How to choose the expression to substitute? We are currently practicing integration using substitution in calculus and I'm having problems choosing the expression to substitute. For example, we have the following solved example:
Solve
$$ \int^r_0 \sqrt{r^2-x^2}dx $$ where $r>0$ using substitution.
Let's choose $ x=r\sin u $ and $ dx=r\cos u du $.
Now $ x=0 \Rightarrow u=0 $ , $ x=r\Rightarrow u=\frac{\pi}{2} $ and
$$ \int^r_0 \sqrt{r^2-x^2}dx=\int^\frac{\pi}{2}_0\sqrt{r^2-r^2\sin^2u} \cdot r \cos udu =r^2\int^\frac{\pi}{2}_0 \cos^2u du=r^2\int^\frac{\pi}{2}_0 \frac{\cos(2u)+1}{2}du=\frac{r^2}{2}\left[ \frac{\sin2u}{2}+u \right]^\frac{\pi}{2}_0=\frac{r^2}{2} \cdot \frac{\pi}{2}=\frac{\pi r^2}{4}.$$
I know there are probably other substitutions that will work out too but how did we know that after choosing $ x=rsinu $ the expression would simplify like that? I mean, whenever I do substitutions I just try out different ways until one of them works out but are there any hints in the original expression that gives away what would be the substitution that will work?
In this particular example I also don't understand how we got 
$$r^2\int^\frac{\pi}{2}_0 \frac{\cos(2u)+1}{2}du = \frac{r^2}{2}\left[ \frac{\sin 2u}{2}+u \right]^\frac{\pi}{2}_0.$$
 A: 1.) I think for these types of problem you have to remember basic formulas.
For example -
$\cos^2x = 1 - \sin^2x$
And second thing you have to identify which variable is constant. 
For example in your question r is constant. Just think it as 1 for trying substitution.
Then you have $1 - x^2$. Now think which basic formulas are of this type i.e $1 - \sin^2x$ and $1 - \cos^2x$. So you able to substitute. And don't forget to include constant with term.
2.) And regarding one step you don't understand.
$\frac{r^2}2$ is constant so we taken it outside. Now integration of $\cos 2u = \frac{\sin 2u}{2}$ and integration of 1 is u. Then substitute limits.
A: We usually choose the substitution to be made based on the form of trigonometric identities. In your example, $r^2-x^2$ might remind us of $\cos^2 x = 1 -  \sin^2 x$, so we choose $u= r \sin x$. Of course we could've been reminded of $\sin^2 x = 1 -  \cos^2 x$ and chosen $u=r \cos x$, which would've worked as well.
Many Calculus books have tables such as this, which tell you what substitution to make based on the form.

As to why $\displaystyle r^2\int^\frac{\pi}{2}_0 \frac{\cos(2u)+1}{2}du = \frac{r^2}{2}\left[ \frac{\sin 2u}{2}+u \right]^\frac{\pi}{2}_0.$
You can breathe easy on this one; it may look a little complex but it's just a basic integral. Here are the steps:
Let $v=2u$. Then $dv=2du$
$\displaystyle \int \dfrac{\cos(2u)+1}{2}du$
$ =\displaystyle\int \frac{\cos(2u)+1}{2}\dfrac {2du}{2} $
$=\displaystyle\int \frac{\cos(v)+1}{4} {dv}$
$=\dfrac 14 \displaystyle\int (\cos(v)+1) dv$
which I believe is straightforward from here.
