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Trig intergration

  1. Mar 30, 2004 #1
    help wat is mean by trigonometric substituations? how do i use it/? thx
     
  2. jcsd
  3. Mar 30, 2004 #2
    hah, another part of calc 2 that i've sorta forgotten.., let me get out my calc book and ill get back :)
     
  4. Mar 30, 2004 #3
    You make a substitution of the form x = some trig function (like sin(u) or tan(u)) and then use trig properties to simplify the integral to a manageable form. Then, after integrating, you make the inverse substitution into the solution to get your final answer.

    Look at the table at the bottom of Mathworld's entry and just make an up example.

    Trigonometric Substitution:
    http://mathworld.wolfram.com/TrigonometricSubstitution.html

    Better yet, here's a sample problem for you. Try it out.

    [tex]\int \frac{dx}{\sqrt{a^2 - x^2}} [/tex]

    cookiemonster
     
  5. Mar 30, 2004 #4
    Figured I should work out my own sample problem.

    [tex]\int \frac{dx}{\sqrt{a^2 - x^2}}[/tex]
    [tex]x = a\sin{\theta}[/tex]
    [tex]dx = a\cos{\theta}d\theta[/tex]
    [tex]\int\frac{a\cos{\theta}d\theta}{\sqrt{a^2(1-\sin^2{\theta})}}[/tex]
    [tex]\int\frac{a\cos{\theta}d\theta}{a\cos{\theta}}[/tex]
    [tex]\int d \theta[/tex]
    [tex]\theta = \arcsin{\frac{x}{a}}[/tex]

    cookiemonster
     
  6. Mar 30, 2004 #5
    i do remember that
    [tex]sin^2 + cos^2 = 1 [/tex]

    so there for, you can rearrange...

    ....... wait, im thinking of trig identities.

    hrm, maybe i shouldnt be doin this calc stuff.... i have too many brain farts.
     
  7. Mar 30, 2004 #6
    Trig identities are good. Keep going.

    You'll notice that that particular trig identity is used to get from the 4th step in my example to the 5th step.

    cookiemonster
     
  8. Mar 30, 2004 #7
    oh cool thx
    this was my problem in solving it

    [tex] \int \frac{1}{x^2\sqrt{1+x^2}} dx [/tex] --->

    [tex] x=tan{\theta} [/tex]
    my step after ur help

    [tex] \int \frac {1}{tan^2{\theta}sec{\theta}}dx [/tex]

    [tex] x=tan{\theta} [/tex]then [tex] dx=sec^2{\theta} d{\theta}[/tex]

    [tex] \int \frac {sec^2{\theta}}{Tan^2{\theta}sec{\theta}}d{\theta}[/tex]

    [tex]\int \frac {sec{\theta}}{tan^2{\theta}} d{\theta} [/tex]

    [tex]\frac{1}{2}\int \frac {2sec{\theta}}{1-sec^2{\theta}} d{\theta} [/tex]


    is this correct ? thx
     
    Last edited: Mar 30, 2004
  9. Mar 30, 2004 #8
    ??? wat happen to my latex graphic?
     
  10. Mar 30, 2004 #9
    You're working too hard.

    Try to simplify

    [tex]\int \frac {sec{\theta}}{tan^2{\theta}} d{\theta} [/tex]

    a bit. Once you do, there's an obvious substitution that completes the integral.

    cookiemonster
     
  11. Mar 31, 2004 #10
  12. Mar 31, 2004 #11
    What do you mean you can't simplify anymore?

    [tex]\int \frac{\sec{\theta}}{\tan^2{\theta}} \, d\theta = \int \frac{1}{\cos{\theta}} \cdot \frac{\cos^2{\theta}}{\sin^2{\theta}} \, d\theta = \int \frac{\cos{\theta}}{\sin^2{\theta}} \, d\theta[/tex]

    There's a substitution that lets you evaluate this integral...

    And which thread are you referring to?

    cookiemonster

    Edit: Gaah! Why won't LaTeX work?

    Edit^2: Guess it does work.
     
    Last edited: Mar 31, 2004
  13. Mar 31, 2004 #12
  14. Mar 31, 2004 #13
    That keeps going to the index of the General Math forum, not a specific thread. Which specific thread in General Math?

    cookiemonster
     
  15. Mar 31, 2004 #14
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