I came across this in Khan Academy Integral Calculus. I am asked to find this integral...
I did the substitution, and it made sense so far...
Then, I was unable to figure how to solve the definite integral. This was their solution...
Where did that sin expression come from? I am so confused!
The derivative of $\sin^{-1}(u)$ is $1 / \sqrt{1 - u^2}$, so the indefinite integral of $1/\sqrt{1 - u^2}$ is $\sin^{-1}(u) + C$. One way to understand where this comes from is to use implicit differentiation. Suppose $$y = \sin^{-1}(u)$$ (and note for later that we're implicitly restricting the values of $u$ and $y$, with $-1 \leq u \leq 1$ and $-\pi/2 \leq y \leq \pi/2$). Then $$\sin(y) = u.$$ Now differentiate on both sides with respect to $u$: $$ \frac{d}{du}\left[\sin(y)\right] = \frac{d}{du}[u] \\ \cos(y) \frac{dy}{du} = 1 \\ \frac{dy}{du} = \frac{1}{\cos(y)}. $$ So now we just need to understand why $\cos(y) = \cos(\sin^{-1}(u)) = \sqrt{1 - u^2}$. If $0 \leq y \leq \pi/2$, then we can imagine that $y$ is one angle of a right triangle with opposite side $u$ and hypotenuse 1:
By the Pythagorean Theorem, the base is $\sqrt{1 - u^2}$, so $$\frac{dy}{du} = \frac{1}{\cos(y)} = \frac{1}{\sqrt{1 - u^2}}.$$ And if instead $-\pi/2 \leq y \leq 0$, then using the fact that sine is odd and cosine is even, we have $\cos(y) = \cos(-y) = \sqrt{1 - (-u)^2} = \sqrt{1 - u^2}$, and we get the same answer.