Using the beta function

Question

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

Answer 1

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.