Baby rudin 8.21

Things to remark before 8.21.

$\Gamma(x) = \int_0^{\infty} t^{x-1}e^{-t}dt. $

If $x>0$ and $y > 0$ , then $\int_0^1 t^{x-1}(1-t)^{y-1}dt $ = $\frac {\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ = $B(x,y)$. ( Mark this equality by $(a)$) The substitution $t$ = $\sin^2(\theta)$ turns $(a)$ into

$2\int_0^{\frac {\pi}{2} } (\sin\theta)^{2x-1} (\cos\theta)^{2y-1}d\theta$ = $B(x,y)$.

I don't understand how did we get that? $\frac {dt}{d\theta} $ = $2\sin\theta\cos\theta$

don't we should have $(\cos\theta)^{3y-1}$ instead of $(\cos\theta)^{2y-1}$ ?

I would be grateful for any kind of help.

Solution 1:

With the substitution $t=\sin^2\theta$, you have $$t^{x-1}=\left(\sin^2\theta\right)^{x-1}=\left(\sin\theta\right)^{2(x-1)}=\left(\sin\theta\right)^{2x-2}\\ \left(1-t\right)^{y-1}=\left(1-\sin^2\theta\right)^{y-1}=\left(\cos^2\theta\right)^{y-1}=\left(\cos\theta\right)^{2(y-1)}=\left(\cos\theta\right)^{2y-2}\\ dt=2\sin\theta\cos\theta\,d\theta$$ Now do the substitution in the integral for $B(x,y)$: $$\int_0^{\pi/2}\left(\sin\theta\right)^{2x-2}\left(\cos\theta\right)^{2y-2}\cdot2\sin\theta\cos\theta\,d\theta= 2\int_0^{\pi/2}\left(\sin\theta\right)^{2x-2+1}\left(\cos\theta\right)^{2y-2+1}\,d\theta\\ =2\int_0^{\pi/2}\left(\sin\theta\right)^{2x-1}\left(\cos\theta\right)^{2y-1}\,d\theta$$