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Trigonomatric substitutions

  1. Jun 15, 2007 #1

    Recently I studied trigonometric substitution way to solve many forms of Integrals. But since I'm new at this I can't get the intuition when to use what. When it goes out of normal "formulas" , I am really lost.

    For example,
    S(----------- )

    I solved it, but had to use too many formulas for sinarctanx and cosarctgx (which are almost exact). anyway, At first I substituted x=tant, and later on I got:

    S sin^-2(t)*cos(t)dt

    and therefore i had to substitute z=sint

    and in the end what I got is -1/sinarctgx+C.

    (I checked it, if we will check what d(-1/sinarctanx) is equal to, you will get the original question, therefore its right)

    And that's NOT A BEAUTIFUL FORM :(.

    If can someone suggest something better out of his experience, Ill appreciate it.

    Thanks in advance,
    But the way, I realized I can't copy from math-type to here (as expected, but that was my only guess).
    Can someone tell me how write the function in more proper way?

    Last edited: Jun 15, 2007
  2. jcsd
  3. Jun 16, 2007 #2
    I came acroos something similar, heres how i got around the problem, i hope this helps:

  4. Jun 16, 2007 #3
    Rofl, I had this question one exercise ago, and it caused me great pain try doing it by integration by parts.
    and now it looks so easy :P

    Anyway I can't take any conclusions from your very nice solution to my problem. (since in my sqrt you can't just transform it to a sin/cos/tan), you can however transform it to cosh and sinh with the sinh2+1=cosh^2, But I am really weak with those functions so I am not touching them.

    Thanks , But it doesn't help me a lot :(

    Any other suggestions?
  5. Jun 16, 2007 #4


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    Staff Emeritus
    Science Advisor
    Gold Member

    Do you know what sin(arctan(x)) is? Try drawing a right triangle, label one side x, the other side 1, then try and find which angle is arctan(x). Finding sin(arctan(x)) becomes pretty simple after that
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