Calculus: Integrate Fraction Containing Arcsin Component I'm struggling with the following problem: 



I searched online for a solution and came upon below.  But I'm confused by the solution itself:



Specifically, how did the problem go from here: 



to here: 



I have a hunch it's just this formula below -- this formula that I just need to remember? ... is this true?
Yes the formula used is just
$$
\int \frac{1}{\sqrt{a^2-x^2}}dx =\int \frac{1}{a\sqrt{1-\frac{x^2}{a^2}}}dx= \sin^{-1}\left(\frac{x} {a}\right) +C
$$
that can be easily recovered from the derivative:
$$
\frac{d}{dx} \sin^{-1}(x)=\frac{1}{\sqrt{1-x^2}}
$$
If this is a question on your test in AP Calculus BC, and if you’re going to need to solve questions of this pattern frequently, then I would suggest memorizing all the derivatives for inverse trig functions. That being said, I was able to solve it using trigonometric substitution. Construct a right triangle triangle with sides $a$, $x$, and $\sqrt{a^{2} - x^{2}}$. Now you can solve it as follows: 

Let $I = \int \frac{dx}{\sqrt{a^{2} - x^{2}}}$

Construct a right triangle with hypotenuse $a$, and legs, $x$, $\sqrt{a^{2} - x^{2}}$

Let $\theta$ be the angle that is opposite to the side $x$

$\sin{\theta} = \frac{x}{a}$

Differentiating both sides:

$a\cos{\theta}d\theta = dx$

We also know: 

$\cos{\theta} = \frac{\sqrt{a^{2} - x^{2}}}{a}$

$a\cos{\theta} = \sqrt{a^{2} - x^{2}}$

Substituting everything into the integral, we get:

$I = \int \frac{a\cos{\theta}}{a\cos{\theta}}d\theta = \int d\theta$

$ = \theta + C$

$ \sin{\theta} = \frac{x}{a}$

$ \arcsin{\frac{x}{a}} = \theta$

$I = \arcsin{\frac{x}{a}} + C$ 

picture of actual solution

Edit: How does this not resolve the question? I showed you how you get from your integral to your solution. I also said that it's better to memorize the formula if these questions appear frequently.