$I = \int_{0}^\infty t^2 e^{-t^2/2} dt$ Q: Evaluate the integral $I = \int_\limits{0}^\infty t^2 e^{-t^2/2} dt$
Hint, write $I^2$ as the following iterated integral and convert to polar coordinates:
\begin{align*}
  I^2 &= \int_\limits{0}^\infty \int_\limits{0}^\infty x^2 e^{-x^2/2} \cdot y^2 e^{-y^2/2} \, dx \, dy \\
\end{align*}
I can see the final answer is $\frac{\pi}{2}$ but I don't see how to get this.
This problem is very similar to the Gaussian Integral: $I = \int_\limits{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi}$. I can follow the derivation to this.
The Gaussian Integral technique of converting to polar coordinates doesn't seem to work as cleanly on this problem.
I can convert to polar coordinates:
\begin{align*}
  I^2 &= \int_\limits{0}^\infty \int_\limits{\pi/2}^\pi r^5 \sin^2 \theta \cos^2 \theta e^{-r^2/2} \, d\theta \, dr \\
\end{align*}
That doesn't look easy to evaluate.
 A: Let T a random variable normally distributed mean 0 , variance 1
$E(T^2)=E(T^21_{T>0})+E(T^21_{T<=0})$. By symmetry of T, we have 
$var(T)=E(T^2)=2E(T^21_{T>0})=1$
Moreover, $E(T^21_{T>0})=\frac{1}{\sqrt{2\pi}}\int_\limits{0}^{\infty}t^2e^{-\frac{t^2}{2}}dt$.
Then we can conclude
A: *

*Why not integrate by parts from the initial integral? One has  $$   
   I=-\frac12\int_0^\infty x \cdot \left(-2xe^{-x^2}   
   \right)dx=-\frac12\int_0^\infty x \cdot \left(e^{-x^2} \right)'dx $$
then one may use the Gaussian integral to conclude.

*Another path is to differentiate the gaussian identity $$
   \int_0^\infty e^{-tx^2}dx=\frac{\sqrt{\pi }}{2 \sqrt{t}} \qquad t>0,
   $$  with respect to $t$ and put $t=1$.

A: 
I thought it might be instructive to present a way forward that relies on "Feynman's Trick" for differentiating under the integral sign.  

Proceeding, we let $I(a)$ be the integral given by
$$\begin{align}
I(a)&=\int_0^\infty e^{-at^2}\,dt\\\\
&=\frac12\sqrt{\frac{\pi}{a}} \tag 1
\end{align}$$
Differentiating $(1)$ reveals
$$-\int_0^\infty t^2 e^{-at^2}\,dt=-\frac14\frac{\sqrt\pi}{a^{3/2}} \tag 2$$
Setting $a=1/2$ in $(2)$ and multiplying by $-1$ yields the coveted result

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty t^2e^{-t^2/2}\,dt=\sqrt{\frac\pi{2}}}$$

And we are done!
A: If you want to use  your original hint, you have:  $$I^2 = \int_\limits{0}^\infty \int_\limits{\pi/2}^\pi r^5 \sin^2 \theta \cos^2 \theta e^{-r^2/2} \, d\theta \, dr $$
First, we can switch the order of integration and start with $$\int_\limits{0}^\infty r^5 e^{-r^2/2}dr=\int_\limits{0}^\infty (r^4)(r e^{-r^2/2})dr $$
That can be integrted by parts twice, using the fact that $\int re^{-r^2/2}dr=-e^{-r^2/2}$, and one final integration yields the result $$\int_\limits{0}^\infty r^5 e^{-r^2/2}dr=8$$
And our integral becomes$I^2 =\int_\limits{\pi/2}^\pi 8\sin^2 \theta \cos^2 \theta d\theta$
Using the identity $\cos^2(x)\sin^2(x)=\frac{1}{4}\sin^2(2\theta)$ we have $I^2 =\int_\limits{\pi/2}^\pi 8 \frac{1}{4}\sin^2(2\theta) d\theta=\int_\limits{\pi/2}^\pi 2\sin^2(2\theta) d\theta$ And so
$$I^2=\int_\limits{\pi/2}^\pi 2\sin^2(2\theta) d\theta=\frac{\pi}{2}$$
$$I=\sqrt\frac{\pi}{2}$$
A: I would use the definition of gamma function
$I=\int_{0}^{\infty}t^2e^{-t^2/2}dt  $
Let   $t^2/2 = s$
then,  $ I  = \sqrt{2}\int_{0}^{\infty}s^{1/2}e^{-s}ds $
$          = \sqrt2 \Gamma(3/2) $
$=\sqrt2 \frac{\sqrt{\pi}}{2} $
$=\sqrt\frac{\pi}{2}$
