对于偶次幂的三角函数，可以尝试用倍角公式降幂

一、题目

$$
\int \sin^{4} x \times \cos^{2} x \mathrm{~d} x = ?
$$

难度评级：

二、解析

由于被积函数的次幂都是偶数，因此，我们可以尝试使用三角函数 $\sin$ 的倍角公式和三角函数 $\cos$ 的倍角公式对原来的被积函数 $\sin^{4} x \cos^{2} x$ 进行降幂，即：

$$
\begin{aligned}
& \textcolor{lightgreen}{ \int \sin^{4} x \times \cos^{2} x \mathrm{~d} x } \\ \\
= \ & \int \left( \frac{1 – \cos 2x}{2} \right)^{2} \times \frac{1 + \cos 2x}{2} \mathrm{~d} x \\ \\
= \ & \int \frac{\left( 1 – \cos 2x \right)^{2}}{4} \times \frac{1 + \cos 2x}{2} \mathrm{~d} x \\ \\
= \ & \int \frac{\left( 1 – \cos 2x \right)^{2} \times \left( 1 + \cos 2x \right)}{8} \mathrm{~d} x \\ \\
= \ & \textcolor{orange}{ \frac{1}{8} \int \left( 1 – \cos 2x \right)^{2} \times \left( 1 + \cos 2x \right) \mathrm{~d} x }
\end{aligned}
$$

对于上面这个多项和乘积的计算，可以参考「荒原之梦考研数学」的《峰图 | 用表格的形式辅助计算多项和的乘积》这篇文章提供的方法，从而得：

$$
\begin{aligned}
& \textcolor{lightgreen}{ \int \sin^{4} x \times \cos^{2} x \mathrm{~d} x } \\ \\
= \ & \textcolor{orange}{ \frac{1}{8} \int \left( 1 – \cos 2x \right)^{2} \times \left( 1 + \cos 2x \right) \mathrm{~d} x } \\ \\
= \ & \frac{1}{8} \int (\cos^{3} 2x – \cos^{2} 2x – \cos 2x + 1) \mathrm{~d} x \\ \\
= \ & \frac{1}{16} \int (1 – \sin^{2} 2x) \mathrm{~d} (\sin 2x) – \frac{1}{8} \int \frac{1 + \cos 4x}{2} \mathrm{~d} x + \frac{1}{8} \int (1 – \cos 2x) \mathrm{~d} x \\ \\
= \ & \frac{1}{16} \left( \sin 2x – \frac{1}{3} \sin^{3} 2x \right) + \frac{1}{16} \int \left( 2 – 2 \cos 2x – 1 – \cos 4x \right) \mathrm{~d} x \\ \\
= \ & \frac{1}{16} \left( \sin 2x – \frac{1}{3} \sin^{3} 2x \right) + \frac{1}{16} \int \left( 1 – 2 \cos 2x – \cos 4x \right) \mathrm{~d} x \\ \\
= \ & \frac{1}{16} \left( \sin 2x – \frac{1}{3} \sin^{3} 2x \right) + \frac{1}{16} \left( \int 1 \mathrm{~d} x – 2 \int \cos 2x \mathrm{~d} x – \int \cos 4x \mathrm{~d} x \right) \\ \\
= \ & \frac{1}{16} \left( \sin 2x – \frac{1}{3} \sin^{3} 2x \right) + \frac{1}{16} \left( x – 2 \times \frac{1}{2} \sin 2x – \frac{1}{4} \sin 4x \right) + C \\ \\
= \ & \frac{1}{16} \left( \sin 2x – \frac{1}{3} \sin^{3} 2x \right) + \frac{1}{16} \left( x – \sin 2x – \frac{1}{4} \sin 4x \right) + C \\ \\
= \ & \frac{1}{16} \left( \sin 2x – \frac{1}{3} \sin^{3} 2x + x – \sin 2x – \frac{1}{4} \sin 4x \right) + C \\ \\
= \ & \textcolor{lightgreen}{ \frac{1}{16} \left( x – \frac{1}{3} \sin^{3} 2x – \frac{1}{4} \sin 4x \right) + C }
\end{aligned}
$$

其中，$C$ 为任意常数.

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