How to find the antiderivative of the $\sqrt{4-x^2}$? 
How to find the $\int \sqrt{4-x^2}$

I know that you can use substitution for x and sub in $2\sin\theta$, but i don't understand how and why you know to do that? That is not something I would think of doing 

$x=2\sin\theta\implies\sqrt{4-(2\sin\theta)^2}=2\sqrt{1-\sin^2\theta}=2\cos\theta$$dx=2\cos\theta\:d\theta$ $$\int\sqrt{4-x^2}\:dx=\int2\cos\theta(2\cos\theta\:d\theta)$$ 
I think rest of work is easy enough. Why choosing $x=2\sin\theta$ is good because it convert your $\sqrt{4-x^2}$ into $2\cos\theta$. You have to remember that because of $4$ you need something before $\sin\theta$ which complete the identity $1-\sin^2\theta$. Hence in general $\sqrt{a-x^2}\implies x=\sqrt{a}\sin\theta$
