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.
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:
$$\int_0^1 dx \: (\arcsin{x})(\sin{k x}) = \frac{\pi}{2 k} [J_0(k)-\cos{k}] $$
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
$$ J_\alpha(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}. $$
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{1}\arcsin\pars{x}\sin\pars{{\pi \over 2}\,x}\,\dd x:\ {\large ?}}$
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.
$$\color{#00f}{\large\int_{0}^{1}\arcsin\pars{x}\sin\pars{{\pi \over 2}\,x}\,\dd x = {\rm J}_{0}\pars{\pi \over 2}} \approx 0.4720 $$