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Math Help - Integral of arbitary powers

  1. #1
    Member akhayoon's Avatar
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    Integral of arbitary powers

    the integral b=1, a=0 of (1/2)[(a^x-a^-x)/(a^x+a^-x)]
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by akhayoon View Post
    the integral b=1, a=0 of (1/2)[(a^x-a^-x)/(a^x+a^-x)]
    do you mean \int_0^1 \frac {a^x - a^{-x}}{a^x + a^{-x}}~dx ?

    using a for the limits and then a again for the constants in the integral is confusing. be specific and clear

    if that is your integral, use the substitution u = a^x + a^{-x}
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  3. #3
    Member akhayoon's Avatar
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    after the substitution will the new bounds be

    lna~ \int_2^\frac {a^2+1} {a} \frac {du} {u} ?

    which then leads to

     lna~ [ln|u|] ? with those bounds I mentioned of course
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by akhayoon View Post
    after the substitution will the new bounds be

    lna~ \int_2^\frac {a^2+1} {a} \frac {du} {u} ?

    which then leads to

     lna~ [ln|u|] ? with those bounds I mentioned of course
    yes, that's one way to do it. or, you could switch back to a function of x and use the old limits if you like.
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