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Integrating a tan function

  1. Mar 29, 2012 #1
    Hello.

    I was trying to solve Lagrangian equation and I manage to reduce second order differential equation that I got:

    [itex]\ddot{\varphi}[/itex]+[itex]\alpha[/itex][itex]\frac{tan\varphi}{cos^{2}\varphi}[/itex]=0;

    where [itex]\alpha[/itex] is a constant,

    to first order differential equation:

    [itex]\dot{\varphi}^{2}[/itex]+ [itex]\alpha[/itex][itex]tan^{2}\varphi[/itex] -C=0;

    where C is integration constant and from starting conditions I calculated it to be:
    C= 8[itex]\alpha[/itex].


    Now all I have left is this integral to solve:

    [itex]\int[/itex] [itex]\frac{d\varphi}{\sqrt{8-tan^{2}\varphi}}[/itex]

    But I can't find the right substitution.


    I did try using some trigonometric identities to make this integral easier to solve or familiar but I didn't manage to get anywhere with it.
     
  2. jcsd
  3. Mar 29, 2012 #2

    MathematicalPhysicist

    User Avatar
    Gold Member

    The integral you want to calculate is [itex]\int\frac{d\phi}{\sqrt{1-tan^2(\phi)}}[/itex] (obviously your integral we can factor out the 8 by suitable change of variables.

    so use the fact that 1-tan^2(phi)= (cos^2()-sin^2())/cos^2()=cos(2phi)/cos^2()
    after taking the sqrt and reciprocal you get:
    cos(phi)dphi/sqrt(cos(2 phi)) = sqrt((cos(2 phi)+1)/(2cos(2 phi))) dphi
    Now put cos(2 phi) = t to get an integral of the form (without the right factors obviously :-)):

    -sqrt((t+1)/t) sqrt(1/(1-t^2) dt = sqrt(1/(t(1-t)) dt = sqrt(1/(-(t-1/2)^2+1/4)) dt

    and that's look like a nice integral of the form:
    ds/sqrt(1-s^2)

    which its integral is?
     
  4. Mar 30, 2012 #3
    Thank you for your answer.
    You explained it very nicely.
    And the solution for that integral is arcussin(s).

    But...
    I tried substituting [itex]\frac{tan\varphi}{\sqrt{8}}[/itex]=tan[itex]\theta[/itex] but I didn't get desired result because I got this:
    d[itex]\varphi[/itex]=[itex]\frac{\sqrt{8}}{1-4cos^{2}\theta}[/itex]d[itex]\theta[/itex]

    But I may have made the mistake somewhere (and I made many stupid mistakes today).
     
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