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Show that $\displaystyle \int_{0}^{\frac{\pi }{2}}\cos^{n} \theta d\theta=\int_{0}^{\frac{\pi }{2}}\sin^{n} \theta d\theta=\frac{\sqrt{\pi}[\frac{(n-1)}{2}]!}{2(\frac{n}{2})!}$

Amzoti
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tweelly
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1 Answers1

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First substitute $u=\cos{\theta}$ in the first integral to get

$$\int_0^1 du \, (1-u^2)^{-1/2} u^n$$

Now sub $u=v^2$ to get

$$\frac12 \int_0^1 dv \, (1-v)^{-1/2} v^{(n-1)/2}$$

Use the definition of a beta function to get

$$\frac{\Gamma{\left(\frac12\right)}\Gamma{\left(\frac{n+1}{2}\right)}}{2\Gamma{\left(\frac{n}{2}+1\right)}}$$

It should be clear that the same substitution holds for the second integral.

Ron Gordon
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  • thanks, its a useful answer, i found another way using $B(m+1,n+1)=2\int_{0}^{\frac{\pi}{2}}\cos^{2m+1}\theta \sin^{2n+1}\theta d\theta = \frac{n!m!} {(m+n+1)!}$ with $n=\frac{n-1}{2}$ and $m=-\frac{1}{2}$ so, we get $\int_{0}^{\frac{\pi}{2}}\sin^n\theta d\theta=\frac{(-\frac{1}{2})!(\frac{n-1}{2})!}{(\frac{n}{2})!}$
    finally $\int_{0}^{\frac{\pi}{2}}\sin^n\theta d\theta=\frac{\sqrt{\pi}(\frac{n-1}{2})!}{2(\frac{n}{2})!}$     – tweelly Jun 03 '13 at 21:06