To find the value of the following definite integral.

$$B(p,q)=\int_0^1t^{p-1}(1-t)^{q-1}dt\;(p>0,q>0)$$ I want to find the value of $$B(\frac1 2,\frac1 2)$$. And I tried to make $$t=\cos ^2\theta$$ so that the integral is equal to \begin{align*}&-\int_{-\infty}^{\infty}\left(\frac{\cos2\theta+1} 2\right)^{-\frac1 2}\left(\frac{1-\cos2\theta} 2\right)^{-\frac1 2}\sin 2\theta \;d\theta\\ &=-2\int_{-\infty}^{\infty}\frac{\sin 2\theta}{\sqrt{(1-\cos 2\theta)(1+\cos 2\theta)}}d\theta\\ &=-2\theta\bigg|_{-\infty}^{\infty}\\&=-\infty\end{align*} Since the negative infinity is not likely to be the correct result, could someone tell me where I went wrong?

• Doesn't $\theta$ vary between $0$ and $\pi/2$ Jul 10, 2020 at 11:03
• So the interval should go from $\pi/2$ to $0$ right? Then the result is $\pi$. It makes sense. Sorry for the stupid mistake. Jul 10, 2020 at 11:41

Recall that for $$x,y>0$$, the following holds:
$$$$\mathrm{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$$$
$$$$\mathrm{B}\left(\frac{1}{2},\frac{1}{2}\right)=\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(1)} =\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{0!} = \Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)$$$$
It is known that $$\Gamma(1/2)=\sqrt{\pi}$$, therefore $$\mathrm{B}\left(\frac{1}{2},\frac{1}{2}\right) = \pi$$