I've been quite bizzy with a exercise from Probabilty Theory. As you know when calculating variation you have to calculate with integrals. My calculus hasn't been that good. So, I asked Wolfram to help me. It's been a great help, but it's making a step I really can't explain. I found some integrals which look like it, but there the step is somewhat easier to understand.
Given the equation:
$\frac{1}{\pi} \int \frac{\left( u + \frac{1}{2} \right)^2}{\sqrt{\frac{1}{4} - u^2}} du $ How does one end up here?
$ \frac{1}{\pi} \int \left(\frac{\sin(v)}{2} + \frac{1}{2}\right)^2 dv $
Wolfram tells me that $ u = \frac{\sin(s)}{2}$ and $du = \frac{\cos(s)}{2} $ And this all makes perfect sense to me, but somehow I can only end up with a slightly different form:
$ \frac{1}{\pi} \int \frac{(\sin(v) + \frac{1}{2})^2}{\frac{\cos(v)}{2}} \frac{1}{2} \cos(v) dv $
How does simplifying this result to the answer Wolfram gives?
You just have to cancel the cosine factor: $$ \frac{(\frac{\sin(v)}2 + \frac{1}{2})^2}{ \color{red}{\frac{\cos(v)}{2}}} \color{red}{\frac{1}{2} \cos(v)}=\left(\frac{\sin(v)}2 + \frac{1}{2}\right)^2. $$