Struggling with Trig Integrals? Get Help Here!

[noblerare]

Homework Statement

\int\frac{cosxdx}{\sqrt{1+cosx}}

Homework Equations

n/a

The Attempt at a Solution

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

Plugging them back into the integral gives me:

-\int\frac{udu}{\sqrt{1-u^2}\sqrt{1+u}}

I don't know where to go from there.

I've also tried:

\int\frac{cosx\sqrt{1-cosx}dx}{sin^2x}

or:

\int\sqrt{\frac{cos^2x}{1+cosx}}dx

But I don't know where to go from any of the above attempts. Any help would be greatly appreciated.

 

[rocomath]

Ok ... now I need a hint, lol.

 

[jhicks]

I figured it out by mere chance. Note that \frac{1}{\sqrt{1+cos(x)}}=\frac{1}{\sqrt{2}cos(x/2)} and that cos(x)=cos^2(x/2)-sin^2(x/2). I used the half angle and double angle formulas, respectively. Wikipedia had the answer to \int sin(ax)tan(ax)dx ( http://en.wikipedia.org/wiki/List_of_integrals_of_trigonometric_functions )

 

[noblerare]

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.

 

[noblerare]

So I'll have:

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

Which I simplify to:

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

Is that what I'm supposed to do?

Where did the sin(ax)tan(ax)dx thing come from?

 

[jhicks]

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.

 

[noblerare]

ah...i see, wow thanks, jhicks!