Results 1 to 2 of 2

Math Help - Integration problem, please help.

  1. #1
    Newbie
    Joined
    Apr 2012
    From
    California
    Posts
    16

    Integration problem, please help.

    the integral from pi/4 to pi/3 of (tanx/(cosx^2(6+tanx^2)^.5)dx
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,964
    Thanks
    1008

    Re: Integration problem, please help.

    If it's \displaystyle \begin{align*} \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}{\frac{\tan{(x  )}}{\cos^2{(x)}\sqrt{6 + \tan^2{(x)}}}} \end{align*}, notice that \displaystyle \begin{align*} \frac{d}{dx} \left[ \tan{(x)} \right] = \frac{1}{\cos^2{(x)}} \end{align*}, so a substitution of the form \displaystyle \begin{align*} u = \tan{(x)} \implies du = \frac{1}{\cos^2{(x)}}\,dx \end{align*} is appropriate. Notice that when \displaystyle \begin{align*} x = \frac{\pi}{4} , \, u = 1 \end{align*} and when \displaystyle \begin{align*} x = \frac{\pi}{3} , \, u = \sqrt{3} \end{align*}, and the integral becomes

    \displaystyle \begin{align*} \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}{\frac{\tan{(x  )}}{\cos^2{(x)}\sqrt{6 + \tan^2{(x)}}}} = \int_1^{\sqrt{3}}{\frac{u}{\sqrt{6 + u^2}}\,du} \end{align*}

    Can you go from here?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. MIT INTEGRATION Bee problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 8th 2012, 02:13 AM
  2. integration problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 22nd 2010, 06:20 PM
  3. Replies: 2
    Last Post: February 19th 2010, 10:55 AM
  4. Integration problem
    Posted in the Calculus Forum
    Replies: 5
    Last Post: January 14th 2010, 12:19 AM
  5. Integration problem?
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 11th 2009, 06:36 AM

Search Tags


/mathhelpforum @mathhelpforum