I have the following derivation for showing that the integral of the standard normal distribution is equal to one.
I don't completely understand the steps so I'll provide the derivation and then write what I think is happening.
$$ \begin{align} 2\pi \int\limits_{-\infty}^{\infty} p(x) dx &= 2\pi \int\limits_{-\infty}^{\infty} p(x) dx \tag{1} \\ &= 2 \pi \int\limits_{-\infty}^{\infty} p(x) dx \int\limits_{-\infty}^{\infty} p(y) dy \tag{2}\\ &= \int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty e^{-(x^2+y^2)/2}dx\ dy \tag{3}\\ &= \int\limits_{\theta=0}^{2\pi} \int\limits_{r=0}^\infty e^{-r^2/2} r\ dr\ d\theta \tag{4} \\ &= 2\pi \tag{5} \end{align} $$
Is what we're trying to integrate (wasn't included in the text but I added it for clarity). I think both sides should be multiplied by $2\pi$ which we do to make the leap in (3) easier so I included that.
We're multiplying by the integral in terms of $y$. I think the argument is that this is similar to multiplying by 1, but I'm not fully satisfied by this because we're supposed to be proving that. I think this fallacy is called "begging the question".
Integration is commutative. I'm comfortable with the inside argument but I'm not sure why the $2\pi $ got consumed.
We converted to polar coordinates. No problem here.
Evaluate the integral. No problem here.
So as was explained earlier I'm worried about begging the question in (2) and I don't know why the $\boldsymbol{2\pi}$ got consumed in (3).
Let $I$ be the value of the integral. Then get rid of line (1), and on the left hand side of equation (2) write $2\pi I^2 = ...$.
You end up with $2\pi I^2 = 2\pi$. Then solve for $I$, using the fact that $I \ge 0$.