# 这道题为啥要设 t=3x 而不是 t=2x ?

## 二、解析

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

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

\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}

\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}

\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}

\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}