Finding $\int_{0}^{2\pi}\int_{0}^{\pi}\sin^3y \ e^{3\cos x\sin y+4\sin x\sin y}\,dy\,dx$ I am working on this double integral
$$\int_{0}^{2\pi}\int_{0}^{\pi}\sin^3y \ e^{3\cos x\sin y+4\sin x\sin y}\,dy\,dx$$
so far, I don't know how to start.
Can someone give a hint?
Thanks
 A: Write $3\cos x + 4\sin x = 5\cos(x+c)$ for some constant $c$. 
Then, change the order of integration to get $\displaystyle\int_{0}^{\pi}\int_{0}^{2\pi}\sin^3 y \ e^{5\sin y \cos(x+c)}\,dx\,dy$. 
Note that $\displaystyle\int_{0}^{2\pi}e^{\alpha \cos x}\,dx = 2\pi I_0(\alpha)$ where $I_0(\alpha)$ is the modified Bessel function of the first kind. 
This reduces the double integral to $2\pi\displaystyle\int_{0}^{\pi}\sin^3 y I_0(5\sin y)\,dy$. 
According to Wolfram Alpha this evaluates to roughly $156.651$. 
I don't think there is a closed form for the antiderivative of $\sin^3 y I_0(5\sin y)$.  
A: You can employ the following formula
$\displaystyle\bbox[#EFF,15px,border:3px solid blue]{\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\sin^3(y)f\big[\sin\left(x\right)\sin\left(y\right)\big]{\mathrm{d}}x{\mathrm{d}}y=\frac{\pi}{4}\int_{0}^{1}\left(t^{2}+1\right)f\left(t\right){\mathrm{d}}t}$
Then
\begin{align*} &\quad\quad \int_{0}^{2\pi}{\mathrm{d}}x\int_{0}^{\pi}e^{3\cos\left(x\right)\sin\left(y\right)+4\sin\left(x\right)\sin\left(y\right)}\,\,\sin^3\left(y\right){\mathrm{d}}y\\ &=\int_{0}^{2\pi}{\mathrm{d}}x\int_{0}^{\pi}e^{5\sin\left(x+\arctan\frac{3}{4}\right)\sin\left(y\right)}\,\sin^3\left(y\right){\mathrm{d}}y\\ &=\int_{0}^{2\pi}{\mathrm{d}}x\int_{0}^{\pi}e^{5\sin\left(x\right)\sin\left(y\right)}\,\sin^3\left(y\right){\mathrm{d}}y\\ &=2\int_{0}^{2\pi}{\mathrm{d}}x\int_{0}^{\frac{\pi}{2}}e^{5\sin\left(x\right)\sin\left(y\right)}\,\sin^3\left(y\right){\mathrm{d}}y\\ &=2\int_{0}^{\frac{\pi}{2}}\sin^3\left(y\right){\mathrm{d}}y\int_{0}^{2\pi}e^{5\sin\left(x\right)\sin\left(y\right)}\,{\mathrm{d}}x\\ &=2\int_{0}^{\frac{\pi}{2}}\sin^3\left(y\right){\mathrm{d}}y\left[\int_{0}^{\pi}e^{5\sin\left(x\right)\sin\left(y\right)}\,{\mathrm{d}}x+\int_{\pi}^{2\pi}e^{5\sin\left(x\right)\sin\left(y\right)}\,{\mathrm{d}}x\right]\\ &=2\int_{0}^{\frac{\pi}{2}}\sin^3\left(y\right){\mathrm{d}}y\left[\int_{0}^{\pi}e^{5\sin\left(x\right)\sin\left(y\right)}\,{\mathrm{d}}x+\int_{0}^{\pi}e^{-5\sin\left(x\right)\sin\left(y\right)}\,{\mathrm{d}}x\right]\\ &=2\int_{0}^{\frac{\pi}{2}}\sin^3\left(y\right){\mathrm{d}}y\int_{0}^{\pi}\left[e^{5\sin\left(x\right)\sin\left(y\right)}+e^{-5\sin\left(x\right)\sin\left(y\right)}\,\right]\!{\mathrm{d}}x\\ &=4\int_{0}^{\frac{\pi}{2}}\sin^3\left(y\right){\mathrm{d}}y\int_{0}^{\frac{\pi}{2}}\left[e^{5\sin\left(x\right)\sin\left(y\right)}+e^{-5\sin\left(x\right)\sin\left(y\right)}\,\right]\!{\mathrm{d}}x\\ &=4\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\left[e^{5\sin\left(x\right)\sin\left(y\right)}+e^{-5\sin\left(x\right)\sin\left(y\right)}\,\right]\sin^3\left(y\right){\mathrm{d}}x{\mathrm{d}}y\\ \end{align*}
Hence
\begin{align*} &\quad\quad \int_{0}^{2\pi}{\mathrm{d}}x\int_{0}^{\pi}e^{3\cos\left(x\right)\sin\left(y\right)+4\sin\left(x\right)\sin\left(y\right)}\,\,\sin^3\left(y\right){\mathrm{d}}y\\ &=4\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\left[e^{5\sin\left(x\right)\sin\left(y\right)}+e^{-5\sin\left(x\right)\sin\left(y\right)}\,\right]\sin^3\left(y\right){\mathrm{d}}x{\mathrm{d}}y\\ &=4\cdot\dfrac{\pi}{4}\int_{0}^{1}(t^2+1)\left(e^{5t}+e^{-5t}\right){\mathrm{d}}t\\ &=\frac{2\pi}{125}\left(21e^{5}-\frac{31}{e^{5}}\right) \end{align*}
