Integrate

Question

Integrate

anonymous · Answer

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

anonymous · Answer

unlikely since it is a perfect square right?

anonymous · Answer

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.

nenadmatematika · Answer

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.

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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