Show that the marginal pdf of $Y_1$ is normal given joint pdf of $Y_1$ and $Y_2$ 
Let $Y_1,Y_2$ be random variables with joint pdf $$f(y_1, y_2) =
\frac{1}{2\pi}\exp\left[{-\frac{1}{2}(y_1^2+y_2^2)}\right] \left(
1+y_1y_2\exp \left[{-\frac{1}{2}(y_1^2+y_2^2-2)}\right]\right)$$
  Show that the marginal pdf of $Y_1$ is normal. 

I am having issues integrating. I know there is some 'trick' to it since it doesn't appear to be possible via traditional techniques. I am unable to find it.  
Hopefully, someone can start me in the right direction.
 A: I believe we would have $\exp(-\frac{1}{2}(y_1^2+y_2^2))$ etc so it integrates.
We integrate the joint distribution with respect to $y_2$ from $-\infty$ to $ \infty$.
We separate it into two integrals, one for each term in the final bracket.
The first one integrates to $\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}y_1^2}$, which is a standard Gaussian.
The second one is: $C(y_1) \int_{-\infty}^{\infty}(e^{-\frac{1}{2}y_2^2})' dy_2 = 0$
A: You have
\begin{align}
\int\limits_{-\infty}^{+\infty} & \frac 1 {2\pi} \exp\left[-\frac{1}{2}(y_1^2+y_2^2)\right] \\
& \left(
1+y_1y_2\exp \left[-\frac{1}{2}(y_1^2+y_2^2-2) \right] \right) \, dy_2 \\[12pt]
= {} & \int\limits_{-\infty}^{+\infty} \frac 1 {2\pi} \exp\left[-\frac{1}{2}(y_1^2+y_2^2)\right] \, dy_2 \tag 1 \\[6pt]
& {} + y_1 e^{-2} \int\limits_{-\infty}^{+\infty} \frac 1 {2\pi} y_2\exp\left[ -(y_1^2+y_2^2) \right] \, dy_2 \tag 2 \\[6pt] 
\end{align}
Line $(1)$ is a familiar integral. Line $(2)$ is $0$ because an odd function is integrated over an interval symmetric about $0,$ and the integral of the absolute value is finite.
