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Homework Help: Trig integral help

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

    [tex]\int[/tex][tex]\frac{dx}{1-tan^2(x)}[/tex]

    2. Relevant equations

    n/a

    3. The attempt at a solution

    Here's what I've tried:
    u=tanx
    x=arctanu

    dx=[tex]\frac{du}{1+u^2}[/tex]

    [tex]\int[/tex][tex]\frac{du}{(1+u^2)(1-u^2)}[/tex]

    I then used partial fractions to get:
    [tex]\int[/tex][tex]\frac{0.5}{1-u^2}[/tex]+[tex]\frac{0.5}{1+u^2}[/tex]du

    [tex]\frac{1}{2}[/tex][tex]\int[/tex][tex]\frac{du}{1+u^2}[/tex]+[tex]\frac{1}{2}[/tex][tex]\int\frac{du}{1-u^2}[/tex]

    The left side is simply arctanu while I used trig substitution to solve the right side
    [tex]\frac{1}{2}[/tex]arctanu-[tex]\frac{1}{2}[/tex][tex]\int[/tex]csc[tex]\theta[/tex]d[tex]\theta[/tex]

    [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:

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

    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

    rock.freak667

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    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

    exk

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    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

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    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

    benorin

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    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

    benorin

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    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]

    Check:

    [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

    exk

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

    [tex]
    \frac{1}{2}\int\frac{du}{1+u^2}+\frac{1}{2}\int\frac{du}{1-u^2}
    [/tex]

    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!
     
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