分部积分法怎么用？？？-灵析社区

$$ \begin{align} & \int_0^{\pi}(1-x^2)\cos nxdx \\\ =& \int_0^{\pi}(1-x^2)\frac{1}{n}d \sin nx \\\ =& \frac{1}{n} \left[ \left . (1-x^2)\sin nx \right |_0^{\pi} -\int_0^{\pi}\sin nx d(1-x^2) \right] \\\ =& -\frac{1}{n} \int_0^{\pi}\sin nx d(1-x^2) \\\ =& -\frac{1}{n} \int_0^{\pi} -2x\sin nx dx \\\ =& \frac{2}{n} \int_0^{\pi}x\sin nx dx \\\ =& \frac{2}{n} \int_0^{\pi}x(-\frac{1}{n})d\cos nx \\\ =& -\frac{2}{n^2} \int_0^{\pi}xd\cos nx \\\ =& -\frac{2}{n^2} \left [ \left . x\cos nx \right | _0^{\pi} -\int_0^{\pi}\cos nxdx \right ] \\\ =& -\frac{2}{n^2} \left [ \pi\cos n\pi -\int_0^{\pi}\cos nxdx \right ] \\\ =& -\frac{2}{n^2} \left [ \pi\cos n\pi -\left . \frac{1}{n}\sin nx \right |_0^{\pi} \right ] \\\ =& -\frac{2\pi\cos n\pi}{n^2} \\\ =& \frac{2\pi(-1)^{(n+1)}}{n^2} \end{align} $$ 用了两次分部积分法