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Math Help - very basic integration problem

  1. #1
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    very basic integration problem

    Hi guys. I am trying to integrate \frac{1}{9-x^2}dx by trig substiution, but I'm not really sure if you can even do trig sub here since we are dealing with square roots. If I'm right, is there another method besides trig sub to evaluate this, even if it's not necessarily easier? This is what I got:

    \int\frac{1}{9-x^2}dx

    Let x = 3sin\theta

    So dx = 3 cos\theta d\theta
    Substitution yields \int \frac{3cos\theta}{9-9sin^2\theta}d\theta

    Simplification yields
     <br />
\frac{1}{3}\int \frac{cos\theta}{1-sin^2\theta}d\theta<br />

    or simply:
     <br /> <br />
\frac{1}{3}\int sec\theta d\theta<br />

    So now we have \frac{1}{3}ln|sec\theta+tan\theta| + C

    Using a right triangle to make substitutions, my final answer is

     <br />
\frac{1}{3}ln\left|\frac{x}{\sqrt{x^2-9}}+\frac{3}{\sqrt{x^2-9}}\right|+C<br />

    So again, even if this right, is there another way to evaluate this?
    Thanks for any help,
    James
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  2. #2
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    Quote Originally Posted by james121515 View Post
    Hi guys. I am trying to integrate \frac{1}{9-x^2}dx by trig substiution, but I'm not really sure if you can even do trig sub here since we are dealing with square roots. If I'm right, is there another method besides trig sub to evaluate this, even if it's not necessarily easier? This is what I got:

    \int\frac{1}{9-x^2}dx

    Let x = 3sin\theta

    So dx = 3 cos\theta d\theta
    Substitution yields \int \frac{3cos\theta}{9-9sin^2\theta}d\theta

    Simplification yields
     <br />
\frac{1}{3}\int \frac{cos\theta}{1-sin^2\theta}d\theta<br />

    or simply:
     <br /> <br />
\frac{1}{3}\int sec\theta d\theta<br />

    So now we have \frac{1}{3}ln|sec\theta+tan\theta| + C

    Using a right triangle to make substitutions, my final answer is

     <br />
\frac{1}{3}ln\left|\frac{x}{\sqrt{x^2-9}}+\frac{3}{\sqrt{x^2-9}}\right|+C<br />

    So again, even if this right, is there another way to evaluate this?
    Thanks for any help,
    James
    If I had to do this question I would use partial fractions:

    \frac1{9-x^2}=-\frac1{x^2-9}=-\left(\frac{A}{x+3} + \frac{B}{x-3} \right)

    which will yield A=\frac16 and B=-\frac16.
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  3. #3
    Super Member 11rdc11's Avatar
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    You could use partial fractions
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