1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Trig integral help

  1. Apr 13, 2008 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations


    3. The attempt at a solution

    Here's what I've tried:



    I then used partial fractions to get:


    The left side is simply arctanu while I used trig substitution to solve the right side

    [tex]\frac{1}{2}[/tex]arctan(tanx)+[tex]\frac{1}{2}[/tex]ln(csc[tex]\theta[/tex]+cot[tex]\theta[/tex]) + C

    [tex]\frac{1}{2}[/tex]x + [tex]\frac{1}{2}[/tex]ln([tex]\frac{1}{\sqrt{1-u^2}}[/tex]+[tex]\frac{u}{\sqrt{1-u^2}}[/tex]) + C

    With more re-substitution, I end up with:


    For some reason, this is incorrect because when I try to take the derivative of that, I do not end up with [tex]\frac{1}{1-tan(x)^2}[/tex]

    Is there something I'm doing wrong? What should I do?
  2. jcsd
  3. Apr 13, 2008 #2
    I'd like to see your work with the partial fractions.

    Did it look like ... [tex]\frac{1}{(1+u^2)(1-u^2)}=\frac{Ax+B}{1+u^2}+\frac{Cx+D}{1-u^2}[/tex] ... ?
  4. Apr 13, 2008 #3


    User Avatar
    Homework Helper

    Shouldn't it be

    [tex]\frac{1}{(1+u^2)(1-u)(1+u)} \equiv \frac{Au+B}{1+u^2}+\frac{C}{1-u}+ \frac{D}{1+u}[/tex]

    or does it not matter about the [itex]1-u^2[/itex]?
  5. Apr 13, 2008 #4
    Good question, I overlooked that part.
  6. Apr 13, 2008 #5


    User Avatar

    I think his decomposition is valid (it certainly combines back correctly).

    Try using this: [tex]\int \frac{1}{a^{2}-u^{2}}du = \frac{1}{2a}\ln{\frac{|u+a|}{|u-a|}}+c[/tex]

    For your sake I hope you are using a calculator when you work it out.
  7. Apr 13, 2008 #6
    To rocomath: yes, my partial fractions work did look like that. I did not try it the way rock.freak667 did it. Does it matter?

    To exk: Thanks for the help, but could you maybe explain how that you got that integral? And yes, I am using a calculator : D
  8. Apr 14, 2008 #7

    Gib Z

    User Avatar
    Homework Helper

    He got that integral through partial fractions again, and then using [itex]\log b - \log a = \log \left( \frac{b}{a}\right)[/itex]
  9. Apr 14, 2008 #8


    User Avatar
    Homework Helper

    To do this integral, use partial fractions and integrate.
  10. Apr 14, 2008 #9
    How do I make

    [tex]\int[/tex][tex]\frac{dx}{1-tan^2x}[/tex] look like [tex]\int[/tex][tex]\frac{du}{a^2-u^2}[/tex] though?

    Do I need to subtitute u=tanx? If so, I don't know how to change the variable of integration so that it matches that the second integral

    Or can I simply take tanx as plug it into that equation?
  11. Apr 14, 2008 #10


    User Avatar
    Homework Helper

    Wow! noblerare, your solution is correct!

    [tex]\int\frac{dx}{1-\tan^2 x}=\frac{1}{2}x+\frac{1}{2}\ln\left( 1+ \tan x\right)-\frac{1}{2}\ln \sqrt{1- \tan^2 x}+C[/tex]


    [tex]\frac{d}{dx}\left[\frac{1}{2}x+\frac{1}{2}\ln\left( 1+ \tan x\right)-\frac{1}{2}\ln \sqrt{1- \tan^2 x}+C\right] = \frac{1}{2}+\frac{\sec^2 x}{2(1+ \tan x)}+\frac{\sec^2 x\tan x}{2(1- \tan^2 x)} = \frac{1}{1- \tan^2 x}[/tex]
  12. Apr 14, 2008 #11


    User Avatar

    You already had it in that form following your substitution. You have:


    If a=1 and u=u, then the integral in the second part is in the form you want.
  13. Apr 15, 2008 #12
    ah...i see,i got it; okay, thanks everybody!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Trig integral help