这道题为啥要设 t=3x 而不是 t=2x ?

一、题目

难度评级：

二、解析

首先，令：

$$
\int_{0}^{\frac{\pi}{2}} f(x) \mathrm{~d} x = K
$$

接着，为了将已知式子中的积分 “$\int_{0}^{\frac{\pi}{6}} f( \textcolor{black}{\colorbox{springgreen}{3x}}) \mathrm{~d} x$” 向未知式子中的积分 “$\int_{0}^{\frac{\pi}{2}} f(x) \mathrm{~d} x$” 转换，我们可以设：

$$
\begin{aligned}
t & = \textcolor{black}{\colorbox{springgreen}{3x}} \\ \\
x & = \frac{1}{3} t
\end{aligned}
$$

Tip

为啥要令 $t$ $=$ $\textcolor{black}{\colorbox{springgreen}{3x}}$ 呢？
因为已知式子中的积分 “$\int_{0}^{\frac{\pi}{6}} f( \textcolor{black}{\colorbox{springgreen}{3x}}) \mathrm{~d} x$” 中含有 “$3x$”, 且已知式子中的积分上限 “$\frac{\pi}{6}$” 乘以 $3$ 刚好是要求解的积分中的积分上限 “$\frac{\pi}{2}$”

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于是，当 $x \in \left( 0, \frac{\pi}{6} \right)$ 的时候，有：

$$
\begin{aligned}
& t \in \left( 0, \quad 3 \times \frac{\pi}{6} \right) \\ \\
\Rightarrow & t \in \left( 0, \quad \frac{\pi}{2} \right)
\end{aligned}
$$

于是：

$$
\begin{aligned}
\int_{0}^{\frac{\pi}{6}} f(3x) \mathrm{~d} x \\ \\
& = \int_{0}^{\frac{\pi}{2}} f(t) \mathrm{~d} \left( \frac{1}{3} t \right) \\ \\
& = \frac{1}{3} \int_{0}^{\frac{\pi}{2}} f(t) \mathrm{~d} t \\ \\
& = \frac{1}{3} K
\end{aligned}
$$

进而：

$$
\begin{aligned}
& f(x) + \sin ^{6} x = \int_{0}^{\frac{\pi}{6}} f(3x) \mathrm{~d} x \\ \\
\Rightarrow & \textcolor{springgreen}{f(x) + \sin ^{6} x = \frac{1}{3} K}
\end{aligned}
$$

对上面的式子 $\textcolor{springgreen}{f(x)}$ $\textcolor{springgreen}{+}$ $\textcolor{springgreen}{\sin ^{6} x}$ $\textcolor{springgreen}{=}$ $\textcolor{springgreen}{\frac{1}{3} K}$ 两端同时做 $0$ 到 $\frac{\pi}{2}$ 的定积分，得：

$$
\begin{aligned}
& \textcolor{yellow}{\int_{0}^{\frac{\pi}{2}} f(x) \mathrm{~d} x} + \int_{0}^{\frac{\pi}{2}} \sin ^{6} x \mathrm{~d} x = \int_{0}^{\frac{\pi}{2}} \frac{1}{3} K \mathrm{~d} x \\ \\
\Rightarrow & \textcolor{yellow}{K} + \int_{0}^{\frac{\pi}{2}} \sin ^{6} x \mathrm{~d} x = \textcolor{pink}{\int_{0}^{\frac{\pi}{2}} \frac{1}{3} K \mathrm{~d} x} \\ \\
\Rightarrow & K + \textcolor{magenta}{\int_{0}^{\frac{\pi}{2}} \sin ^{6} x \mathrm{~d} x} = \textcolor{pink}{\frac{\pi}{6} K} \\ \\
\Rightarrow & K + \textcolor{magenta}{\frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} } = \frac{\pi}{6} K \\ \\
\Rightarrow & K + \frac{5}{32} \pi = \frac{\pi}{6} K \\ \\
\Rightarrow & \left( 1 – \frac{\pi}{6} \right) K = \frac{-5}{32} \pi \\ \\
\Rightarrow & \left( \frac{\pi}{6} – 1 \right) K = \frac{5}{32} \pi \\ \\
\Rightarrow & K = \frac{\frac{5}{32} \pi}{\frac{\pi}{6} – 1} \\ \\
\Rightarrow & \textcolor{springgreen}{\boldsymbol{ K = \frac{15 \pi}{16 (\pi – 1) } }}
\end{aligned}
$$

Note

上面对 “$\textcolor{magenta}{\int_{0}^{\frac{\pi}{2}} \sin ^{6} x \mathrm{~d} x}$” 的计算用到了“点火公式”，详细内容可以查阅荒原之梦考研数学的文章《偶次幂点火公式》和《奇次幂点火公式》。

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