Cannot understand trig manipultion $a$ is a radius in a origin centered circle.
Any point on a circle circumference has coordinates $(a\ \cos\theta, a \ \sin\theta)$. 
I do not understand the last transformation in this expression $x = \sqrt{a^2 - y^2} = \sqrt{a^2 - a^2 \sin^2\theta } = a \cos\theta $ 
How do we get $ \sqrt{a^2 - a^2 \sin^2\theta } = a \cos\theta $ ???
Note: this is just a part of calc lecture in MIT ocw
 A: $$ \sqrt{ a^2 - a^2 \sin^2 \theta } $$
$$ = \sqrt{ a^2 ( 1 - \sin^2 \theta ) } $$
$$ = a \sqrt{ \cos^2 \theta } $$
$$ = a \cos \theta $$
We can simplify $1-\sin^2\theta$, because of the pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$.
A: Important note:
$$\sqrt{a^2-a^2\sin^2\theta} = |a\cos\theta |$$
as $\sqrt{\Delta^2} = |\Delta|$ and not $\Delta$.
Typically for indefinite integrals it's assumed the substitution is done over an appropriate interval (sometimes this is hand-waved away for whatever reason), as in your case it is required that $-\dfrac{\pi}{2}\le\theta\le\dfrac{\pi}{2}$, or similar, so that $\cos\theta\ge 0$. For definite integrals, this changes things.$$\\$$
For example, in integrating
$$\int_{-\pi}^{-3\pi/2} \dfrac{1}{\sqrt{2^2-2^2\sin^2(\theta)}} d\theta$$
One would obtain $\sqrt{2^2-2^2\sin^2\theta} = |2\cos\theta| = -2\cos\theta$ on this interval.
A: It's from the Pythagorean identity $\cos^2\theta + \sin^2\theta = 1$.
Multiply both sides by $a^2$.  Then $a^2\cos^2\theta + a^2\sin^2\theta = a^2$.  Now subtract $a^2\sin^2\theta$ from both sides.  Then take square roots.
