TITLE: To evaluate integral using Beta function - Which substitution should i use?
QUESTION [6 upvotes]: $$\int_{0}^{1} \frac{x^{m-1}(1-x)^{n-1}}{(a+bx)^{m+n}}dx = \frac{B(m,n)}{(a+b)^ma^n}$$
I have to use some kind of substitution but i do not understand what i use and why ?
Thanks

REPLY [3 votes]: Let's try the substitution 
$$x=\frac{1-y}{1+cy}$$
so that $1-x=(1+c)\frac{y}{1+cy}$.
Then, when $x=0$, $y=1$ and when $x=1$, $y=0$.  We also have $dx=-(1+c)\frac{1}{(1+cy)^2}\,dy$.  Then, we can write
$$\begin{align}\int_0^1\frac{x^{m-1}(1-x)^{n-1}}{(a+bx)^{m+n}}\,dx&=\int_0^1 \frac{\left(\frac{1-y}{1+cy}\right)^{m-1}\left((1+c)\frac{y}{1+cy}\right)^{n-1}}{\left(a+b\frac{1-y}{1+cy}\right)^{m+n}}\,(1+c)\frac{1}{(1+cy)^2}\,dy\\\\
&=\frac{(1+c)^n}{(a+b)^{n+m}}\int_0^1\frac{y^{n-1}(1-y)^{m-1}}{\left(1-\frac{b-ac}{a+b}y\right)^{m+n}}
\end{align}$$
Choosing $c=b/a$, we obtain
$$\begin{align}
\int_0^1\frac{x^{m-1}(1-x)^{n-1}}{(a+bx)^{m+n}}\,dx&=\frac{1}{a^n(a+b)^m}\int_0^1 y^{n-1}(1-y)^{m-1}\,dy\\\\
&=\frac{1}{a^n(a+b)^m}B(m,n)
\end{align}$$
And we are done!