沃利斯积分公式是求解形如∫ 0π2sin⁡ nx d x \int_0^{\frac\pi2}\sin^nx\text{d}x∫02π​​sinnxdx这种积分的公式。

记I n =∫ 0π 2 sin ⁡n xdxI_n=\int_0^{\frac\pi2}\sin^nx\text{d}xIn​=∫02π​​sinnxdx。当n≥2n\ge2n≥2时，应用凑微分和分部积分得 I n=∫ 0π2sin⁡ nx d x = −∫ 0π2sin⁡n−1x d ( cos ⁡ x )= −[ sin ⁡ n − 1xcos⁡x∣0π2 −∫0π2 cos ⁡ x d (sin⁡n−1 x ) ] = −[ sin ⁡ n − 1xcos⁡x∣0π2 − ( n − 1 )∫0π2cos⁡2 xsin⁡n−2 x d x ] = −[ 0 − 0 − ( n − 1 )∫0π2 ( 1 −sin⁡2 x )sin⁡n−2 x d x ] = − ( n − 1 )[ −∫0π2sin⁡n−2 x d x +∫0π2sin⁡n x d x ] = ( n − 1 )In−2− ( n − 1 )I n\begin{aligned}I_n&=\int_0^{\frac\pi2}\sin^nx\text{d}x=-\int_0^{\frac\pi2}\sin^{n-1}x\text{d}(\cos x)\\&=-\left[\left.\sin^{n-1}x\cos x\right|_0^{\frac\pi2}-\int_0^{\frac\pi2}\cos x\text{d}(\sin^{n-1}x)\right]\\&=-\left[\left.\sin^{n-1}x\cos x\right|_0^{\frac\pi2}-(n-1)\int_0^{\frac\pi2}\cos^2x\sin^{n-2}x\text{d}x\right]\\&=-\left[0-0-(n-1)\int_0^{\frac\pi2}(1-\sin^2x)\sin^{n-2}x\text{d}x\right]\\&=-(n-1)\left[-\int_0^{\frac\pi2}\sin^{n-2}x\text{d}x+\int_0^{\frac\pi2}\sin^nx\text{d}x\right]\\&=(n-1)I_{n-2}-(n-1)I_n\end{aligned}In​​=∫02π​​sinnxdx=−∫02π​​sinn−1xd(cosx)=−[sinn−1xcosx∣∣​02π​​−∫02π​​cosxd(sinn−1x)]=−[sinn−1xcosx∣∣​02π​​−(n−1)∫02π​​cos2xsinn−2xdx]=−[0−0−(n−1)∫02π​​(1−sin2x)sinn−2xdx]=−(n−1)[−∫02π​​sinn−2xdx+∫02π​​sinnxdx]=(n−1)In−2​−(n−1)In​​即I n= ( n − 1 )In−2− ( n − 1 )I nI_n=(n-1)I_{n-2}-(n-1)I_nIn​=(n−1)In−2​−(n−1)In​ nI n= ( n − 1 )In−2nI_n=(n-1)I_{n-2}nIn​=(n−1)In−2​I n= n−1 n In−2I_n=\frac{n-1}{n}I_{n-2}In​=nn−1​In−2​当nnn为奇数时，I n= n−1 n In−2= n−1 n⋅ n−3n−2 In−4= ⋯ = n−1 n⋅ n−3n−2⋅ ⋯ ⋅2 3 I 1I_n=\frac{n-1}{n}I_{n-2}=\frac{n-1}{n}\cdot\frac{n-3}{n-2}I_{n-4}=\cdots=\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\cdots\cdot\frac23I_1In​=nn−1​In−2​=nn−1​⋅n−2n−3​In−4​=⋯=nn−1​⋅n−2n−3​⋅⋯⋅32​I1​其中I 1 =∫ 0π 2sin⁡xdx=− cos ⁡ x ∣0π 2=1I_1=\int_0^{\frac{\pi}{2}}\sin x\text{d}x=-\left.\cos x\right|_0^{\frac\pi2}=1I1​=∫02π​​sinxdx=−cosx∣02π​​=1，故此时I n = n − 1n ⋅ n − 3 n − 2⋅⋯⋅2 3 I_n=\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\cdots\cdot\frac23In​=nn−1​⋅n−2n−3​⋅⋯⋅32​； 当nnn为偶数时，I n= n−1 n⋅ n−3n−2⋅ ⋯ ⋅1 2 I 0I_n=\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\cdots\cdot\frac12I_0In​=nn−1​⋅n−2n−3​⋅⋯⋅21​I0​而∫ 0π 2 sin ⁡0 xdx=∫ 0π 2dx=π 2 \int_0^{\frac\pi2}\sin^0x\text{d}x=\int_0^{\frac\pi2}\text{d}x=\frac\pi2∫02π​​sin0xdx=∫02π​​dx=2π​，故此时I n = n − 1n ⋅ n − 3 n − 2⋅⋯⋅1 2 ⋅π 2 I_n=\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\cdots\cdot\frac12\cdot\frac\pi2In​=nn−1​⋅n−2n−3​⋅⋯⋅21​⋅2π​。 综上所述，I n={n − 1 n⋅n − 3 n − 2 ⋅⋯⋅23,n为奇数，n − 1 n⋅n − 3 n − 2 ⋅⋯⋅12⋅π2,n为偶数。I_n=\begin{cases}\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\cdots\cdot\frac23,\quad&n\text{为奇数，}\\\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\cdots\cdot\frac12\cdot\frac\pi2,\quad&n\text{为偶数。}\end{cases}In​={nn−1​⋅n−2n−3​⋅⋯⋅32​,nn−1​⋅n−2n−3​⋅⋯⋅21​⋅2π​,​n为奇数，n为偶数。​记忆时，我们只需牢记递推公式I n= n−1 n In−2I_n=\frac{n-1}{n}I_{n-2}In​=nn−1​In−2​即可。

在I n =∫ 0π 2 sin ⁡n xdxI_n=\int_0^{\frac\pi2}\sin^nx\text{d}xIn​=∫02π​​sinnxdx中，令u=π 2 −xu=\frac\pi2-xu=2π​−x，则cos⁡u=sin⁡x\cos u=\sin xcosu=sinx，du=−dx\text{d}u=-\text{d}xdu=−dx，I n=∫π2 0cos⁡ nu ⋅ − d u =∫ 0π2cos⁡ nu d u I_n=\int_\frac\pi2^0\cos^n u\cdot-\text{d}u=\int_0^{\frac\pi2}\cos^nu\text{d}uIn​=∫2π​0​cosnu⋅−du=∫02π​​cosnudu所以余弦函数和正弦函数的公式是完全一样的：I n=∫ 0π2sin⁡ nx d x =∫ 0π2cos⁡ nx d x ={n − 1 n⋅n − 3 n − 2 ⋅⋯⋅23,n为奇数，n − 1 n⋅n − 3 n − 2 ⋅⋯⋅12⋅π2,n为偶数。I_n=\int_0^{\frac\pi2}\sin^nx\text{d}x=\int_0^{\frac\pi2}\cos^nx\text{d}x=\begin{cases}\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\cdots\cdot\frac23,\quad&n\text{为奇数，}\\\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\cdots\cdot\frac12\cdot\frac\pi2,\quad&n\text{为偶数。}\end{cases}In​=∫02π​​sinnxdx=∫02π​​cosnxdx={nn−1​⋅n−2n−3​⋅⋯⋅32​,nn−1​⋅n−2n−3​⋅⋯⋅21​⋅2π​,​n为奇数，n为偶数。​

对于∫ 0 πsin ⁡n xdx\int_0^\pi\sin^nx\text{d}x∫0π​sinnxdx，因为sin⁡x\sin xsinx关于x=π 2 x=\frac\pi2x=2π​对称，所以∫ 0 πsin ⁡n xdx=2∫ 0π 2 sin ⁡n xdx=2I n \int_0^\pi\sin^nx\text{d}x=2\int_0^\frac\pi2\sin^nx\text{d}x=2I_n∫0π​sinnxdx=2∫02π​​sinnxdx=2In​。 对于∫ 0 πcos ⁡n xdx\int_0^\pi\cos^nx\text{d}x∫0π​cosnxdx： (1) 当nnn为奇数时， cos ⁡n x\cos^nxcosnx在x=π 2 x=\frac\pi2x=2π​两边互为相反数，所以积分值为000； (2) 当nnn为偶数时， cos ⁡n x\cos^nxcosnx关于x=π 2 x=\frac\pi2x=2π​对称，所以∫ 0 πcos ⁡n xdx=2∫ 0π 2 cos ⁡n xdx\int_0^\pi\cos^nx\text{d}x=2\int_0^\frac\pi2\cos^nx\text{d}x∫0π​cosnxdx=2∫02π​​cosnxdx。