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Math Help - Trig substitution integrals

  1. #1
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    [Solved]Trig substitution integrals

    I solved the problem the 3's are part of the constant c thats why they arent in the answer.

    Hello,
    I have the integral


    dy / sqrt( 9 + y^2)

    I have worked it all the way down to

    ln| sec theta + tan theta| + C

    I need to now put it in terms of y.

    from my previous work during the problem I have
    y= 3 tan theta

    so using right triangle trig
    tan theta= y/3

    hypotenuse is sqrt(9+y^2)

    the answer I get is ln|(sqrt(9+y^2)/3) + (y/3)| + C
    the book answer is ln|sqrt(9 + y^2) + y| + C

    could someone please explain. I think I am having a misunderstanding with my substitutions.
    Last edited by ur5pointos2slo; April 5th 2009 at 09:16 AM. Reason: had to change my answer typed it wrong
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  2. #2
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    \ln\left|\frac{\sqrt{9+y^2}}{3} + \frac{y}{3}\right| + C

    \ln\left|\frac{\sqrt{9+y^2} + y}{3}\right| + C

    \ln\left|\sqrt{9+y^2} + y\right| - \ln{3} + C



    ... -\ln{3} + C is just another C

    \ln\left|\sqrt{9+y^2} + y\right| + C
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