# Help on integral with Trig

1. Homework Statement

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

2. Homework Equations

n/a

3. 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.
1. Homework Statement

2. Homework Equations

3. The Attempt at a Solution

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

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 )

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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.

So I'll have:

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

Which I simplify to:

$$\frac{1}{\sqrt{2}}$$$$\int$$cos(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?

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.

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