求不定积分 $\displaystyle\int\sin^4 x\mathrm{d}x$, $\displaystyle\int\cos^4 x\mathrm{d}x$.
求不定积分 $\displaystyle\int\sin^6 x\mathrm{d}x$, $\displaystyle\int\cos^6 x\mathrm{d}x$.
1
(1)
\[
\sin^4 x=(\sin^2 x)^2=\bigl(\frac{1-\cos 2x}{2}\bigr)^2=\frac{1}{4}(1-2\cos 2x+\cos^2 2x),
\]
因此,
\[
\begin{split}
\int\sin^4 x\mathrm{d}x&=\frac{1}{4}\int\bigl(1-2\cos 2x+\cos^2 2x\bigr)\mathrm{d}x\\
&=\frac{1}{4}x-\frac{1}{2}\int\cos 2x\mathrm{d}x+\frac{1}{4}\int\cos^2 2x\mathrm{d}x\\
&=\frac{1}{4}x-\frac{1}{4}\sin 2x+\frac{1}{4}\int\frac{1+\cos 4x}{2}\mathrm{d}x\\
&=\frac{1}{4}x-\frac{1}{4}\sin 2x+\frac{1}{8}\bigl(x+\frac{1}{4}\sin 4x\bigr)+C\\
&=\frac{3}{8}x-\frac{1}{4}\sin 2x+\frac{1}{32}\sin 4x+C.
\end{split}
\]
验证:
\[
\begin{split}
\Bigl(\frac{3}{8}x-\frac{1}{4}\sin 2x+\frac{1}{32}\sin 4x\Bigr)' &=\frac{3}{8}-\frac{1}{2}\cos 2x+\frac{1}{8}\cos 4x\\
&=\frac{3}{8}-\frac{1}{2}\bigl(1-2\sin^2 x\bigr)+\frac{1}{8}\bigl(1-2\sin^2 2x\bigr)\\
&=\frac{3}{8}-\frac{1}{2}+\sin^2 x+\frac{1}{8}-\frac{1}{4}\sin^2 2x\\
&=\sin^2 x-\frac{1}{4}\sin^2 2x\\
&=\sin^2 x-\frac{1}{4}\bigl(1-\cos^2 2x\bigr)\\
&=\sin^2 x-\frac{1}{4}+\frac{1}{4}\Bigl(1-2\sin^2 x\Bigr)^2\\
&=\sin^2 x-\frac{1}{4}+\frac{1}{4}\Bigl(1-4\sin^2 x+4\sin^4 x\Bigr)\\
&=\sin^4 x\ .
\end{split}
\]
(2)
\[
\cos^4 x=(\cos^2 x)^2=\bigl(\frac{1+\cos 2x}{2}\bigr)^2=\frac{1}{4}(1+2\cos 2x+\cos^2 2x),
\]
因此,
\[
\begin{split}
\int\cos^4 x\mathrm{d}x&=\frac{1}{4}\int\bigl(1+2\cos 2x+\cos^2 2x\bigr)\mathrm{d}x\\
&=\frac{1}{4}x+\frac{1}{2}\int\cos 2x\mathrm{d}x+\frac{1}{4}\int\cos^2 2x\mathrm{d}x\\
&=\frac{1}{4}x+\frac{1}{4}\sin 2x+\frac{1}{4}\int\frac{1+\cos 4x}{2}\mathrm{d}x\\
&=\frac{1}{4}x+\frac{1}{4}\sin 2x+\frac{1}{8}\bigl(x+\frac{1}{4}\sin 4x\bigr)+C\\
&=\frac{3}{8}x+\frac{1}{4}\sin 2x+\frac{1}{32}\sin 4x+C.
\end{split}
\]
验证:
\[
\begin{split}
\Bigl(\frac{3}{8}x+\frac{1}{4}\sin 2x+\frac{1}{32}\sin 4x\Bigr)' &=\frac{3}{8}+\frac{1}{2}\cos 2x+\frac{1}{8}\cos 4x\\
&=\frac{3}{8}+\frac{1}{2}\bigl(2\cos^2 x-1\bigr)+\frac{1}{8}\bigl(2\cos^2 2x-1\bigr)\\
&=\frac{3}{8}+\cos^2 x-\frac{1}{2}+\frac{1}{4}\cos^2 2x-\frac{1}{8}\\
&=\cos^2 x+\frac{1}{4}\cos^2 2x-\frac{1}{4}\\
&=\cos^2 x+\frac{1}{4}\bigl(2\cos^2 x-1\bigr)^2-\frac{1}{4}\\
&=\cos^2 x+\frac{1}{4}\Bigl(4\cos^4 x-4\cos^2 x+1\Bigr)-\frac{1}{4}\\
&=\cos^4 x\ .
\end{split}
\]
(2)
\[
\cos^4 x=(\cos^2 x)^2=\bigl(\frac{1+\cos 2x}{2}\bigr)^2=\frac{1}{4}(1+2\cos 2x+\cos^2 2x),
\]
因此,
\[
\begin{split}
\int\cos^4 x\mathrm{d}x&=\frac{1}{4}\int\bigl(1+2\cos 2x+\cos^2 2x\bigr)\mathrm{d}x\\
&=\frac{1}{4}x+\frac{1}{2}\int\cos 2x\mathrm{d}x+\frac{1}{4}\int\cos^2 2x\mathrm{d}x\\
&=\frac{1}{4}x+\frac{1}{4}\sin 2x+\frac{1}{4}\int\frac{1+\cos 4x}{2}\mathrm{d}x\\
&=\frac{1}{4}x+\frac{1}{4}\sin 2x+\frac{1}{8}\bigl(x+\frac{1}{4}\sin 4x\bigr)+C\\
&=\frac{3}{8}x+\frac{1}{4}\sin 2x+\frac{1}{32}\sin 4x+C.
\end{split}
\]
验证:
\[
\begin{split}
\Bigl(\frac{3}{8}x+\frac{1}{4}\sin 2x+\frac{1}{32}\sin 4x\Bigr)' &=\frac{3}{8}+\frac{1}{2}\cos 2x+\frac{1}{8}\cos 4x\\
&=\frac{3}{8}+\frac{1}{2}\bigl(2\cos^2 x-1\bigr)+\frac{1}{8}\bigl(2\cos^2 2x-1\bigr)\\
&=\frac{3}{8}+\cos^2 x-\frac{1}{2}+\frac{1}{4}\cos^2 2x-\frac{1}{8}\\
&=\cos^2 x+\frac{1}{4}\cos^2 2x-\frac{1}{4}\\
&=\cos^2 x+\frac{1}{4}\bigl(2\cos^2 x-1\bigr)^2-\frac{1}{4}\\
&=\cos^2 x+\frac{1}{4}\Bigl(4\cos^4 x-4\cos^2 x+1\Bigr)-\frac{1}{4}\\
&=\cos^4 x\ .
\end{split}
\]