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Thread: Integrals w/ trig substitutions

  1. #1
    Member
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    Aug 2007
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    Integrals w/ trig substitutions

    I'm having some trouble with a few problems, any hints or help would be great!

    1) antiderivative of 1/(81-x^2)
    i ended up getting this down to (1/9)integral of (1/cos(theta)), and i think the integral of that is (1/9)(ln(sec(theta)+tan(theta))), but it was wrong. any ideas? i may have just integrated the wrong thing, im not sure


    2) antiderivative of 1/((16x^2+1)^2)
    i have no idea where to begin with this one, haven't seen something of this type before

    3)antiderivative of 1/((2x^2+2x+7)^2)
    again, unsure of where to begin. should i multiply that all out, or what?
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  2. #2
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
    Posts
    397
    1. no need for trigonometric substituions here:


    $\displaystyle \frac{1}
    {{81 - x^2 }} = \frac{1}
    {{\left( {9 - x} \right)\left( {9 + x} \right)}} = \frac{{\frac{1}
    {{18}}}}
    {{9 - x}} + \frac{{\frac{1}
    {{18}}}}
    {{9 + x}}
    $


    2. try the following substitution:

    $\displaystyle
    x = \frac{1}
    {4}\tan u
    $

    3. $\displaystyle
    \frac{1}
    {{\left( {2x^2 + 2x + 7} \right)^2 }} = \frac{1}
    {{4\left( {x^2 + x + \frac{7}
    {2}} \right)^2 }} = \frac{1}
    {{4\left[ {\left( {x + \frac{1}
    {2}} \right)^2 + \frac{{13}}
    {4}} \right]^2 }}
    $

    now try the following substitution:

    $\displaystyle
    x = \sqrt {\frac{{13}}
    {4}} \tan u - \frac{1}
    {2}
    $
    Last edited by Peritus; Feb 27th 2008 at 02:04 PM.
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