$\int_{-1}^2 (1+x)^{p-1}(1-x)^{q-1} dx$ = $2^{p+q-1}\beta{(p,q)}.$

Question

Prove that:

$\int_{-1}^2 (1+x)^{p-1}(1-x)^{q-1} dx$ = $2^{p+q-1}\beta{(p,q)}.$

I tried converting the integral into the standard forms of beta function:

$\beta{(p,q)}=\int_{0}^1 (x)^{p-1}(1-x)^{q-1} dx$

$\beta(p, q) = \int_{0}^1 \frac{x^{p-1} + x^{q-1}}{(1+x)^{p+q}}dx$

using various substitutions like ${(1+x)}=t$, ${(1-x)}=t$, $(1-x)/(1+x)=t$ but they all failed eventually .

Answer 1

Hint

Focus on the antiderivative first and change variable $x=2y-1$.

This gives $$\int (1+x)^{p-1}(1-x)^{q-1}\, dx=2^{p+q-1}\int y^{p-1} (1-y)^{q-1}\,dy$$