问题及解答

计算定积分

\[
\int_0^1 \frac{x^2\arcsin x}{\sqrt{1-x^2}}dx
\]

令 $t=\arcsin x$, 则 $x=\sin t$. 因此原积分等于

\[
\begin{split}
\int_0^{\frac{\pi}{2}}t\cdot\sin^2 t dt&=\int_0^{\frac{\pi}{2}}t\cdot\frac{1-\cos(2t)}{2}dt\\
&=\frac{1}{2}\biggl[\int_0^{\frac{\pi}{2}}tdt-\int_0^{\frac{\pi}{2}}t\cos(2t)dt\biggr]\\
&=\frac{1}{4}t^2\biggr|_{0}^{\frac{\pi}{2}}-\frac{1}{4}\int_0^{\frac{\pi}{2}}td\sin(2t)\\
&=\frac{\pi^2}{16}-\frac{1}{4}\biggl[t\sin(2t)\biggr|_{0}^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}\sin(2t)dt\biggr]\\
&=\frac{\pi^2}{16}+\frac{1}{4}
\end{split}
\]