The derivative of $\int_0^{\sin x} \sqrt{1-t^2} dt$ is given as $\lvert \cos x\rvert \cos x$. But why the absolute value? The given problem: Find the derivative of $$ F(x)=\int_0^{\sin x}\sqrt{1-t^2}dt $$ I know the answer is supposed to be $\lvert \cos x\rvert \cos x$, but I don't understand where the absolute value comes in. 
Using the fundamental theorem, my working out was as follows:
\begin{align}
F'(x) & =\cos x \sqrt{1-sin^2x}\\
& = \cos x \sqrt{\cos^2x} \\
& = \cos x (\pm \cos x)\\
& = \pm \cos^2x
\end{align}
Which was evidently wrong, but I don't understand why the sqrt of $\cos^2x$ must be an absolute value. It is a square, so what would be the problem if it were negative? 
Am I missing something very simple? I'm a long distance student so I can't exactly raise my hand in class and ask what's what. 
Afterthought: Perhaps I'm grabbing the wrong end of the stick and it's the other cos term that's been made an absolute value, not the sqrt...even if that were the case though I'd still be confused. 
 A: Note that $\sqrt{x^2} = |x|$. This is somewhat better than writing $\sqrt{x^2} = \pm x$ as the latter doesn't explicitly tell you when looking at the right hand side when to take the plus sign and when to take the minus. The absolute value takes care of that.
A: What you have done is somewhat correct if you consider cases. $|x|$ takes one of two values: $x$ or $-x$ (for real $x$). The way to see this is by cases. If $x>0$ then $|x|=x$ since it is already positive. If $x<0$, then $|x| = - x$ to make it positive. So if you consider the $x$ such that $\cos $ is either positive or negative, you'll see that what you have is implicitly the same thing. 
A: We can use Fundamental Theorem of Calculus and chain rule. We know that
$$ \frac{\mathrm{d}}{\mathrm{d}x}\int_{0}^{x}{f{(t)} \, \mathrm{d}t} = f{(x)}. $$
So,
$$
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x}\int_{0}^{\sin{x}}{\sqrt{1 - t^{2}} \, \mathrm{d}t} &= \cos{x} \cdot \sqrt{1 - \sin^{2}{x}} \\
&= \cos{x} \cdot \sqrt{\cos^{2}{x}}
\end{align*}
$$
Because $ \sqrt{z^{2}} = z $ for every real number $ z $, so
$$ \cos{x} \cdot \sqrt{\cos^{2}{x}} = \left|\cos{x}\right| \cos{x}. $$
