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Math Help - Strange integration by substitution

  1. #1
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    Strange integration by substitution

    Hi there,

    I've been given a question to do as homework, it seems pretty weird to me:

    Use the substitution x = 4cosθ to evaluate the following definite integral:

    \int^2_0\frac{x}{16-x^2}  dx
    Any help will be greatly appreciated.

    Thank you.
    Last edited by ecopetition; December 7th 2008 at 08:03 AM.
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  2. #2
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    Quote Originally Posted by title
    Strange integration by substitution
    I would paraphrase that to:
    Quote Originally Posted by title
    Integration by strange substitution
    <br /> <br />
\int^2_0\frac{x}{16-x^2}  dx<br />

    Using the sub x = 4\cos{\theta} \implies dx = -4\sin{\theta}~d\theta yields:

    -16 \int \frac{\sin{\theta}\cos{\theta}}{16(1-\cos^2{\theta})}~d\theta = -\int \frac{\cos{\theta}}{\sin{\theta}}~d\theta = -\ln{|\sin{\theta}|} + C

    Now back substitute and evaluate
    Last edited by Chop Suey; December 7th 2008 at 08:20 AM.
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  3. #3
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    Quote Originally Posted by Chop Suey View Post
    I would paraphrase that to:


    <br /> <br />
\int^2_0\frac{x}{16-x^2} dx<br />

    Using the sub x = 4\cos{\theta} \implies dx = -4\sin{\theta}~d\theta yields:

    -16 \int \frac{\sin{\theta}\cos{\theta}}{16(1-\cos^2{\theta})}~d\theta = -\int \frac{\cos{\theta}}{\sin{\theta}}~d\theta = -\ln{|\sin{\theta}|} + C

    Now back substitute and evaluate
    The substitution u = 16 - x^2 makes for an easier calculation.
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    The substitution u = 16 - x^2 makes for an easier calculation.
    Indeed, but the question asked to use x = 4cosθ, so...
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  5. #5
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    Quote Originally Posted by Chop Suey View Post
    Indeed, but the question asked to use x = 4cosθ, so...
    Indeed. A pity the question didn't also ask that I wear glasses while reading it.
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