Here, I have a little unpleasant way to calculate the following double integral
$$\iint_{D}{\sin(x)e^{\sin(x)\sin(y)}}\text{d}A$$
where $D$ is the square area $D=\{(x,y)\in\mathbb{R}^2: 0 \le x \le \pi/2, 0 \le y \le \pi/2\}$
my attempt:
with the symmetry of the area we know
$$I=\iint_{D}{\sin(x)e^{\sin(x)\sin(y)}}\text{d}A=\iint_{D}{\sin(y)e^{\sin(x)\sin(y)}}\text{d}A$$
thus
$$I=\frac1{2}\iint_{D}{(\sin(x)+\sin(y))e^{\sin(x)\sin(y)}}\text{d}A$$
with the substitution $u=\sin(x)$ and $v=\sin(y)$, we have
$$I=\frac1{2}\iint_{D^{*}}{\frac{u+v}{\sqrt{(1-u^2)(1-v^2)}}e^{uv}}\text{d}A$$
and now the integral area is $D^{*}=\{(u,v)\in\mathbb{R}^2: 0 \le u \le 1, 0 \le v \le 1\}$
notice that the function $\frac{u+v}{\sqrt{(1-u^2)(1-v^2)}}e^{uv}$ also holds a symmetric form, thus
$$I=\iint_{D^{*}\cap(u\le v)}{\frac{u+v}{\sqrt{(1-u^2)(1-v^2)}}e^{uv}}\text{d}A=\iint_{D^{*}\cap(u\ge v)}{\frac{u+v}{\sqrt{(1-u^2)(1-v^2)}}e^{uv}}\text{d}A$$
so I can only foucs on area $D^{*}\cap(u\ge v)$, set another substitution
$$u+v=\alpha$$
$$uv=\beta$$
under the condition $u\ge v$, the determinant of Jacobian matrix writes
$$|J(\alpha,\beta)|=\frac1{\sqrt{\alpha^2-4\beta}}$$
and
$$\frac1{\sqrt{(1-u^2)(1-v^2)}}=\frac1{\sqrt{(1+\beta)^2-\alpha^2}}$$
the integral area change to $\{(\alpha,\beta)\in\mathbb{R}^2: 2\sqrt{\beta} \le \alpha \le 1+\beta, 0 \le \beta \le 1\}$, so
$$\begin{align} I&=\int_{0}^{1}\int_{2\sqrt{\beta}}^{1+\beta}{\frac{\alpha}{\sqrt{((1+\beta)^2-\alpha^2)(\alpha^2-4\beta)}}e^{\beta}}\text{d}\alpha \text{d}\beta \\&=\int_{0}^{1}\left(-\tan^{-1}\sqrt{\frac{(1+\beta)^2-\alpha^2}{\alpha^2-4\beta}}\right)\biggr|_{\alpha=2\sqrt{\beta}}^{1+\beta} e^{\beta}\text{d}\beta \\&=\frac{\pi}{2}(e-1) \end{align}$$
obviously, I choose an inconvenient way in the second half of this calculation, but I still can not come up with another good method.
thanks in advance for any suggestion!
Using the Taylor expansion of the exponential $$ \iint_{D}{\sin(x)e^{\sin(x)\sin(y)}}\text{d}A=\sum_{k\geq 0}\frac{1}{k!}\int_0^{\pi/2}\mathrm{d}x~(\sin x)^{k+1}\int_0^{\pi/2}\mathrm{d}y~(\sin y)^{k}=\sum_{k\geq 0}\frac{1}{k!}I(k)I(k+1)\ , $$ where $$ I(k)=\int_0^{\pi/2}\mathrm{d}x~(\sin x)^{k}=\int_0^1\mathrm{du}~u^k(1-u^2)^{-1/2}=\frac{\sqrt{\pi}~\Gamma \left(\frac{k+1}{2}\right)}{2 \Gamma \left(\frac{k}{2}+1\right)}\ . $$ Therefore $$ \frac{1}{k!}I(k)I(k+1)=\frac{\left(\sqrt{\pi } \Gamma \left(\frac{k+1}{2}\right)\right) \left(\sqrt{\pi } \Gamma \left(\frac{k}{2}+1\right)\right)}{k! \left(2 \Gamma \left(\frac{k}{2}+1\right)\right) \left(2 \Gamma \left(\frac{k+3}{2}\right)\right)}=\frac{\pi }{2 \Gamma (k+2)} $$ after simplifications, which then leads to $$ \sum_{k\geq 0}\frac{1}{k!}I(k)I(k+1)=\frac{1}{2} (e-1) \pi\ . $$