问题及解答

求 $\int_0^{\pi}(x\sin x)^2dx$

\[
\begin{split}
\int_0^{\pi}(x\sin x)^2dx&=\int_0^{\pi}x^2\sin^2 xdx=\int_0^{\pi}x^2\cdot\frac{1-\cos 2x}{2}dx\\
&=\frac{1}{2}\int_0^{\pi}x^2dx-\frac{1}{2}\int_0^{\pi}x^2\cos 2xdx\\
&=\frac{1}{6}x^3\biggr|_{0}^{\pi}-\frac{1}{4}\int_0^{\pi}x^2d\sin(2x)\\
&=\frac{1}{6}\pi^3-\frac{1}{4}\biggl[x^2\sin(2x)\biggr|_{0}^{\pi}-\int_0^{\pi}\sin(2x)dx^2\biggr]\\
&=\frac{1}{6}\pi^3+\frac{1}{2}\int_0^{\pi}x\sin(2x)dx\\
&=\frac{1}{6}\pi^3-\frac{1}{4}\int_0^{\pi}xd\cos(2x)\\
&=\frac{1}{6}\pi^3-\frac{1}{4}\biggl[x\cos(2x)\biggr|_{0}^{\pi}-\int_0^{\pi}\cos(2x)dx\biggr]\\
&=\frac{1}{6}\pi^3-\frac{\pi}{4}.
\end{split}
\]