$\int \limits_0^{\infty} x^2 \exp(-2x^2) dx$ How to evaluate this integral? 
$$\int \limits_0^{\infty} x^2 \exp(-2x^2) dx$$
I found similar problem, but don't know how to apply them here.
What do I have to substitute?

There are often several ways to evaluate some integrals. Here is one of the less standard ways. 
Consider the integral
\begin{align}
I(a) = \int_{0}^{\infty} x^2 \, e^{- a x^2} \, dx
\end{align}
for which 
\begin{align}
I(a) &= - \partial_{a} \, \int_{0}^{\infty} e^{-a x^2} \, dx 
= - \frac{1}{2} \, \partial_{a} \left( \sqrt{\frac{\pi}{a}} \right) 
= \frac{1}{4a} \, \sqrt{\frac{\pi}{a}}. 
\end{align}
Where $a=2$ the result desired is obtained. 
Integral evaluation for clarity:
\begin{align}
J(a) &= \int_{0}^{\infty} e^{-a x^{2}} \, dx \\
&= \frac{1}{2 \sqrt{a}} \, \int_{0}^{\infty} e^{-t} \, t^{-1/2} \, dt \hspace{10mm} t= a x^{2} \\
&= \frac{1}{2\sqrt{a}} \, \sqrt{\frac{\pi}{a}}.  
\end{align}
Further calculations:
\begin{align}
\int_{0}^{\infty} x^{2n} \, e^{-a x^{2}} \, dx &= (-1)^{n} \partial_{a}^{n} \left( \frac{1}{2} \sqrt{\frac{\pi}{a}} \right) \\
&= \frac{(2n)!}{2^{2n+1} \, n! \, a^{n}} \, \sqrt{\frac{\pi}{a}} 
\end{align}
