证明积分：$$int_{-pi/2}^{pi/2} (sin(x))^n dx = frac{n-1}{n}int_{-pi/2}^{pi/2} (sin(x))^{n-2} dx$$
证明：
egin{align}int_{-pi/2}^{pi/2} sin^nx \, dx& = -sin^{n-1}x cos xigg|_{-pi/2}^{pi/2} +int_{-pi/2}^{pi/2} cos x cdot (n-1)sin^{n-2} xcos x\, dx\ &= (n-1)int_{-pi/2}^{pi/2}sin^{n-2} x cos^2 x\, dx\ &= (n-1)int_{-pi/2}^{pi/2}(sin^{n-2}x - sin^n x)\, dx\ &= (n-1)int_{-pi/2}^{pi/2} sin^{n-2}x - (n-1)int_{-pi/2}^{pi/2} sin^n x\, dx end{align}
因此
$$nint_{-pi/2}^{pi/2} sin^n x\, dx = (n-1)int_{-pi/2}^{pi/2}sin^{n-2}x\, dx,$$
If n is odd,then
$$int_{-pi/2}^{pi/2}sin^{n}xdx=int_{-pi/2}^{pi/2}sin^{n-2}xdx=0$$
If n is even,then
$$int_{-pi/2}^{pi/2}sin^{n}xdx=2int_{0}^{pi/2}sin^{n}xdx$$
$$2int_0^{pi/2}sin^{2m-1}xcos^{2n-1}xdx=B(m,n)$$
Proof:
$$int_0^{pi/2}sin^{2m-1}xcos^{2n-1}xdx=int_0^{pi/2}sin^{2m-2}xcos^{2n-2}xsin xcos xdx$$
By substituting $t=sin^2x$ and $dt=2sin xcos x$,
$$int_0^{pi/2}sin^{2m-2}xcos^{2n-2}sin xcos xxdx=frac12int_0^1 t^{m-1}(1-t)^{n-1}dt=frac12B(m,n)$$
$$2int_{0}^{pi/2}sin^{n}xdx=B(frac{1+n}{2},frac12)=frac{Gamma(frac{1+n}{2})Gamma(frac12)}{Gamma(1+frac{n}2)}=frac{frac{n-1}{2}}{frac{n}{2}}frac{Gamma(frac{n-1}{2})Gamma(frac12)}{Gamma(frac{n}2)}$$
$$ hereforeint_{-pi/2}^{pi/2}sin^{n}xdx=frac{n-1}{n}int_{-pi/2}^{pi/2}sin^{n-2}xdx$$