Computing the values of $W_\alpha(n):=\int_0^\pi x^{\alpha-1}\sin(x)^{2n}$

Define the following integral as $$W_\alpha(n):=\int_0^\pi x^{\alpha-1}\sin(x)^{n}\,,\quad V_\alpha(n):=\int_0^\pi x^{\alpha-1}\cos(x)^{n}$$ where $n \in\mathbb{N}$. Now in the base case $W_1(n)=W(n)$ this integral simplifies into the well know Wallis integrals, and $V(n)=W(n)\,\forall\,n$. Which has been shown to be

$$\color{black}{W(n) = 2\int_0^{\pi/2}\sin(x)^n\mathrm{d}x = \left\{ \begin{array}{ll} \frac{\pi}{2^{2p}} \binom{2p}{p} & \text{if} \ \ n = 2p \\ \frac{2^{2p+1}}{p+1} \big/ \binom{2p+1}{n} & \text{if} \ \ n = 2p + 1 \end{array} \right.}$$

Through simmilar means one can also show that $V_2(2n)=W_2(2n)=\pi/2 W(2n)$. By using $$\color{black}{ \int_0^\pi x R(\sin x,\cos^2x)\,\mathrm{d}x=\frac{\pi}{2}\int_0^\pi R(\sin x,\cos^2x)\,\mathrm{d}x }$$

This gives directly that $$W_2(2n) = \int_0^{\pi} x \sin^{2n}(x)\,\mathrm{d}x = \frac{\pi}{2} \int_0^{\pi} \sin^{2n}(x)\,\mathrm{d}x = \pi \int_0^{\pi/2} \sin^{2n}(x)\,\mathrm{d}x = \frac{\pi}{2}W(2n)$$ as wanted, and exactly the same can be done for $V(n)$. My question is:

• Are there any more cases one can solve? When are they alike? Is there a general formula for $W_\alpha(n)$, $V_\alpha(n)$
• Note that $\sin^{2n}x$ and $\cos^{2n}x$ can be written as a finite sum of terms of type $\sin k x$, $\cos k x$ with $k=0,\ldots, 2n$. The remaining integrals can be expressed in terms of the hypergeometric function $_1F_2$. For example: $$\int_0^{\pi}x^{\alpha-1}\cos kx\,dx=\frac{\pi^{\alpha}}{\alpha}{}_1F_2\left[ \begin{array}{c} \alpha/2 \\ 1/2,1+\alpha/2 \end{array};-\frac{\pi^2 k^2}{4}\right]$$ So a general formula exist for both $W_{\alpha}(n)$, $V_{\alpha}(n)$. – Start wearing purple Apr 19 '14 at 12:14
• @N3buchadnezzar I am also started to look into these... I was doing this by using a recurrent relation. Did you figured it out already? What if you have something like [ \int_{0}^\pi x^{\alpha}cos^m(x)sin^n(x) dx ]? This is something I am trying to figure out... Let me know if you have already done it. – user209663 May 21 at 20:52
• @N3buchadnezzar Actually the formula turned out pretty nice after you convert everything into exponential!! – user209663 May 25 at 23:16