Results 1 to 5 of 5

Math Help - Trig integral

  1. #1
    Senior Member
    Joined
    Apr 2010
    Posts
    487

    Trig integral

    The question:

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

    My attempt:
    It looks similar to the derivative of inverse tan, so dividing the denominator by 9:

    \frac{1}{9}\int{\frac{1}{(\frac{x}{3})^2 + 1} \ dx}
    = \frac{1}{9}tan(\frac{x}{3})^{-1} + C

    However it doesn't seem to match what Wolfram is telling me. I'm sure I've made a silly mistake, but I'm not seeing it. Any help would be great. Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2010
    From
    Staten Island, NY
    Posts
    451
    Thanks
    2
    u=x/3. So du=1/3 dx.

    So the coefficient of arctan in the solution should be 1/3, not 1/9.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Apr 2010
    Posts
    487
    Thank you.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,835
    Thanks
    1604
    \displaystyle \int{\frac{1}{x^2+9}\,dx}.

    Make the substitution \displaystyle x = 3\tan{\theta} so that \displaystyle dx = 3\sec^2{\theta}\,d\theta and the integral becomes

    \displaystyle \int{\left[\frac{1}{(3\tan{\theta})^2 + 9}\right]\,3\sec^2{\theta}\,d\theta}

    \displaystyle = \int{\frac{3\sec^2{\theta}}{9\tan^2{\theta} + 9}\,d\theta}

    \displaystyle = \int{\frac{3\sec^2{\theta}}{9(\tan^2{\theta}+1)}\,  d\theta}

    \displaystyle = \int{\frac{\sec^2{\theta}}{3\sec^2{\theta}}\,d\the  ta}

    \displaystyle = \int{\frac{1}{3}\,d\theta}

    \displaystyle = \frac{\theta}{3} + C

    \displaystyle = \frac{\arctan{\left(\frac{x}{3}\right)}}{3} + C.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Apr 2010
    Posts
    487
    Ahh, I didn't consider that. Thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. A Trig Integral
    Posted in the Calculus Forum
    Replies: 5
    Last Post: January 17th 2011, 04:08 PM
  2. Trig Integral
    Posted in the Calculus Forum
    Replies: 7
    Last Post: January 1st 2011, 02:06 PM
  3. Help with a trig integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 2nd 2010, 01:43 PM
  4. trig integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 25th 2009, 05:35 AM
  5. trig integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 18th 2009, 09:51 PM

Search Tags


/mathhelpforum @mathhelpforum