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Math Help - [SOLVED] integration no.7

  1. #1
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    [SOLVED] integration no.7

    integration no.7
    Attached Thumbnails Attached Thumbnails [SOLVED] integration no.7-p207-re.ex21-q47.gif  
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    Quote Originally Posted by afeasfaerw23231233 View Post
    integration no.7
    \int \frac{1 - 4u^2}{u^2 + 1} du

    \int \frac{1}{u^2 + 1}du - 4 \int \frac{u^2}{u^2 + 1} du

    Does this help?

    -Dan
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    i know how to do it now , thanks!
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  4. #4
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    The denominator (1 +sin^2(theta)) is not (1 +u^2) because u = cos(theta).

    1 +sin^2(theta)
    = 1 +1 -cos^2(theta)
    = 2 -cos^2(theta)
    = 2 -u^2

    So it must be
    INT.[(1 -4u^2) / (2 -u^2)]du
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