How do you integrate this without using separable integrals? I have to integrate
$\ \int_{-\pi/2}^{\pi/2} \int_{0}^{2}e^{-r^{2}}r \text{d}r\text{d}\theta$
The problem is the only example I can find simplifies the problem to which is some sort of "separable integrals" trick that I do not understand how it works.
$\ (\int_{-\pi/2}^{\pi/2}\text{d}\theta)(\int_{0}^{2} re^{-r^{2}}dr)$
I have heard that this is possible to do without using the trick they showed but I can not figure out how to actually integrate it without it.
How would you do this problem without separable integrals? How is does the separable integrals trick work?
 A: First, recall the following:
$$\int c\cdot f(x)\text{d}x=c\cdot\int f(x)\text{d}x$$
for any constant $c$. Then you only need to notice that $$\int_0^2e^{-r^2}r\text{d}r$$
does not depend on $\theta$ and is a constant, so you can apply the rule above to achieve:
\begin{align}
\int_{-\pi/2}^{\pi/2}\int_0^2e^{-r^2}r\text{d}r\text{d}\theta
&=\left(\int_0^2e^{-r^2}r\text{d}r\right)\left(\int_{-\pi/2}^{\pi/2}\text{d}\theta\right)\\
&=\left[-\frac{e^{-r^2}}{2}\right]_{r=0}^2\cdot\left[\theta\right]_{\theta=-\pi/2}^{\pi/2}\\
&=\left(-\frac{e^{-4}}{2}+\frac{1}{2}\right)\pi\end{align}
Without using the constant factor rule, you can simply solve the inner integral first:
\begin{align}\int_{-\pi/2}^{\pi/2}\int_0^2e^{-r^2}r\text{d}r\text{d}\theta&=\int_{-\pi/2}^{\pi/2}\left[-\frac{e^{-r^2}}{2}\right]_{r=0}^2\text{d}\theta\\
&=\left[\left(-\frac{e^{-4}}{2}+\frac{1}{2}\right)\theta\right]_{\theta=-\pi/2}^{\pi/2}\\
&=\left(-\frac{e^{-4}}{2}+\frac{1}{2}\right)\pi
\end{align}
A: Without separating the integrals, you can use u-substitution on $e^{-r^2}r 
 d\theta$. Set $u=r^2$, then $du = 2r d\theta$, therefore $r d\theta = du / 2$. Proceed as before. As a previous commenter pointed out, the integrand depends only on one variable, so that simplifies matters greatly.
