sin x和cos x的若干次方的定积分

sin x和cos x的若干次方的定积分

$${\int }_0^{\frac{\pi }{2}}{\sin }^{n}x={\int }_0^{\frac{\pi }{2}}{\cos }^{n}x=\{\begin{array}{l}\frac{n-1}{n}\times \frac{n-3}{n-2}\times \cdots \times \frac{1}{2}\times \frac{\pi }{2}n为偶数\\ \\ \frac{n-1}{n}\times \frac{n-3}{n-2}\times \cdots \times \frac{1}{2}\times 1n为奇数\end{array}$$

证明：

\qquad 根据$\sin x$和$\cos x$在$[0,\frac{\pi }{2}]$上的对称性可得${\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx={\int }_0^{\frac{\pi }{2}}{\cos }^{n}xdx$

\qquad 令$I_n={\int }_0^{\frac{\pi }{2}}{\sin }^{n}x={\int }_0^{\frac{\pi }{2}}\cos xdx$，则

$$I_n={\int }_0^{\frac{\pi }{2}}\sin x\cdot {\sin }^{n-1}xdx={\int }_0^{\frac{\pi }{2}}{\sin }^{n-1}xd(-\cos x)=-\cos x\cdot {\sin }^{n-1}x{|}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}(-\cos x)d({\sin }^{n-1}x)$$

$$=(n-1){\int }_0^{\frac{\pi }{2}}{\cos }^{2}x\cdot {\sin }^{n-2}xdx=(n-1)({\int }_0^{\frac{\pi }{2}}{\sin }^{n-2}xdx-{\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx)=(n-1)I_{n-2}-(n-1)I_n$$

\qquad 等式两边同时加$(n-1)I_n$得$n{I}_n=(n-1)I_{n-2}$，即$I_n=\frac{n-1}{n}{I}_{n-2}$。由数学归纳法即可得证。