Q:

calculate $ F(x)= \int_{0}^{\sin x}\sqrt{1-t^2}\,dt $

Calculate $F'(x)$

I have this exercise in my worksheet, I am having a problem obtaining the correct answer which is as listed on the answer sheet $-\cos^2(x)$.

A:

You have $F(x)=G(\sin x)$ where $G(y)=\int_0^y\sqrt{1-t^2}dt$ satisfies, by the fundamental theorem of calculus, $G'(y)=\sqrt{1-y^2}$. So by the chain rule
$$F'(x)=G'(\sin x)\cos x=\sqrt{1-\sin^2x}\cdot \cos x=|\cos x|\cos x.$$
That's $\cos^2x$ on $[-\pi/2+2k\pi,\pi/2+2k\pi]$, and $-\cos^2 x$ on $[\pi/2+2k\pi,3\pi/2+2k\pi]$.

