\[ I_n=\begin{cases} \frac{(2k-1)!!}{(2k)!!}\cdot\frac{\pi}{2},&\text{if}\ n=2k;\\ \frac{(2k)!!}{(2k+1)!!},&\text{if}\ n=2k+1.\\ \end{cases} \]
类似的, 求
\[
I_n=\int_0^{\pi}x\cos^n xdx,\quad I_n=\int_0^{\pi}x\sin^n xdx.
\]
1
设
\[
I_n=\int_0^{\pi}x\cos^n xdx,
\]
解:
\[
\begin{split}
I_n&=\int_0^{\pi}x\cos^n xdx\\
&=\int_0^{\pi}x\cos^{n-1}xd\sin x\\
&=x\cos^{n-1}x\sin x\biggr|_{0}^{\pi}-\int_0^{\pi}\sin xd(x\cos^{n-1}x)\\
&=0-\int_0^{\pi}\sin x\biggl(\cos^{n-1}x+x(n-1)\cos^{n-2}x\cdot(-1)\sin x\biggr)dx\\
&=-\int_0^{\pi}\sin x\cos^{n-1}xdx+(n-1)\int_0^{\pi}x\cos^{n-2}x\sin^2 xdx\\
&=\int_0^{\pi}\cos^{n-1}xd\cos x+(n-1)\int_0^{\pi}x\cos^{n-2}x(1-\cos^2 x)dx
\end{split}
\]