Solving an integral using beta function

Question

Using the fact that $$ \int_{0}^{1}dx \ x^{\alpha-1}(1-x)^{\beta-1}=\frac{\Gamma(\alpha)\Gamma( \beta)}{\Gamma(\alpha+\beta)} $$ where $\Gamma(x)$ is the Euler gamma function. I want to show that $$ \int_0^{\pi}d\theta \ \sin^k(\theta)=\frac{\Gamma(\frac{1}{2})\Gamma(\frac{1}{2}(k+1))}{\Gamma(1+\frac{1}{2}k)}$$ but I can't seem to find the right substitution. A hint or the answer would be much appreciated.

Answer 1

Let $ k\in\mathbb{N} : $

$$ \int_{0}^{\pi}{\sin^{k}{\theta}\,\mathrm{d}\theta}=\int_{0}^{\frac{\pi}{2}}{\sin^{k}{\theta}\,\mathrm{d}\theta}+\int_{\frac{\pi}{2}}^{\pi}{\sin^{k}{\theta}\,\mathrm{d}\theta} $$

Then, making the substitution $ \left\lbrace\begin{aligned}\theta &=\pi-x \\ \mathrm{d}\theta &=-\,\mathrm{d}x \end{aligned}\right. $ in the second term, we get that : $$ \int_{0}^{\pi}{\sin^{k}{\theta}\,\mathrm{d}\theta}=2\int_{0}^{\frac{\pi}{2}}{\sin^{k}{\theta}\,\mathrm{d}\theta} $$

Using the substitution $ \left\lbrace\begin{aligned}u&=\sin^{2}{\theta}\\ \mathrm{d}\theta &=\frac{\mathrm{d}x}{2\sqrt{x}\sqrt{1-x}}\end{aligned}\right. $, we get the following : $$ \int_{0}^{\pi}{\sin^{k}{\theta}\,\mathrm{d}\theta}=\int_{0}^{1}{x^{\frac{k-1}{2}}\left(1-x\right)^{-\frac{1}{2}}\,\mathrm{d}x}=\beta\left(\frac{k+1}{2},\frac{1}{2}\right)=\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{k+1}{2}\right)}{\Gamma\left(\frac{k}{2}+1\right)} $$