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anonymous

  • 4 years ago

Integrate

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  1. anonymous
    • 4 years ago
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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.

  2. anonymous
    • 4 years ago
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    unlikely since it is a perfect square right?

  3. anonymous
    • 4 years ago
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    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.

  4. nenadmatematika
    • 4 years ago
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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.

  5. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    Nvm, I made a mistake. I should have integrated each piece seperately THanks.

  7. anonymous
    • 4 years ago
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    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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