證明積分：$$\int_{-\pi/2}^{\pi/2} (\sin(x))^n dx = \frac{n-1}{n}\int_{-\pi/2}^{\pi/2} (\sin(x))^{n-2} dx$$

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證明：

\begin{align}\int_{-\pi/2}^{\pi/2} \sin^nx \, dx& = -\sin^{n-1}x \cos x\bigg|_{-\pi/2}^{\pi/2} +\int_{-\pi/2}^{\pi/2} \cos x \cdot (n-1)\sin^{n-2} x\cos 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}

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因此

$$n\int_{-\pi/2}^{\pi/2} \sin^n x\, dx = (n-1)\int_{-\pi/2}^{\pi/2}\sin^{n-2}x\, dx,$$

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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
$$\int_{-\pi/2}^{\pi/2}\sin^{n}xdx=2\int_{0}^{\pi/2}\sin^{n}xdx$$
$$2\int_0^{\pi/2}\sin^{2m-1}x\cos^{2n-1}xdx=B(m,n)$$
Proof:
$$\int_0^{\pi/2}\sin^{2m-1}x\cos^{2n-1}xdx=\int_0^{\pi/2}\sin^{2m-2}x\cos^{2n-2}x\sin x\cos xdx$$
By substituting $t=\sin^2x$ and $dt=2\sin x\cos x$,
$$\int_0^{\pi/2}\sin^{2m-2}x\cos^{2n-2}\sin x\cos xxdx=\frac12\int_0^1 t^{m-1}(1-t)^{n-1}dt=\frac12B(m,n)$$
$$2\int_{0}^{\pi/2}\sin^{n}xdx=B(\frac{1+n}{2},\frac12)=\frac{\Gamma(\frac{1+n}{2})\Gamma(\frac12)}{\Gamma(1+\frac{n}2)}=\frac{\frac{n-1}{2}}{\frac{n}{2}}\frac{\Gamma(\frac{n-1}{2})\Gamma(\frac12)}{\Gamma(\frac{n}2)}$$
$$\therefore\int_{-\pi/2}^{\pi/2}\sin^{n}xdx=\frac{n-1}{n}\int_{-\pi/2}^{\pi/2}\sin^{n-2}xdx$$

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轉(zhuǎn)載于:https://www.cnblogs.com/ymshuibingcheng/p/4298542.html

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