How to calculate $\int_{0}^{1}(\arcsin{x})(\sin{\frac{\pi}{2}x})dx$? How to find the follwing integral's value ?
$$\int_{0}^{1}(\arcsin{x})(\sin{\frac{\pi}{2}x})dx$$
Actually, I don't know it can be represented as closed form.
 A: Here is another approach based on power series and beta function 
$$ \int_{0}^{1}(\arcsin{x})(\sin{\frac{\pi}{2}x})dx = \frac{2}{\pi}\,\int_{0}^{1}\!{\frac{\cos \left( \frac{\pi x }{2} \right) }{\sqrt {1-
{x}^{2}}}}{dx}$$
$$ = \frac{2}{\pi} \sum_{k=0}^{\infty}\frac{(-1)^k\left(\frac{\pi}{2}\right)^{2k}}{(2k)!}\int_{0}^{1}\frac{x^{2k}}{\sqrt{1-x^2}}dx $$
$$ = \frac{2}{\pi} \sum_{k=0}^{\infty}\frac{(-1)^k\left(\frac{\pi}{2}\right)^{2k} }{(2k)!}\frac{1}{2}\beta\left(k+\frac{1}{2},\frac{1}{2}\right)$$ 
$$ = \frac{1}{\pi} \sum_{k=0}^{\infty}\frac{(-1)^k\left(\frac{\pi}{2}\right)^{2k} }{(2k)!}{\frac {\Gamma\left(\frac{1}{2}\right) \Gamma\left( k+\frac{1}{2} \right) }{\Gamma \left( k+1 \right) }}$$
$$ = \frac{1}{\sqrt{\pi}} \sum_{k=0}^{\infty}\frac{(-1)^k \left(\frac{\pi}{2}\right)^{2k} }{(2k)!}{\frac { \Gamma\left( k+\frac{1}{2} \right) }{\Gamma \left( k+1 \right) }}$$
$$ = \frac{1}{\sqrt{\pi}} \sum_{k=0}^{\infty}\frac{(-1)^k \left(\frac{\pi}{2}\right)^{2k} }{{\frac {{2}^{2k}\Gamma  \left( k +1\right) \Gamma 
\left( k+1/2 \right)}{\sqrt{\pi}}}}{\frac{\Gamma\left(k+\frac{1}{2}\right)}{\Gamma\left(k+1\right)}} $$
$$ = {{J_{0}}\left(\frac{\pi}{2} \right)}, $$
where 

$$ (2k)! = \frac {{2}^{2k}\Gamma  \left(k+1\right) \Gamma\left( k+1/2 \right) }{\sqrt {\pi }} ,$$ 

and $J_{\alpha}(x)$ is the Bessel function

$$ J_\alpha(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}.  $$

A: Integrate by parts to get
$$\begin{align}\int_{0}^{1} dx \:(\arcsin{x})(\sin{\frac{\pi}{2}x}) = \underbrace{-\frac{2}{\pi} \left [ \cos{\left ( \frac{\pi}{2} x\right)} \arcsin{x} \right ]_0^1}_{\text{this}=0} + \frac{2}{\pi} \int_0^1 \frac{dx}{\sqrt{1-x^2}} \cos{\left ( \frac{\pi}{2} x\right)}\end{align}$$
Now use the Fourier transform relationship:
$$\int_{-1}^1 dx \: \frac{e^{i k x}}{\sqrt{1-x^2}} = \pi J_0(k)$$
where $J_0$ is the Bessel function of the first kind.  The integral is then
$$\int_{0}^{1} dx \:(\arcsin{x})(\sin{\frac{\pi}{2}x}) = J_0{\left(\frac{\pi}{2}\right)}$$
EDIT
In case some of you want to see why that FT relation holds, plug the integral into the differential expression defining the Bessel function of zero order:
$$k y''+y'+k y=0$$
$$y(0)=1$$
We then get
$$k y''+y'+k y=k \int_{-1}^1 dx \; \sqrt{1-x^2} e^{i k x} + i \int_{-1}^1 dx \; \frac{x}{\sqrt{1-x^2}} e^{i k x}$$
Integrate the second integral by parts and the above expression is zero.  Evaluating the integral
$$\frac{1}{\pi} \int_{-1}^1 dx \: \frac{1}{\sqrt{1-x^2}} = 1$$
verifies that the integral is in fact the Bessel function as stated.
BONUS
It turns out that the factor of $\pi/2$ - normally crucial in order to evaluate an integral like this - is nothing special at all.  Using the same technique I summarized above, I get the following, more general result:
$$\int_0^1 dx \: (\arcsin{x})(\sin{k x}) = \frac{\pi}{2 k} [J_0(k)-\cos{k}] $$
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$\ds{\int_{0}^{1}\arcsin\pars{x}\sin\pars{{\pi \over 2}\,x}\,\dd x:\ {\large ?}}$

With $\ds{x \equiv \sin\pars{\theta}\quad\imp\quad\theta = \arcsin\pars{x}}$
  \begin{align}
&\color{#c00000}{\int_{0}^{1}\arcsin\pars{x}\sin\pars{{\pi \over 2}\,x}\,\dd x}
=\half\int_{-\pi/2}^{\pi/2}
\theta\sin\pars{{\pi \over 2}\,\sin\pars{\theta}}\cos\pars{\theta}\,\dd\theta
\\[3mm]&=\half\int_{0}^{\pi}\pars{\theta - {\pi \over 2}}
\bracks{-\sin\pars{{\pi \over 2}\,\cos\pars{\theta}}}\sin\pars{\theta}\,\dd\theta
\\[3mm]&=-\,{1 \over \pi}\int_{\theta\ =\ 0}^{\theta\ =\ \pi}
\pars{\theta - {\pi \over 2}}\dd\bracks{\cos\pars{{\pi \over 2}\,\cos\pars{\theta}}}
\\[3mm]&=\underbrace{\left.-\,{1 \over \pi}
\pars{\theta - {\pi \over 2}}\cos\pars{{\pi \over 2}\,\cos\pars{\theta}}
\right\vert_{0}^{\pi}}_{\ds{=\ 0}}\
+\
\underbrace{%
{1 \over \pi}\int_{0}^{\pi}\cos\pars{{\pi \over 2}\,\cos\pars{\theta}}\,\dd\theta}
_{\ds{=\ {\rm J}_{0}\pars{\pi \over 2}}}
\end{align}
  where $\ds{{\rm J}_{\nu}\pars{z}}$ is a
  First Kind Bessel Function. See ${\bf 9.1.18}$ in
  this link.

$$\color{#00f}{\large\int_{0}^{1}\arcsin\pars{x}\sin\pars{{\pi \over 2}\,x}\,\dd x = {\rm J}_{0}\pars{\pi \over 2}} \approx 0.4720
$$
