm{~d} x$

这是因为 $\arcsin x$ 的取值范围为 $(-\frac{\pi}{2}, \frac{\pi}{2})$, $\arcsin (\sin x)$ 的取值范围自然也是 $(-\frac{\pi}{2}, \frac{\pi}{2})$. 有关该部分的内容可以查看《$\arcsin(\sin x)$ 一定等于 $x$ 吗？不一定哦！》这篇文章——

同样由这篇文章可知，当 $x \in (\frac{\pi}{2}, \pi)$ 时，$\arcsin (\sin x)$ $=$ $\arcsin [\sin (\pi – x)]$ $=$ $\pi – x$

继续：

$$
\begin{aligned}
A \\ \\
& = \int_{\textcolor{orangered}{0}}^{\textcolor{orangered}{\frac{\pi}{2}}} \textcolor{orangered}{x} \sin ^{2} x \mathrm{~d} x + \int_{\textcolor{orangered}{\frac{\pi}{2}}}^{\textcolor{orangered}{\pi}} x \sin ^{2} \textcolor{orangered}{\arcsin (\sin x)} \mathrm{~d} x – \pi A \\ \\
& = \int_{0}^{\frac{\pi}{2}} x \sin ^{2} x \mathrm{~d} x + \int_{\textcolor{orangered}{\frac{\pi}{2}}}^{\textcolor{orangered}{\pi}} \textcolor{orangered}{(\pi – x)} \sin ^{2} x \mathrm{~d} x – \pi A
\end{aligned}
$$

根据三角函数“奇变偶不变”的原则，$y = \sin^{2} x$ 与 $y = \sin ^{2} (\pi – x)$ 的函数图象是完全重合的，且当 $x \in (\frac{\pi}{2}, \pi)$ 时，$(\pi – x) \in (\frac{\pi}{2}, 0)$, 因此：

$$
\begin{aligned}
A \\ \\
& = \int_{0}^{\frac{\pi}{2}} x \sin ^{2} x \mathrm{~d} x + \int_{\textcolor{orangered}{\frac{\pi}{2}}}^{\textcolor{orangered}{\pi}} (\pi – x) \sin ^{2}(\pi – x) \mathrm{~d} x – \pi A \\ \\
& = \int_{0}^{\frac{\pi}{2}} x \sin ^{2} x \mathrm{~d} x + \int_{\textcolor{yellow}{\frac{\pi}{2}}}^{\textcolor{yellow}{0}} \textcolor{red}{ (\pi – x) \sin ^{2}(\pi – x) \mathrm{~d}(\pi – x) } – \pi A
\end{aligned}
$$

继续：

$$
\begin{aligned}
A \\ \\
& = \int_{0}^{\frac{\pi}{2}} x \sin ^{2} x \mathrm{~d} x+\int_{0}^{\frac{\pi}{2}} t \sin ^{2} t \mathrm{~d} t-\pi A \\ \\
& = 2 \int_{0}^{\frac{\pi}{2}} x \sin ^{2} x \mathrm{~d} x-\pi A \\ \\
& = \int_{0}^{\frac{\pi}{2}} x(1-\cos 2 x) \mathrm{~d} x-\pi A \\ \\
& = \int_{0}^{\frac{\pi}{2}} x \mathrm{~d} x – \int_{0}^{\frac{\pi}{2}} x \cos 2 x \mathrm{~d} x – \pi A \\ \\
& = \frac{\pi^{2}}{8} – \int_{0}^{\frac{\pi}{2}} x \cos 2 x \mathrm{~d} x – \pi A \\ \\
& = \frac{\pi^{2}}{8} – \frac{1}{2} \int_{0}^{\frac{\pi}{2}} x \cos 2 x \mathrm{~d} (2x) – \pi A \\ \\
& = \frac{\pi^{2}}{8} – \frac{1}{2} \int_{0}^{\frac{\pi}{2}} x \mathrm{~d} (\sin 2x) – \pi A \\ \\
& = \frac{\pi^{2}}{8} – \frac{1}{2} \left[ x \sin 2x \Big|_{0}^{\frac{\pi}{2}} – \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin 2 x \mathrm{~d} (2x) \right] – \pi A \\ \\
& = \frac{\pi^{2}}{8} + \frac{1}{2} – \pi A \\ \\
& = \frac{\pi^{2} + 4}{8} – \pi A
\end{aligned}
$$

综上可知：

$$
A = \frac{\pi^{2} + 4}{8}-\pi A \Rightarrow
$$

$$
\textcolor{springgreen}{
A=\frac{\pi^{2}+4}{8(\pi+1)}
}
$$

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