Evaluate $\int_0^{\pi/2}\int_0^{\pi/2} \sin x \sin^{-1}(\sin x \sin y) \ dx \ dy$ 
Evaluate $$\int_0^{\pi/2}\int_0^{\pi/2} \sin x \sin^{-1}(\sin x \sin y) \ dx \ dy$$

My attempt. Taking $\sin x \sin y=t$, $\iint\sin x  \ dx \ dy$.
I am not sure but I think the limits would change to $x=0$ to $\pi/2$ and $t=0$ to $1$.
I do not know how to change the elemental area i.e $dx\, dy$. Can I apply Jacobian where I am only changing one variable (i.e $y$ to $t$)?
Is this the correct way. Is there any easier way to go about it?
 A: Here is an approach using power series expansions.
It is known that for $t\in (-1,1)$
$$\frac{1}{\sqrt{1-t}} =(1-t)^{-1/2}=\sum_{n=0}^\infty 
\binom{-1/2}{n}  (-t)^{n}=
\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n}  t^{n},$$
and for $t\in [-1,1]$,
$$\arcsin(t) =\int_0^t\frac{ds}{\sqrt{1-s^2}}= \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{ t^{2n+1}}{2n+1}.$$
Hence, the given integral is equal to
\begin{align*}I&=\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{2n+1}\int_0^{\pi/2}\int_0^{\pi/2} \sin x (\sin x \sin y)^{2n+1} \ dx \ dy\\
&=\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{2n+1}W_{2n+2} W_{2n+1} \\
&=\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{2n+1}\cdot \frac{\pi}{4(n+1)}\\
&=\frac{\pi}{4}\left(2\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{2n+1}-\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{n+1}\right)\\
&=\frac{\pi}{4}\left(2\arcsin(1)-\int_0^1\frac{dt}{\sqrt{1-t}}\right)\\
&=\frac{\pi(\pi-2)}{4}.
\end{align*}
where $W_r=\int_{0}^{\pi/2}\sin^r(t)dt $ denotes the Wallis integral of order $r$.
A: I understand I am a bit late to the party, but there is a more rudimentary method to solve this.
$$I = \int_0^{\pi/2}\int_0^{\pi/2} \sin x \sin^{-1}(\sin x \sin y) dx dy$$
Substituting $\sin x\sin y=\sin\theta$, we get
$$I = \int_0^{\pi/2}\int_0^{x} \sin x \frac{\theta \cos \theta}{\sin x\sqrt{1-\frac{\sin^2\theta}{\sin^2x}}} d\theta dx$$
We change order of integration to get
$$I = \int_0^{\pi/2}\int_{\theta}^{\pi /2} \frac{\theta \cos \theta}{\sqrt{1-\frac{\sin^2\theta}{\sin^2x}}} dx d\theta = \int_0^{\pi/2}\int_{\theta}^{\pi /2} \frac{\theta \cos \theta \sin x}{\sqrt{\cos^2 \theta- \cos^2 x}} dx d\theta$$
This is now integrable:
$$I = \int_0^{\pi/2} \theta \cos \theta [-\sin^{-1}(\frac{\cos x}{\cos \theta})]_\theta ^{\pi /2} d\theta = \frac \pi 2 \int_0^{\pi/2}\theta \cos \theta d\theta = \frac \pi 2 (\frac \pi 2 -1)$$
A: First consider the following identity: $$\int_0^{\pi/2} \sin^n(x)\,\mathrm{d}x = \frac{\sqrt{\pi}\operatorname{\Gamma}\left(\frac{n+1}{2}\right)}{2\operatorname{\Gamma}\left(\frac{n}{2}+1\right)}$$ Now consider the Taylor expansion of $a\sin^{-1}(ab)$ around $b = 0$: $$a\sin^{-1}(ab) = \sum_{k=0}^{\infty} \frac{(1/2)_k^2a^{2k+2}b^{2k+1}}{(3/2)_kk!}$$ where $(x)_k$ is the rising factorial (this is used in the theory of hypergeometric functions). Using the above identity, $$\int_0^{\pi/2}\int_0^{\pi/2} \sin^{2k+2}(x)\sin^{2k+1}(y)\,\mathrm{d}x\,\mathrm{d}y = \frac{\sqrt{\pi}\operatorname{\Gamma}\left(k+\frac{3}{2}\right)}{2\operatorname{\Gamma}(k+2)}\cdot \frac{\sqrt{\pi}\operatorname{\Gamma}(k+1)}{2\operatorname{\Gamma}\left(k+\frac{3}{2}\right)} = \frac{\pi}{4(k+1)}$$ Therefore, by dominated convergence, \begin{align*} \int_0^{\pi/2}\int_0^{\pi/2} \sin(x)\sin^{-1}(\sin(x)\sin(y))\,\mathrm{d}x\,\mathrm{d}y &= \int_0^{\pi/2}\int_0^{\pi/2} \sum_{k=0}^{\infty} \frac{(1/2)_k^2\sin^{2k+2}(x)\sin^{2k+1}(y)}{(3/2)_kk!} \\ &= \frac{\pi}{4}\sum_{k=0}^{\infty} \frac{(1/2)_k^2}{(3/2)_k(k+1)!} \end{align*} Now, we note that $$\frac{(1/2)_k^2}{(3/2)_k(k+1)!} = \frac{2(1/2)_k^2}{(3/2)_kk!}-\frac{2k+1}{(k+1)!}\frac{(1/2)_k^2}{(3/2)_k} = \frac{2(1/2)_k^2}{(3/2)_kk!}-\frac{(1/2)_k}{(k+1)!}$$ We know that $$\sum_{k=0}^{\infty} \frac{2(1/2)_k^2x^{2k+1}}{(3/2)_kk!} = 2\sin^{-1}(x)$$ so $$\sum_{k=0}^{\infty} \frac{2(1/2)_k^2}{(3/2)_kk!} = 2\sin^{-1}(1) = \pi$$ and we can figure out that $$\sum_{k=0}^{\infty} \frac{(1/2)_kx^k}{(k+1)!} = \frac{2}{1+\sqrt{1-x}}$$ by checking this, so $$\sum_{k=0}^{\infty} \frac{(1/2)_k}{(k+1)!} = 2$$ Therefore, the value of the integral is $$\int_0^{\pi/2}\int_0^{\pi/2} \sin(x)\sin^{-1}(\sin(x)\sin(y))\,\mathrm{d}x\,\mathrm{d}y = \frac{\pi(\pi-2)}{4}$$
A: Robert Z' approach of expanding $\arcsin z$ as a Taylor series, then exploiting $\int_{0}^{\pi/2}\sin(x)^{2k}\,dx\int_{0}^{\pi/2}\sin(x)^{2k+1}\,dx=\frac{\pi}{4k+2}$ is very slick, but there is an interesting alternative approach. The original integral clearly equals
$$ \mathcal{I}=\int_{0}^{1}\int_{0}^{1}\frac{x\arcsin(xy)}{\sqrt{(1-x^2)(1-y^2)}}\,dx\,dy \tag{1}$$
and
$$ \int_{0}^{1}\frac{\arcsin(xy)}{\sqrt{1-y^2}}\,dy = \frac{4}{\pi}\sum_{n\geq 0}\frac{1}{(2n+1)^2}\int_{0}^{1}\frac{T_{2n-1}(xy)}{\sqrt{1-y^2}}\,dy\tag{2}$$
by the Fourier-Chebyshev series for the arcsin function (page 33 here). This leads to:
$$\begin{eqnarray*} \mathcal{I}&=& \frac{4}{\pi}\sum_{n\geq 1}\frac{1}{(2n-1)^2}\int_{0}^{1}\frac{x}{\sqrt{1-x^2}}\int_{0}^{x}\frac{T_{2n-1}(y)}{\sqrt{x^2-y^2}}\,dy\,dx\\&=&\frac{4}{\pi}\sum_{n\geq 1}\frac{1}{(2n-1)^2}\int_{0}^{1}T_{2n-1}(y)\int_{y}^{1}\frac{x}{\sqrt{(1-x^2)(x^2-y^2)}}\,dx\,dy\\ 
&=&\frac{2}{\color{blue}{\pi}}\sum_{n\geq 1}\frac{1}{(2n-1)^2}\int_{0}^{1}T_{2n-1}(y)\color{blue}{\int_{y^2}^{1}\frac{1}{\sqrt{(1-x)(x-y^2)}}\,dx}\,dy\\
&=&2\sum_{n\geq 1}\frac{1}{(2n-1)^2} \int_{0}^{1}T_{2n-1}(y)\,dy
\\&=&2\Big[\sum_{\substack{n\geq 1 \\ n\text{ odd}}}\frac{1}{(2n)(2n-1)^2}-\sum_{\substack{n\geq 1 \\ n\text{ even}}}\frac{1}{(2n-2)(2n-1)^2}\Big]\tag{3}\end{eqnarray*} $$
by the wonderful cancellation of the blue terms. The last series are straightforward to compute in terms of $\frac{\pi}{4},\zeta(2),\log(2)$ and Catalan's constant by partial fraction decomposition.
A: Perform integration by parts,
$\begin{align}J&=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arcsin(\sin x \sin y) \ dx \ dy\\
&=\int_0^{\frac{\pi}{2}}\left(\Big[-\cos x\arcsin(\sin x \sin y)\Big]_0^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}}\frac{\cos^2x\sin y}{\sqrt{1-\sin^2 x\sin^2 y }}\ dx\right)\ dy\\
&=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\frac{\cos^2x\sin y}{\sqrt{1-\sin^2 x\sin^2 y }}\ dx \ dy\\
&=\int_0^{\frac{\pi}{2}}\left(\Big[-\frac{\cos x\text { arcsinh} \left(\tan x\cos y\right)}{\tan x}\Big]_{y=0}^{y=\frac{\pi}{2}}\right)\ dx\\
&=\int_0^{\frac{\pi}{2}} \frac{\cos x\text{ arcsinh}(\tan x)}{\tan x}\ dx
\end{align}$
Perform the change of variable $u=\tan x$,
$\begin{align}J&=\int_0^{\infty}\frac{\text{ arcsinh } x}{x\left(1+x^2\right)^{\frac{3}{2}}}\ dx
\end{align}$
Perform integration by parts,
$\begin{align}J&=\left[\left(\frac{1}{\sqrt{1+x^2}}-\text{arcsinh}\left(\frac{1}{x}\right)\right)\text{arcsinh }x\right]_0^{\infty}-\int_0^{\infty}\left(\frac{1}{\sqrt{1+x^2}}-\text{arcsinh}\left(\frac{1}{x}\right)\right)\frac{1}{\sqrt{1+x^2}}\ dx\\
&=\int_0^{\infty}\frac{\text{arcsinh}\left(\frac{1}{x}\right)}{\sqrt{1+x^2}}\ dx-\int_0^{\infty}\frac{1}{1+x^2}\ dx\\
&=\int_0^{\infty}\frac{\text{arcsinh}\left(\frac{1}{x}\right)}{\sqrt{1+x^2}}\ dx-\frac{\pi}{2}
\end{align}$
In the following integral perform the change of variable $u=\frac{1}{x}$,
$\begin{align}
K&=\int_0^{\infty}\frac{\text{arcsinh}\left(\frac{1}{x}\right)}{\sqrt{1+x^2}}\ dx\\
&=\int_0^{\infty}\frac{\text{arcsinh }x}{x\sqrt{1+x^2}}\ dx\\
\end{align}$
Perform the change of variable $u=\text{arcsinh }x$,
$\begin{align}
K&=\int_0^{\infty}\frac{x}{\sinh x}\ dx
\end{align}$
and according to Contour integral of $\int_{0}^{\infty}\frac{x}{\sinh x}\operatorname{dx}$ ,
$\displaystyle K=\frac{\pi^2}{4}$
Therefore,
$\boxed{\displaystyle J=\frac{\pi^2}{4}-\frac{\pi}{2}}$
Addenda:
1)Another way to compute $K$.
Performing integration by parts,
$\begin{align}K&=\left[x\ln\left(\tanh\left(\frac{x}{2}\right)\right)\right]_0^{\infty}-\int_0^{\infty}\ln\left(\tanh\left(\frac{x}{2}\right)\right)\ dx\\
&=-\int_0^{\infty}\ln\left(\tanh\left(\frac{x}{2}\right)\right)\ dx\\
\end{align}$
Perform the change of variable $y=\tanh\left(\frac{x}{2}\right)$,
$\begin{align}K&=2\int_0^1 \frac{\ln x}{x^2-1}\ dx\\
&=\int_0^1 \frac{\ln x}{x-1}\ dx-\int_0^1 \frac{\ln x}{x+1}\ dx
\end{align}$
$\begin{align}L&=\int_0^1 \frac{\ln x}{x-1}\ dx+\int_0^1 \frac{\ln x}{x+1}\ dx\\
&=\frac{1}{2}\int_0^1 \frac{2x\ln\left(x^2\right)}{x^2-1}\ dx
\end{align}$
Perform the change of variable $u=x^2$,
$\begin{align}L&=\frac{1}{2}\int_0^1 \frac{\ln x}{x-1}\ dx
\end{align}$
Therefore,
$\begin{align}\int_0^1 \frac{\ln x}{x+1}\ dx=-\frac{1}{2}\int_0^1 \frac{\ln x}{x-1}\ dx
\end{align}$
Therefore,
$\begin{align}K&=\frac{3}{2}\int_0^1 \frac{\ln x}{x-1}\ dx
\end{align}$
Knowing that,
$\begin{align}\int_0^1 \frac{\ln x}{x-1}\ dx=\dfrac{\pi^2}{6}\end{align}$
Therefore,
$\boxed{\displaystyle K=\dfrac{\pi^2}{4}}$
2)Playing around with lindep, intnum, functions of GP-PARI, i suspect that the following formulae are true:
$\begin{align}
\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arcsin(\cos x \cos y) \ dx \ dy&=\frac{1}{8}\pi^2-\ln 2\\
\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arccos(\sin x \sin y) \ dx \ dy&=\frac{1}{2}\pi\\
\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arccos(\cos x \cos y) \ dx \ dy&=\frac{1}{8}\pi^2+\ln 2\\
\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arctan(\sin x \sin y) \ dx \ dy&=\frac{1}{8}\pi^2-\frac{1}{4}\pi\ln 2\\
\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arctan^2(\sin x \sin y) \ dx \ dy&=\frac{1}{32}\pi^3+\frac{1}{8}\pi^2\ln 2-\frac{1}{2}\text{G}\pi\\
\end{align}$
$\text{G}$ is the Catalan constant.
