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Math Help - another tirg integral

  1. #1
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    another tirg integral

    ok what am i doing rong =/

    integral of xsinx
    i am using integration by parts and what i get is cosx-xcosx
    but the answer is sinx-xcosx
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by xslim12 View Post
    ok what am i doing rong =/

    integral of xsinx
    i am using integration by parts and what i get is cosx-xcosx
    but the answer is sinx-xcosx
    IBP is
    \int udv = uv - \int vdu
    what was your u? dv? and so your du and v?
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  3. #3
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    u=x du=1
    dv=sinx v=-cosx

    and i still get the same answer i got before =/
    Last edited by xslim12; November 4th 2007 at 09:52 PM.
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  4. #4
    MHF Contributor kalagota's Avatar
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    \int_{-1}^1 \sqrt{1-t^2}dt \cdot \frac{\sqrt{1-t^2}}{\sqrt{1-t^2}} = \int_{-1}^1 \frac{1}{\sqrt{1-t^2}}dt - \int_{-1}^1 \frac{t^2}{\sqrt{1-t^2}}dt

    = sin^{-1} (1) - sin^{-1} (-1) - \int_{-1}^1 \frac{t^2}{\sqrt{1-t^2}}dt

    for
    \int_{-1}^1 \frac{t^2}{\sqrt{1-t^2}}dt

    set sin \theta = \sqrt{1-t^2} \implies t = cos \theta
    cos \theta d\theta = \frac{-t}{\sqrt{1-t^2}}d\theta

    so,
    \int \frac{t^2}{\sqrt{1-t^2}}dt = -\int cos^2\theta d\theta = - \frac{1}{2}\int (1+cos2\theta) d\theta

    = -\frac{1}{2}\theta - \frac{1}{4} sin 2\theta

    = -\frac{1}{2} cos^{-1} t - \frac{1}{4} sin 2(cos^{-1} t)

    so original is = sin^{-1} (1) - sin^{-1} (-1) + \frac{1}{2}cos^{-1} t + \frac{1}{4} sin 2(cos^{-1} t) = \frac{\pi}{2}
    Last edited by kalagota; November 4th 2007 at 11:18 PM.
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  5. #5
    Math Engineering Student
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    Quote Originally Posted by kalagota View Post
    \int_{-1}^1 \sqrt{1-t^2}dt
    Of course, over the simetric interval [-1,1], we're integratin' an even function, so we can set

    \int_{-1}^1\sqrt{1-t^2}\,dt=2\int_0^1\sqrt{1-t^2}\,dt.
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