Integral $\int_{0}^{\pi}\frac{\sin^px}{1+\cos x} dx $

Using $1+\cos(x)=2\cos^2(x/2)$ we see that

$$\begin{align} \int_0^\pi \frac{\sin^p(x)}{1+\cos(x)}\,dx&=\int_0^\pi \frac{\sin^p(x)}{2\cos^2(x/2)}\,dx\\\\ &=\int_0^{\pi/2} \frac{\sin^p(2x)}{\cos^2(x)}\,dx\\\\ &=2^p\int_0^{\pi/2} \sin^p(x)\cos^{p-2}(x)\,dx\\\\ &=2^{p-1}\text{B}\left(\frac{p+1}{2},\frac{p-1}{2}\right)\\\\ &=2^{p-1}\text{B}\left(\frac{p-1}{2}+1,\frac{p-1}{2}\right)\tag1\\\\ &=\text{B}\left(\frac12,\frac{p-1}{2}\right)\tag2 \end{align}$$

where in going from $(1)$ to $(2)$ we used the identities $\text{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$, $\Gamma(x+1)=x\Gamma(x)$, and

$$\frac1{\Gamma(2x)}=2^{1-2x}\frac{\Gamma(1/2)}{\Gamma(x)\Gamma(x+1/2)}=2^{1-2x}\frac{\text{B}\left(\frac12,x\right)}{ \Gamma^2(x)}$$