Given $\int_0^{\infty} e^{-x^2}dx = \frac{\sqrt\pi}{2}$, find $\int_0^{\infty} e^{-\frac{x^2}{2}} dx$ Given $\int_0^{\infty} e^{-x^2}dx = \frac{\sqrt\pi}{2}$, find $\int_0^{\infty} e^{-\frac{x^2}{2}} dx$ and $\int_0^{\infty} x^2e^{-x^2}dx$
I tried to figure out a way to get the integral in the form of the original integral to solve it, but so far I'm incredibly stuck. I assume that's the way to go to. First intuition was to try using a u substitution, but which substitution should I use?
Here is what I attempted:
$\int_0^{\infty} e^{-\frac{x^2}{2}} dx$
$u =\frac{x}{\sqrt{2}}$
$u^2 = \frac{x^2}{2}$
$du = \frac{1}{\sqrt{2}} dx$
$dx = \sqrt{2} du$
$\int_0^{\infty} e^{-u^2} \sqrt{2} du = \sqrt{2}\int_0^{\infty} e^{-u^2} du$

Following @Nebulae's comment:
Let the standard integral $$\int_0^{\infty} e^{-u^2}\, du = \frac{\sqrt{\pi}}2$$
Next, let $u = \frac{x}{\sqrt2}$, then $du = \frac{dx}{\sqrt2}$
The integrals will be equal:
$$\int_0^{\infty} e^{-u^2}\, du = \int_0^{\infty} \frac1{\sqrt2}e^{\frac{-x^2}2}\, dx = \frac{\sqrt{\pi}}2$$
Lastly, $\int_0^{\infty} \frac1{\sqrt2}e^{\frac{-x^2}2}\, dx = \frac{\sqrt{\pi}}2$
Multiply both sides by root 2
$$\int_0^{\infty} e^{\frac{-x^2}2}\, dx = \sqrt{\frac\pi2}$$

Next one! $$\int_0^\infty x^2 e^{-x^2}\, dx$$
This one is a really cool by-parts trick. One would think to separate it by the $x^2$ term and the $e^{[::]}$ term, but no...
Let $$u = x,\ \ du = dx,\ \ dv = xe^{-x^2},\ \ v = -\frac12e^{-x^2}$$
Now use $\int u\, dv = uv - \int v\, du$
Then we have $$-\frac{x}2e^{-x^2} \Bigr|_0^\infty+ \int_0^\infty \frac12e^{-x^2}\, dx$$
$$=\frac12 \int_0^\infty e^{-x^2}\, dx = \frac{\sqrt{\pi}}4$$
