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Homework Help: Help on integral with Trig

  1. Mar 25, 2008 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    n/a

    3. The attempt at a solution

    Well I tried:
    u=cosx
    x=arccosx
    dx=[tex]\frac{-du}{\sqrt{1-u^2}}[/tex]

    Plugging them back into the integral gives me:

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

    I don't know where to go from there.

    I've also tried:

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

    or:

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

    But I don't know where to go from any of the above attempts. Any help would be greatly appreciated.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 25, 2008 #2
    Ok ... now I need a hint, lol.
     
  4. Mar 25, 2008 #3
    I figured it out by mere chance. Note that [tex]\frac{1}{\sqrt{1+cos(x)}}=\frac{1}{\sqrt{2}cos(x/2)}[/tex] and that [tex]cos(x)=cos^2(x/2)-sin^2(x/2)[/tex]. I used the half angle and double angle formulas, respectively. Wikipedia had the answer to [tex]\int sin(ax)tan(ax)dx[/tex] ( http://en.wikipedia.org/wiki/List_of_integrals_of_trigonometric_functions )
     
    Last edited: Mar 25, 2008
  5. Mar 25, 2008 #4
    wow cool! i'm glad to see that you figured it out, and I understand how you got the two equations you wrote but I don't see what I should do to solve the problem.
     
  6. Mar 25, 2008 #5
    So I'll have:

    [tex]\int[/tex][tex]\frac{cos^2(x/2)-sin^2(x/2)}{\sqrt{2}cos(x/2)}[/tex]

    Which I simplify to:

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

    Is that what I'm supposed to do?

    Where did the sin(ax)tan(ax)dx thing come from?
     
  7. Mar 25, 2008 #6
    So the left integral is easily solvable, but the right integral needs a little work, and I just pointed out that you can rewrite the right side's terms as the product of a sine and tangent.
     
  8. Mar 25, 2008 #7
    ah...i see, wow thanks, jhicks!
     
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