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Math Help - Evaluating Definite Integrals

  1. #1
    Member purplec16's Avatar
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    Evaluating Definite Integrals

    \int_{-2}^{2} \frac{3dt}{4+3t^2} \, dt

    is my u=4+3t^2?
    Last edited by mr fantastic; June 18th 2011 at 08:24 PM. Reason: Fixed the latex.
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  2. #2
    Senior Member BAdhi's Avatar
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    Arrow A different method

    that won't work because, then du=6t dt.

    Since there is no t on the upper side of the fraction (sorry I can't remember the name that is called) it will be a dead end.

    try using the identity,
    \frac{d\tan{^{-1}}x}{dx}=\frac{1}{1+x^2}
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  3. #3
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    Re: Evaluating Definite Integrals

    Quote Originally Posted by purplec16 View Post
    [tex]int_{-2^2}{\frac{3dt}{4+3t^2}}[\tex]

    is my u=4+3t^2?
    Why not try t=\frac{2}{\sqrt{3}}\tan\theta
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  4. #4
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    Re: A different method

    Quote Originally Posted by BAdhi View Post
    that won't work because, then du=6t dt.

    Since there is no t on the upper side of the fraction (sorry I can't remember the name that is called) it will be a dead end.

    try using the identity,
    \frac{d\tan{^{-1}}x}{dx}=\frac{1}{1+x^2}
    Yes, but using t's instead of x's, of course .....

    @OP: And since the integrand is even, you can integrate from 0 to 2 and then double the resulting answer.
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  5. #5
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    Re: A different method

    Quote Originally Posted by BAdhi View Post
    that won't work because, then du=6t dt.

    Since there is no t on the upper side of the fraction (sorry I can't remember the name that is called) it will be a dead end.
    In English, the "numerator" (and the lower side is the "denominator").

    try using the identity,
    \frac{d\tan{^{-1}}x}{dx}=\frac{1}{1+x^2}
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