问题及解答

计算

\[\int\frac{1}{\sin^4(x)+\cos^4(x)}dx\]

References:

对三大数学软件 Mathematica 、Maple 、MATLAB 的小测试比较

https://blog.csdn.net/iteye_18800/article/details/82294214?utm_medium=distribute.pc_relevant.none-task-blog-BlogCommendFromMachineLearnPai2-2.channel_param&depth_1-utm_source=distribute.pc_relevant.none-task-blog-BlogCommendFromMachineLearnPai2-2.channel_param

\[
\begin{split}
\sin^4 x+\cos^4 x&=(\sin^2 x+\cos^2 x)^2-2\sin^2 x\cdot\cos^2 x\\
&=1-\frac{1}{2}(2\sin x\cdot\cos x)^2\\
&=1-\frac{1}{2}\sin^2(2x)
\end{split}
\]

因此,

\[
\begin{split}
\frac{1}{\sin^4 x+\cos^4 x}&=\frac{1}{1-\frac{1}{2}\sin^2(2x)}=\frac{2}{2-\sin^2(2x)}\\
&=\frac{2}{2-\frac{1-\cos(4x)}{2}}=\frac{4}{3+\cos(4x)}
\end{split}
\]

于是,

\[
\int\frac{1}{\sin^4 x+\cos^4 x}\mathrm{d}x=\int\frac{4}{3+\cos(4x)}\mathrm{d}x
\]

若令 $u=4x$, 则变为

\[
\int\frac{1}{3+\cos u}\mathrm{d}u
\]

然后用万能变换, 令 $t=\tan\frac{u}{2}$, 则 $\cos u=\frac{1-t^2}{1+t^2}$, $\mathrm{d}u=\frac{2}{1+t^2}\mathrm{d}t$. 于是原积分变为

\[
\begin{split}
\int\frac{1}{3+\frac{1-t^2}{1+t^2}}\cdot\frac{2}{1+t^2}\mathrm{d}t&=\int\frac{1}{t^2+2}\mathrm{d}t=\frac{1}{\sqrt{2}}\arctan\frac{t}{\sqrt{2}}+C\\
&=\frac{1}{\sqrt{2}}\arctan\frac{\tan(2x)}{\sqrt{2}}+C.
\end{split}
\]