Some inquisitive students have asked me before, how the MF15 integration formulas come about. I thought I should share it too then.

So we want to $\int\frac {1}{\sqrt{4-x^2}}dx$ and yes, we know the formula can be plugged directly… but what if we want to avoid the formula. Now actually, the trick here involves your trigonometry identities, along with substitution methods. We have a $sin$ here, so we only know one trigo identity that involves $sin$ and that is $sin^{2}x + cos^{2}x = 1$. Hmmm, so my approach here will be to let $x=2cost$. We will see shortly why I chose to put a 2 and that using $cost$ or $sint$ will make no difference. You can try them yourself!

lets first find that $dx=2sint dt$

$\int\frac {1}{\sqrt{4-x^2}}dx= \int \frac {1}{\sqrt{4-4cos^{2}t}}(2sint dt)$

If you notice, this explains why there is a need for us to introduce $2cost$ instead of just $cost$.

Having $4-4cos^{2}t$ allows us to simplify it to $4sin^{2}t$

We have $\int \frac{1}{\sqrt{4sin^{2}t}}(2sint dt)=\int\frac{1}{2sint}(2sint)dt=\int1dt$

Finally, $\int1dt=t+C = sin^{-1}(\frac{x}{2}) + C$

That was long! But i hope it give you some insights to the formulas.