Integrate

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\[\int\limits_{0}^{\pi/2} (2(1-\sin \theta))^2 d \theta\] I'm getting a negative number for some reason.
unlikely since it is a perfect square right?
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.

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

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