What's the derivative of $\int_1^x\sin(t)dt$? What's the derivative of the integral $$\int_1^x\sin(t) dt$$
Any ideas? I'm getting a little confused.

You can use the fundamental theorem of calculus, but if you have not yet covered that theorem, in short, you'll be taking the derivative - with respect to x - of the integral of $\sin(t)dt$ when the integral is evaluated from $1$ to $x$:
$$\frac{d}{dx}\left(\int_1^x \sin(t) \text{d}t\right) = \frac{d}{dx} [-\cos t]_1^x = \frac{d}{dx}\left(-\cos(x) - (-\cos(1))\right) = \sin(x).$$
and you'll no doubt be encountering the Fundamental Theorem of Calculus very, very soon:
For any integrable function $f$, and constant $a$: $$\frac{d}{dx} \left(\int_a^x f(t)dt \right)= f(x),$$ 
(provided $f$ is continuous at $x$).
