The definite integral of several powers of sin x and cos x
∫ 0 π 2 sin nx = ∫ 0 π 2 cos nx = { n − 1 n × n − 3 n − 2 × ⋯ × 1 2 × π 2 n is an even number n − 1 n × n − 3 n − 2 × ⋯ × 1 2 × 1 n is an odd number\int_0^{\frac{\pi}{2}}\sin^nx=\int_0^{\frac{\pi}{2}}\cos^nx= \begin {cases} \dfrac{n-1}{n}\times \dfrac{n-3}{n-2}\times \cdots\times \dfrac 12\times \dfrac{\pi}{2}\qquad n is an even number\\ \qquad \\ \dfrac{n-1}{n}\times \dfrac{n-3}{n-2}\times \cdots\times \dfrac 12\times 1\qquad n is an odd number\ end{cases}∫02psinnx=∫02pcosnx=⎩ ⎨ ⎧nn−1×n−2n−3×⋯×21×2pn is an even numbernn−1×n−2n−3×⋯×21×1n is an odd number
prove:
\qquad According to sin x \sin xsinx和 cos x \cos x cosx在[ 0 , π 2 ] [0,\dfrac{\pi}{2}][0,2p] on∫ 0 π 2 sin nxdx = ∫ 0 π 2 cos nxdx \int_0^{\frac{\pi}{2}}\sin^n xdx=\int_0^{\frac{\ pi}{2}}\cos^n xdx∫02psinnx d x=∫02pcosnx d x
\qquad 令 I n = ∫ 0 π 2 sin n x = ∫ 0 π 2 cos x d x I_n=\int_0^{\frac{\pi}{2}}\sin^nx=\int_0^{\frac{\pi}{2}}\cos xdx In=∫02psinnx=∫02pcosx d x , then
I n = ∫ 0 π 2 sin x ⋅ sin n − 1 xdx = ∫ 0 π 2 sin n − 1 xd ( − cos x ) = − cos x ⋅ sin n − 1 x ∣ 0 π 2 − ∫ 0 π 2 ( − cos x ) d ( sin n − 1 x ) 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\bigg\vert_0 ^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}(-\cos x)d(\sin^{n-1}x)In=∫02psinx⋅sinn−1x d x=∫02psinn−1xd(−cosx)=−cosx⋅sinn−1x 02p−∫02p(−cosx)d(sinn−1x)
= ( n − 1 ) ∫ 0 π 2 cos 2 x ⋅ sin n − 2 x d x = ( n − 1 ) ( ∫ 0 π 2 sin n − 2 x d x − ∫ 0 π 2 sin n x d x ) = ( n − 1 ) I n − 2 − ( n − 1 ) I n =(n-1)\int_0^{\frac{\pi}{2}}\cos^2x\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 =(n−1)∫02pcos2x⋅sinn−2x d x=(n−1)(∫02psinn−2x d x−∫02psinnx d x )=(n−1)In−2−(n−1)In
\qquad Add ( n − 1 ) I n (n-1)I_n to both sides of the equation(n−1)InObtained n I n = ( n − 1 ) I n − 2 nI_n=(n-1)I_{n-2}I _n=(n−1)In−2，即I n = n − 1 n I n − 2 I_n=\dfrac{n-1}{n}I_{n-2}In=nn−1In−2. It can be proved by mathematical induction.