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