Wallis 공식

부분 적분을 통해 다음과 같이 쓸 수 있다.

$$\begin{array}{rl}& \\ {\int }_0^{\pi /2}{\sin }^{n}xdx& ={\int }_0^{\pi /2}{\sin }^{n-1}x\sin xdx\\ & =[{\sin }^{n-1}x(-\cos x){]}_0^{\pi /2}-{\int }_0^{\pi /2}(n-1){\sin }^{n-2}x\cos x(-\cos x)dx\\ & =-[{\sin }^{n-1}x\cos x{]}_0^{\pi /2}+(n-1){\int }_0^{\pi /2}{\sin }^{n-2}x{\cos }^{2}xdx\\ & =-[{\sin }^{n-1}x\cos x{]}_0^{\pi /2}+(n-1){\int }_0^{\pi /2}{\sin }^{n-2}x(1-{\sin }^{2}x)dx\\ & =-[{\sin }^{n-1}x\cos x{]}_0^{\pi /2}+(n-1){\int }_0^{\pi /2}{\sin }^{n-2}x-{\sin }^{n}x)dx\\ n{\int }_0^{\pi /2}{\sin }^{n}xdx& =-[{\sin }^{n-1}x\cos x{]}_0^{\pi /2}+(n-1){\int }_0^{\pi /2}{\sin }^{n-2}xdx\\ \therefore {\int }_0^{\pi /2}{\sin }^{n}xdx& =\frac{n-1}{n}{\int }_0^{\pi /2}{\sin }^{n-2}xdx\end{array}$$ $$\begin{array}{rl}{I}_n& =\frac{n-1}{n}{I}_{n-2}\end{array}$$