(3)
\[
\begin{split}
\int\cos^6 x\mathrm{d}x&=\int\cos^5 x\mathrm{d}\sin x\\
&=\sin x\cdot\cos^5 x-\int\sin x\mathrm{d}\cos^5 x\\
&=\sin x\cdot\cos^5 x-\int\sin x\cdot 5\cos^4 x\cdot(-1)\sin x\mathrm{d}x\\
&=\sin x\cdot\cos^5 x+5\int\sin^2 x\cdot\cos^4 x\mathrm{d}x\\
&=\sin x\cdot\cos^5 x+5\int(1-\cos^2 x)\cos^4 x\mathrm{d}x\\
&=\sin x\cdot\cos^5 x+5\int\cos^4 x\mathrm{d}x-5\int\cos^6 x\mathrm{d}x,
\end{split}
\]
这推出
\[
6\int\cos^6 x\mathrm{d}x=\sin x\cdot\cos^5 x+5\int\cos^4 x\mathrm{d}x.
\]
若记 $I_n=\int\cos^n x\mathrm{d}x$, 则
\[
I_6=\frac{1}{6}\sin x\cdot\cos^5 x+\frac{5}{6}I_4.
\]
而
\[
I_4=\int\cos^4 x\mathrm{d}x=\frac{3}{8}x+\frac{1}{4}\sin 2x+\frac{1}{32}\sin 4x+C,
\]
故
\[
\begin{split}
I_6&=\frac{1}{6}\sin x\cdot\cos^5 x+\frac{5}{6}\Bigl(\frac{3}{8}x+\frac{1}{4}\sin 2x+\frac{1}{32}\sin 4x\Bigr)+C\\
&=\frac{1}{6}\sin x\cdot\cos^5 x+\frac{5}{16}x+\frac{5}{24}\sin 2x+\frac{5}{192}\sin 4x+C.
\end{split}
\]
如果 $I_4$ 使用另一表达式
\[
I_4=\frac{3}{8}x+\frac{1}{4}\sin x\cdot\cos^3 x+\frac{3}{8}\sin x\cdot\cos x+C,
\]
则
\[
\begin{split}
I_6&=\frac{1}{6}\sin x\cdot\cos^5 x+\frac{5}{6}\Bigl(\frac{3}{8}x+\frac{1}{4}\sin x\cdot\cos^3 x+\frac{3}{8}\sin x\cdot\cos x\Bigr)+C\\
&=\frac{1}{6}\sin x\cdot\cos^5 x+\frac{5}{24}\sin x\cdot\cos^3 x+\frac{5}{16}\sin x\cdot\cos x+\frac{5}{16}x+C.
\end{split}
\]
还可以这样算.
\[
\begin{split}
I_6=\int\cos^6 x\mathrm{d}x&=\int(\cos^2 x)^3\mathrm{d}x=\int\Bigl(\frac{1+\cos 2x}{2}\Bigr)^3\mathrm{d}x\\
&=\frac{1}{8}\int\Bigl[1+3\cos(2x)+3\cos^2(2x)+\cos^3(2x)\Bigr]\mathrm{d}x\\
&=\frac{1}{8}\Bigl[x+3\int\cos(2x)\mathrm{d}x+3\int\cos^2(2x)\mathrm{d}x+\int\cos^2(2x)\cdot\cos(2x)\mathrm{d}x\Bigr]\\
&=\frac{1}{8}\Bigl[x+\frac{3}{2}\sin(2x)+3\int\frac{1+\cos(4x)}{2}\mathrm{d}x+\frac{1}{2}\int(1-\sin^2(2x))\mathrm{d}\sin(2x)\Bigr]\\
&=\frac{1}{8}\Bigl[x+\frac{3}{2}\sin(2x)+\frac{3}{2}\bigl(x+\frac{1}{4}\sin(4x)\bigr)+\frac{1}{2}\sin(2x)-\frac{1}{2}\int\sin^2(2x)\mathrm{d}\sin(2x)\Bigr]\\
&=\frac{1}{8}x+\frac{3}{16}\sin(2x)+\frac{3}{16}x+\frac{3}{64}\sin(4x)+\frac{1}{16}\sin(2x)-\frac{1}{48}\sin^3(2x)+C\\
&=\frac{5}{16}x+\frac{1}{4}\sin(2x)+\frac{3}{64}\sin(4x)-\frac{1}{48}\sin^3(2x)+C.
\end{split}
\]
验证:
\[
\begin{split}
&\Bigl(\frac{5}{16}x+\frac{1}{4}\sin(2x)+\frac{3}{64}\sin(4x)-\frac{1}{48}\sin^3(2x)\Bigr)'\\
=&\frac{5}{16}+\frac{1}{2}\cos(2x)+\frac{3}{16}\cos(4x)-\frac{1}{16}\sin^2(2x)\cdot 2\cos(2x)\\
=&\frac{5}{16}+\frac{1}{2}\bigl(2\cos^2 x-1\bigr)+\frac{3}{16}\bigl(2\cos^2(2x)-1\bigr)-\frac{1}{8}\sin^2(2x)\cdot\bigl(2\cos^2 x-1\bigr)\\
=&\frac{5}{16}+\cos^2 x-\frac{1}{2}+\frac{3}{8}\cos^2(2x)-\frac{3}{16}-\frac{1}{8}\cdot 4\sin^2 x\cdot\cos^2 x\cdot\bigl(2\cos^2 x-1\bigr)\\
=&-\frac{3}{8}+\cos^2 x+\frac{3}{8}\cos^2(2x)-\frac{1}{2}\sin^2 x\cdot\cos^2 x\cdot\bigl(2\cos^2 x-1\bigr)\\
=&-\frac{3}{8}+\cos^2 x+\frac{3}{8}(2\cos^2 x-1)^2-\sin^2 x\cdot\cos^4 x+\frac{1}{2}\sin^2 x\cdot\cos^2 x\\
=&-\frac{3}{8}+\cos^2 x+\frac{3}{8}\bigl[4\cos^4 x-4\cos^2 x+1\bigr]-(1-\cos^2 x)\cos^4 x+\frac{1}{2}(1-\cos^2 x)\cos^2 x\\
=&\cos^2 x+\frac{3}{2}\cos^4 x-\frac{3}{2}\cos^2 x-\cos^4 x+\cos^6 x+\frac{1}{2}\cos^2 x-\frac{1}{2}\cos^4 x\\
=&\cos^6 x\ .
\end{split}
\]
(4)
类似地,
\[
\begin{split}
\int\sin^6 x\mathrm{d}x&=\int(\sin^2 x)^3\mathrm{d}x=\int\Bigl(\frac{1-\cos 2x}{2}\Bigr)^3\mathrm{d}x\\
&=\frac{1}{8}\int\Bigl[1^3-3\cdot 1^2\cdot\cos(2x)+3\cdot 1\cdot\cos^2(2x)-\cos^3(2x)\Bigr]\mathrm{d}x\\
&=\frac{1}{8}x-\frac{3}{8}\int\cos(2x)\mathrm{d}x+\frac{3}{8}\int\cos^2(2x)\mathrm{d}x-\frac{1}{8}\int\cos^3(2x)\mathrm{d}x\\
&=\frac{1}{8}x-\frac{3}{16}\sin(2x)+\frac{3}{8}\int\frac{1+\cos(4x)}{2}\mathrm{d}x-\frac{1}{16}\int\cos^2(2x)\mathrm{d}\sin(2x)\\
&=\frac{1}{8}x-\frac{3}{16}\sin(2x)+\frac{3}{16}x+\frac{3}{16}\cdot\frac{1}{4}\sin(4x)-\frac{1}{16}\int(1-\sin^2(2x))\mathrm{d}\sin(2x)\\
&=\frac{5}{16}x-\frac{3}{16}\sin(2x)+\frac{3}{64}\sin(4x)-\frac{1}{16}\Bigl[\sin(2x)-\frac{1}{3}\sin^3(2x)\Bigr]+C\\
&=\frac{5}{16}x-\frac{1}{4}\sin(2x)+\frac{3}{64}\sin(4x)+\frac{1}{48}\sin^3(2x)+C.
\end{split}
\]
或者这样计算,
\[
\begin{split}
\int\sin^6 x\mathrm{d}x&=\int(\sin^2 x)^3\mathrm{d}x=\int\Bigl(\frac{1-\cos 2x}{2}\Bigr)^3\mathrm{d}x\\
&=\frac{1}{8}\int\Bigl[1^3-3\cdot 1^2\cdot\cos(2x)+3\cdot 1\cdot\cos^2(2x)-\cos^3(2x)\Bigr]\mathrm{d}x\\
&=\frac{1}{8}x-\frac{3}{8}\int\cos(2x)\mathrm{d}x+\frac{3}{8}\int\cos^2(2x)\mathrm{d}x-\frac{1}{8}\int\cos^3(2x)\mathrm{d}x\\
&=\frac{1}{8}x-\frac{3}{16}\sin(2x)+\frac{3}{8}\int\frac{1+\cos(4x)}{2}\mathrm{d}x-\frac{1}{8}\int\frac{1+\cos(4x)}{2}\cdot\cos(2x)\mathrm{d}x,
\end{split}
\]
其中
\[
\int\frac{1+\cos(4x)}{2}\mathrm{d}x=\frac{1}{2}x+\frac{1}{8}\sin(4x)+C,
\]
\[
\begin{split}
\int\frac{1+\cos(4x)}{2}\cdot\cos(2x)\mathrm{d}x&=\frac{1}{2}\int\cos(2x)\mathrm{d}x+\frac{1}{2}\int\cos(4x)\cdot\cos(2x)\mathrm{d}x\\
&=\frac{1}{4}\sin(2x)+\frac{1}{2}\int\frac{1}{2}\bigl(\cos(6x)+\cos(2x)\bigr)\mathrm{d}x\\
&=\frac{1}{4}\sin(2x)+\frac{1}{4}\Bigl[\frac{1}{6}\sin(6x)+\frac{1}{2}\sin(2x)\Bigr]+C\\
&=\frac{3}{8}\sin(2x)+\frac{1}{24}\sin(6x)+C.
\end{split}
\]
这里用到了三角函数的积化和差公式(见问题2023)
\[
\cos\alpha\cdot\cos\beta=\frac{1}{2}\Bigl[\cos(\alpha+\beta)+\cos(\alpha-\beta)\Bigr].
\]
因此,
\[
\begin{split}
\int\sin^6 x\mathrm{d}x&=\frac{1}{8}x-\frac{3}{16}\sin(2x)+\frac{3}{8}\int\frac{1+\cos(4x)}{2}\mathrm{d}x-\frac{1}{8}\int\frac{1+\cos(4x)}{2}\cdot\cos(2x)\mathrm{d}x\\
&=\frac{1}{8}x-\frac{3}{16}\sin(2x)+\frac{3}{8}\Bigl[\frac{1}{2}x+\frac{1}{8}\sin(4x)\Bigr]-\frac{1}{8}\Bigl[\frac{3}{8}\sin(2x)+\frac{1}{24}\sin(6x)\Bigr]+C\\
&=\frac{5}{16}x-\frac{15}{64}\sin(2x)+\frac{3}{64}\sin(4x)-\frac{1}{192}\sin(6x)+C.
\end{split}
\]
利用已知结果(见问题3441)
\[
\begin{aligned}
\cos(2x)&=2\cos^2 x-1,\\
\cos(4x)&=8\cos^4 x-8\cos^2 x+1,\\
\cos(6x)&=32\cos^6 x-48\cos^4 x+18\cos^2 x-1,
\end{aligned}
\]
对上面所得式子求导,
\[
\begin{split}
&\Bigl(\frac{5}{16}x-\frac{15}{64}\sin(2x)+\frac{3}{64}\sin(4x)-\frac{1}{192}\sin(6x)\Bigr)'\\
=&\frac{5}{16}-\frac{15}{32}\cos(2x)+\frac{3}{16}\cos(4x)-\frac{1}{32}\cos(6x)\\
=&\frac{5}{16}-\frac{15}{32}(2\cos^2 x-1)+\frac{3}{16}\Bigl(8\cos^4 x-8\cos^2 x+1\Bigr)-\frac{1}{32}\Bigl(32\cos^6 x-48\cos^4 x+18\cos^2 x-1\Bigr)\\
=&\frac{5}{16}-\frac{15}{16}\cos^2 x+\frac{15}{32}+\frac{3}{2}\cos^4 x-\frac{3}{2}\cos^2 x+\frac{3}{16}-\cos^6 x+\frac{3}{2}\cos^4 x-\frac{9}{16}\cos^2 x+\frac{1}{32}\\
=&1-\cos^6 x-3\cos^2 x+3\cos^4 x\\
=&(1-\cos^2 x)^3\\
=&\sin^6 x.
\end{split}
\]
