<s>[INST] 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.
 [/INST] 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: $$\lim_{a\to\infty}I(a)\sqrt{a}=(\pi/2)^{3/2}\implies\lim_{n\to\infty}I(a+2n)\sqrt{n}=\frac{\pi^{3/2}}4.$$
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