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 
 A: $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 $\sqrt{a-x^2}\implies x=\sqrt{a}\sin\theta$
A: The "how you do it" is in the answer from @emonHR . There are several ways to think about "why".


*

*The function suggests the way you  rewrite the identity $$ \sin^2 x + 
   \cos^2 x = 1  $$ to find $\cos x$ in terms of $\sin x$.

*You recognize the function as solving for $y$ in terms of $x$ in the
equation of a circle, so it's no surprise that trigonometry may help.

*You've practiced integration for a while so you're familiar with this
trick. Soon it becomes a technique, not a trick, and you have a good
intution for when it will work.
A: Here is a geometric approach. If you let $y=\sqrt{4-x^2}$ then $x^2+y^2=4$ so the point $(x,y)$ lies on a circle centred on the origin with radius $2$.
The definite integral $\int^a_0 y \space dx$ is the area of the region bounded by the circle, the x axis, and the lines $x=0$ and $x=a$.
If you divide this region into a segment of the circle with angle $\theta$ and a right triangle with sides $a=2\sin \theta$ and $\sqrt {4-a^2}=2\cos \theta$ then you have 
$\int_0^a y \space dx = 2 \theta + 2 \sin \theta \cos \theta= 2\theta + \sin 2\theta$.
