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Help on integral with Trig

  • Thread starter noblerare
  • Start date
47
0
1. Homework Statement

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

2. Homework 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. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 

Answers and Replies

1,750
1
Ok ... now I need a hint, lol.
 
334
0
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:
47
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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.
 
47
0
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?
 
334
0
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.
 
47
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ah...i see, wow thanks, jhicks!
 

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