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Homework Help: The substitution u=tan(x/2)

  1. Mar 29, 2008 #1
    I'm looking to solve this integral with the tan(x/2) substitution but so far, I don't know what to do.

    1. The problem statement, all variables and given/known data


    2. Relevant equations


    3. The attempt at a solution

    Well, using the tan(x/2) substitution, I get that:


    So I get:

    Now I don't know what to do.

    Another method I tried is multiplying by conjugate:








    2[tex]\sqrt{sinx+1}[/tex]+ C

    Obviously, this is the wrong answer but I don't see where I went wrong.

    Which way should I approach this problem?
  2. jcsd
  3. Mar 29, 2008 #2


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    Hi noblerare! :smile:

    Actually, you've gone too far.

    tan(x/2) can be useful just for simplifying a formula … you don't necessarily have to change the vatiable of integration also.

    In this case, just stop at √(1 - sinx) = √(1 - u)^2/√(1 + u^2).

    Now translate it back to x … it's cos(x/2) - sin(x/2).

    (to check, note that 1 - sinx = cos^2(x/2) - 2cos(x/2)sin(x/2) + sin^2(x/2) !).

    So ∫√(1 - sinx)dx = √(cos(x/2) - sin(x/2))dx. :smile:
  4. Mar 29, 2008 #3
    Why do u think that this is the wrong answer, it looks fine to me?


    If the answer in the book is different from this one, they should be equivalent, because nothing is wrong with this.
    Last edited: Mar 29, 2008
  5. Mar 29, 2008 #4
    I'm sorry, tiny-tim, I still don't quite understand what you're trying to say

    I understand this
    So far, I left my integral as so...

    how do I change it back to x? I also don't see how cos^2(x/2) -2cos(x/2)sin(x/2) + sin^2(x/2) comes out to 1-sinx
  6. Mar 29, 2008 #5

    Your answer is correct, what else are you looking for??
  7. Mar 30, 2008 #6


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    … standard trig formulas …

    Because √(1 - u)^2/√(1 + u^2) = (1 - u)/√(1 + u^2)
    = (1 - tan(x/2))cos(x/2)
    = cos(x/2) - sin(x/2). :smile:
    Hi noblerare! :smile:

    Well, cos^2 + sin^2 = 1.

    And sin2x = 2.sinx.cosx. :smile:

    You really need to learn these formulas: :wink:

    sin2x = 2.sinx.cosx = 2tanx.cos^2(x)
    cos2x = cos^2(x) - sin^2(x)
    1 + cos2x = 2cos^2(x)
    1 - cos2x = 2sin^2(x)​
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