Evaluate $\int_{0}^{\pi/2}x\sin^a (x) dx$, $a>0$ 
I want to evaluate $$\int_{0}^{\pi/2}x\sin^a (x)\, dx$$ where $a>0$ is a real number.

I tried: $$I(a)= \int_{0}^{\pi/2}x\sin^a(x)\,dx = \int_{0}^{1}\frac{\arcsin x}{\sqrt{1-x^2}}x^a\,dx$$
$$ I(a)=\sum_{m\geq 1}\frac{4^m}{2m\left(2m+a\right)\binom{2m}{m}}$$
$$I(a)=\frac{1}{a+2}\cdot\phantom{}_3 F_2\left(1,1,1+\tfrac{a}{2};\tfrac{3}{2},2+\tfrac{a}{2};1\right)$$
Any other method please.
Any help will be appreciated. Thank you.
edit The series expansion of $ \arcsin(x)^2$ is
$$2\;\arcsin(x)^2=\sum_{n=1}^\infty \frac{(2x)^{2n}}{n^2\binom{2n}{n}}$$
Differentiating the above series we get the formula used in the question.
 A: Another idea leads to a different formula (still for $\Re a>-1$): $$\boxed{I(a)=2^{-a}\sum_{n=0}^\infty\binom{a}{n}\frac1{(2n-a)^2}-\frac{\pi^{3/2}}{4}\frac{\Gamma\left(\frac{a+1}2\right)}{\Gamma\left(\frac{a+2}2\right)}\cot^2\frac{a\pi}2}$$ (interpreted as a limit if $a$ is an even integer). It is based on the fact that $$I(a)=\left.\frac{\partial}{\partial b}\int_0^{\pi/2}\sin^a x\sin bx\,dx\right|_{b=0},$$ and the approach I've taken here.
This approach might also be followed directly. Assume (temporarily) $\color{blue}{-1<\Re a<0}$. Consider $f(z)=z^{-a-1}(1-z^2)^a\log z$ (principal values everywhere) and the contours $\gamma_+$, $\gamma_-$ from the linked answer. Then, manipulating $0=\int_{\gamma_\pm}f(z)\,dz$ the same way, we obtain
\begin{align}
2^a I(a)&=J_-(a)\cos\frac{a\pi}2-J_+(a),
\\J_\pm(a)&=\int_0^1 x^{-a-1}(1\pm x^2)^a\log x\,dx.
\end{align}
Now $J_-(a)$ reduces to $$\int_0^1 t^{\lambda-1}(1-t)^{\mu-1}\log t\,dt=\frac{\partial}{\partial\lambda}\mathrm{B}(\lambda,\mu)=\mathrm{B}(\lambda,\mu)\big(\psi(\lambda)-\psi(\lambda+\mu)\big)$$ at $\mu=a+1$ and $\lambda=-a/2$; with the reflection formulae for $\Gamma$ and $\psi$, this yields the gamma quotient term in the formula stated at the beginning. And, using the binomial expansion, $$J_+(a)=\sum_{n=0}^\infty\binom{a}{n}\int_0^1 x^{2n-a-1}\log x\,dx=-\sum_{n=0}^\infty\binom{a}{n}\frac1{(2n-a)^2}$$ gives the remaining piece. Finally, the result holds for $\Re a>-1$ by analytic continuation.
A: An elementary (enough) deduction of the (already mentioned) equality $$\boxed{I(a)=\frac{\pi^{3/2}}{4}\frac{\Gamma\left(\frac{a+1}2\right)}{\Gamma\left(\frac{a+2}2\right)}-\sum_{n=1}^\infty\frac{\prod_{k=1}^{n-1}(a+2k)^2}{\prod_{k=1}^{2n}(a+k)}.\quad(\Re a>-1)}$$
We note that $$(a+1)\big(I(a)-I(a+2)\big)=\int_0^{\pi/2}x\cos x\,(\sin^{a+1}x)'\,dx$$ and integrate by parts; this yields the recurrence $$I(a)=\frac{a+2}{a+1}I(a+2)-\frac1{(a+1)(a+2)}.$$
Reusing it with $a+2$ in place of $a$, we get $$I(a)=\frac{(a+2)(a+4)}{(a+1)(a+3)}I(a+4)-\frac1{(a+1)(a+2)}-\frac{a+2}{(a+1)(a+3)(a+4)}$$ and then, by induction (I'm writing it expanded intentionally),
\begin{align}
I(a)&=\frac{(a+2)(a+4)\cdots(a+2n)}{(a+1)(a+3)\cdots(a+2n-1)}I(a+2n)
\\&-\frac{1}{(a+1)(a+2)}-\frac{a+2}{(a+1)(a+3)(a+4)}
\\&-\dots-\frac{(a+2)(a+4)\cdots(a+2n-2)}{(a+1)(a+3)\cdots(a+2n-1)(a+2n)}.
\end{align}
And now we take $n\to\infty$. The fact that $$\lim_{n\to\infty}\frac{(a+2)(a+4)\cdots(a+2n)}{(a+1)(a+3)\cdots(a+2n-1)}\frac1{\sqrt n}=\frac{\Gamma\left(\frac{a+1}2\right)}{\Gamma\left(\frac{a+2}2\right)}$$ may be obtained using infinite product representations of $\Gamma$, or just the known limit $$\lim_{x\to\infty}\frac{\Gamma(x+a)}{x^a\,\Gamma(x)}=1.$$
And Laplace's method gives the remaining piece: $$\lim_{a\to\infty}I(a)\sqrt{a}=(\pi/2)^{3/2}\implies\lim_{n\to\infty}I(a+2n)\sqrt{n}=\frac{\pi^{3/2}}4.$$
A: I prefer to add a separate answer.
In this post, @Aaron Hendrickson gave the complete expansion
$$\left(\frac{\sin (x)}{x}\right)^a=\sum_{m=0}^\infty c_m x^{2m}$$ where $c_0=1$ and
$$c_m=\frac{1}{m}\sum_{k=1}^m (-1)^k\,\frac{k(a+1)-m}{(2k+1)!}\,c_{m-k}$$
$$\int_{0}^{\frac \pi 2} x\sin^a (x)\, dx=\sum_{m=0}^\infty c_m \int_{0}^{\frac \pi 2} x^{2 m+a+1}= \sum_{m=0}^\infty \frac {c_m}{a+2 (m+1) } \left(\frac{\pi}{2}\right)^{a+2 (m+1)}$$
