How to evaluate double integral with sines and exponentials As part of showing that $\displaystyle f(x,y) = \frac{1+\sin x \sin y}{2 \pi} e^{-\left( \frac{x^{2}+y^{2}}{2}\right)}$ is a probability density function, I need to show that $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y) dy dx = 1$.
Normally, when I see integrals like this, I convert to polar coordinates, so I have that $$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y) dy dx = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\frac{1+\sin x \sin y}{2 \pi} e^{-\left( \frac{x^{2}+y^{2}}{2}\right)} dydx \\ = \int_{0}^{2\pi} \int_{0}^{\infty}\frac{1+\sin(r\cos \theta)\sin(r \sin(\theta))}{2\pi} e^{-\left( \frac{r^{2}}{2}\right)}rdr d \theta \\=  \int_{0}^{2\pi} \int_{0}^{\infty}\frac{1}{2 \pi} e^{-\left( \frac{r^{2}}{2}\right)}rdr d \theta +  \int_{0}^{2\pi} \int_{0}^{\infty}\frac{\sin(r\cos \theta)\sin(r \sin(\theta))}{2\pi}e^{-\left( \frac{r^{2}}{2}\right)}rdr d \theta $$
Now, after this last equals sign, the first integral is easy, but I have absolutely no idea how to do the second integral.
Is this even the correct way to show that the integral of $f(x,y) = 1$, or is there an easier way? I'm extremely stuck and need help! Thank you!
 A: The value of the first double integral is one. Now we claim that the second double integral is zero, simply split the outer integral to $\displaystyle\int_{0}^{\pi}+\int_{\pi}^{2\pi}$ and a change of variable $\theta=\theta'-\pi$ will lead to $\displaystyle\int_{\pi}^{2\pi}=\int_{0}^{\pi}$ with the same integrand, so they double: 
\begin{align*}
\displaystyle\int_{0}^{2\pi}\sin(r\cos\theta)\sin(r\sin\theta)d\theta&=2\int_{0}^{\pi}\sin(r\cos\theta)\sin(r\sin\theta)d\theta,
\end{align*}
now split again to 
\begin{align*}
\int_{0}^{\pi/2}\sin(r\cos\theta)\sin(r\sin\theta)d\theta+\int_{\pi/2}^{\pi}\sin(r\cos\theta)\sin(r\sin\theta)d\theta
\end{align*}
for 
\begin{align*}
\int_{\pi/2}^{\pi}\sin(r\cos\theta)\sin(r\sin\theta)d\theta
\end{align*}
use $\theta'=\theta-\pi/2$, then
\begin{align*}
\int_{\pi/2}^{\pi}\sin(r\cos\theta)\sin(r\sin\theta)d\theta=-\int_{0}^{\pi/2}\sin(r\cos\theta)\sin(r\sin\theta)d\theta.
\end{align*}
A: You can write the integral (leaving $2\pi$ aside) as
$$
\int_{{\Bbb R}^2}e^{-\frac{x^2+y^2}{2}}\,dxdy+\int_{{\Bbb R}^2}\sin x\, e^{-\frac{x^2}{2}}\cdot \sin y\,e^{-\frac{y^2}{2}}\,dxdy.
$$
Now the second integral reduces to iterated integration
$$
\int_{\Bbb R}\int_{\Bbb R}\sin x\, e^{-\frac{x^2}{2}}\cdot \sin y\,e^{-\frac{y^2}{2}}\,dxdy=\int_{\Bbb R}\sin y\, e^{-\frac{y^2}{2}}\Bigg(\underbrace{\int_{\Bbb R}\sin x\,e^{-\frac{x^2}{2}}\,dx}_{=I}\Bigg)\,dy=I\cdot \underbrace{\int_{\Bbb R}\sin y\, e^{-\frac{y^2}{2}}\,dy}_{=I}=I^2.
$$
The integrand of $I$ is an odd function, so $I=0$.
