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Math Help - Integrals w/ trig substitutions

  1. #1
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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:


    \frac{1}<br />
{{81 - x^2 }} = \frac{1}<br />
{{\left( {9 - x} \right)\left( {9 + x} \right)}} = \frac{{\frac{1}<br />
{{18}}}}<br />
{{9 - x}} + \frac{{\frac{1}<br />
{{18}}}}<br />
{{9 + x}}<br />


    2. try the following substitution:

    <br />
x = \frac{1}<br />
{4}\tan u<br />

    3. <br />
\frac{1}<br />
{{\left( {2x^2  + 2x + 7} \right)^2 }} = \frac{1}<br />
{{4\left( {x^2  + x + \frac{7}<br />
{2}} \right)^2 }} = \frac{1}<br />
{{4\left[ {\left( {x + \frac{1}<br />
{2}} \right)^2  + \frac{{13}}<br />
{4}} \right]^2 }}<br />

    now try the following substitution:

    <br />
x = \sqrt {\frac{{13}}<br />
{4}} \tan u - \frac{1}<br />
{2}<br />
    Last edited by Peritus; February 27th 2008 at 02:04 PM.
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