Evaluate the indefinite integral.
I'm having trouble how to start. Do I factor out the x to make it ∫√x√16-x dx?
Complete the square: $\displaystyle 16x-x^2=64-64+16x-x^2=64-(x-8)^2$
So you have $\displaystyle \int\sqrt{64-(x-8)^2}\,dx$
Now let $\displaystyle x=8+8\sin t, dx=8\cos t\,dt$, so you get:
$\displaystyle \int\sqrt{64-64\sin^2t}\cdot8\cos t\,dt=64\int\cos^2t\,dt=64\int\frac{\cos(2t)+1}{2} \,dt$
Integrate and plug back in for $\displaystyle t$ using $\displaystyle t=\sin^{-1}\left(\frac{x-8}{8}\right)$.
You can check your answer by differentiating it. You can also use this website: Wolfram|Alpha