I know I have my integral correct:

$\displaystyle V = 41\pi \int\limits^{\pi}_{\frac{\pi}{2}} sin^2x dx$

So then I did this:

$\displaystyle

41\pi \int\limits^{\pi}_{\frac{\pi}{2}} \frac{1}{2} (1-cos (2x))$

$\displaystyle

= \frac{41\pi}{2} \int\limits^{\pi}_{\frac{\pi}{2}} 1-cos (2x)$

$\displaystyle

= \frac{41\pi}{2} \int\limits^{\pi}_{\frac{\pi}{2}} 1 - cos (u) du$

$\displaystyle = \frac{41\pi}{4} \cdot (x - \frac{sin2x}{2}) \|^{\pi}_{\frac{\pi}{2}}$

After I evaluate I get $\displaystyle \frac{41\pi^2}{4}$ which is the wrong answer.

I know this is a lot of work, but can someone work this and just point out where my error is? Don't need all the steps, unless I screwed something up.

Thanks!!

Molly