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Math Help - Integral question:

  1. #1
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    Integral question:

    Hello, sorry I can't really use the math function well yet I need to learn it

    My question is about this:

    Integral (1/t^2)cos(1/t - 1)dt

    I substituted with u = 1/t - 1 and du = 1/t^2

    For a final answer I get cos(1/t - 1) + c but the book says it's
    -sin(1/t - 1) + c.

    I don't see when the cosine was derived?..

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  2. #2
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    Quote Originally Posted by Kaln0s View Post
    Hello, sorry I can't really use the math function well yet I need to learn it

    My question is about this:

    Integral (1/t^2)cos(1/t - 1)dt

    I substituted with u = 1/t - 1 and du = 1/t^2

    For a final answer I get cos(1/t - 1) + c but the book says it's
    -sin(1/t - 1) + c.

    I don't see when the cosine was derived?..

    SO the original problem (I think) was \int \frac{1}{t^2} \cos\left(\frac{1}{t}-1\right)\,dt.

    Then if you make your substitution, u = \frac{1}{t}-1, we have du = -\frac{1}{t^2}\,dt.

    (Then dt = -t^2 du)

    Substituting back gives:

    \int \frac{1}{t^2}\cos(u)\,(-t^2)\,du

    reducing to:

    \int -\cos(u)\,du = -\int \cos(u) \,du.

    Since the antiderivative of cosine is sine, this is:

    -\sin(u)+C.

    Now substitute away the u back to \frac{1}{t}-1.
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  3. #3
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    Recall the integral of cos is sin
    Last edited by treetheta; March 25th 2009 at 05:26 PM. Reason: Beat me to it.
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