点火公式

递归边界

$$\begin{array}{r}{\int }_0^{\pi /2}\sin xdx={\int }_0^{\pi /2}\cos x=1\end{array}$$

$$\begin{array}{r}{\int }_0^{\pi /2}{\sin }^{0}xdx={\int }_0^{\pi /2}{\cos }^{0}x=\pi /2\end{array}$$

对称性等式

$${\int }_0^{\pi /2}{\sin }^{n}xdx={\int }_0^{\pi /2}{\cos }^{n}xdx$$

递归

$$\begin{array}{rl}{\int }_0^{\pi /2}{\sin }^{n}xdx& =-{\int }_0^{\pi /2}{\sin }^{n-1}xd\cos x\\ & =-(si{n}^{n-1}x\cos x{|}_0^{\pi /2}-{\int }_0^{\pi /2}\cos xd({\sin }^{n-1}x))\\ & =(n-1){\int }_0^{\pi /2}{\sin }^{n-2}x{\cos }^{2}xdx\\ & =(n-1){\int }_0^{\pi /2}{\sin }^{n-2}xdx-(n-1){\int }_0^{\pi /2}{\sin }^{n}xdx\\ & \Downarrow \\ {\int }_0^{\pi /2}{\sin }^{n}xdx& =\frac{n-1}{n}{\int }_0^{\pi /2}{\sin }^{n-2}xdx\end{array}$$

递推式

$$\begin{gathered} {\int }_0^{\pi /2}{\sin }^{2n}xdx=\frac{2n-1}{2n}\frac{2n-3}{2n-2}...\frac{1}{2}\frac{\pi }{2} \\ {\int }_0^{\pi /2}{\sin }^{2n+1}xdx=\frac{2n}{2n+1}\frac{2n-2}{2n-1}...\frac{2}{3}1 \end{gathered}$$

变体

$${\int }_0^{\pi }{\sin }^{n}xdx={\int }_0^{\pi /2}{\sin }^{n}xdx+{\int }_{\pi /2}^{\pi }{\sin }^{n}xdx$$