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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

    [tex]\int[/tex][tex]\sqrt{1-sinx}dx[/tex]

    2. Relevant equations

    u=tan(x/2)

    3. The attempt at a solution

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

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

    So I get:
    [tex]\int\frac{2(u-1)du}{(1+u^2)^(3/2)}[/tex]

    Now I don't know what to do.

    Another method I tried is multiplying by conjugate:

    [tex]\int[/tex][tex]\sqrt{1-sinx}[/tex][tex]\frac{\sqrt{1+sinx}}{\sqrt{1+sinx}}[/tex]dx

    [tex]\int[/tex][tex]\frac{\sqrt{1-sin^2(x)}}{\sqrt{1+sinx}}[/tex]dx

    [tex]\int[/tex][tex]\frac{cosxdx}{\sqrt{1+sinx}}[/tex]dx

    u=sinx
    du=cosxdx

    [tex]\int[/tex][tex]\frac{du}{\sqrt{1+u}}[/tex]

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

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

    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

    tiny-tim

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    Homework Helper

    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?

    [tex](2\sqrt{sinx+1}+C)'=2\frac{1}{2}\frac{cosx}{\sqrt{1+sinx}}=\frac{cosx}{\sqrt{1+sinx}}*\frac{\sqrt{1-sinx}}{\sqrt{1-sinx}}=\frac{cosx\sqrt{1-sinx}}{\sqrt{1-sin^{2}x}}=\frac{cosx\sqrt{1-sinx}}{cosx}=\sqrt{1-sinx}[/tex]

    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...

    but
    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
    Noblerare,

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

    tiny-tim

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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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