## anonymous 4 years ago Integrate

1. anonymous

$\int\limits_{0}^{\pi/2} (2(1-\sin \theta))^2 d \theta$ I'm getting a negative number for some reason.

2. anonymous

unlikely since it is a perfect square right?

3. anonymous

Yeah, I took out the 2^2 out and got sin^2 theta -2 sin theta +1. I then used the half angle formula and got 1/2 + 1/2 cos 2 theta -2 sin theta. I then integrated and pluged back the 4 and I got 2 theta + sin 2 theta + 8 cos theta. But that's wrong.

yes, because the function is always positive, so the area must be positive...check if you made some trigonomety mistakes like sin or cos of some ange etc.

5. anonymous

first i would write $4\int (1-\sin(x))^2dx$ then $4\int \sin^2(x)+1-2\sin(x) dx$ then integrates each piece separately

6. anonymous

Nvm, I made a mistake. I should have integrated each piece seperately THanks.

7. anonymous

only annoying part is the first one, but it looks like you got it right. should get $2x-2\sin(x)\cos(x)+4\cos(x)$ as the "anti derivative" oh and add $2\pi$ at the end