Integral of sqrt(1+cos x)
The radical has stumped me on this. Can anybody help?
THE BEST way :
we will multiply and devide $\displaystyle \sqrt{1-cosx}$ into the integral .
$\displaystyle \int \sqrt{1+cosx} * \frac{\sqrt{1-cosx}}{\sqrt{1-cosx}} dx
$
and now from twofold we have :
$\displaystyle \int \frac{|sinx|}{\sqrt{1-cosx}} dx $
and then u = 1-cosx and du = sinx and :
$\displaystyle \int \frac{du}{\sqrt{u}} = 2\sqrt{u} = 2\sqrt{1-cosx} + C$
We have indeterminate Integral . so for |sinx| we can suppose that
$\displaystyle 0<x<\pi$ .
and for $\displaystyle \pi<x<2\pi$ we should multiply -1 to answer .
Please be careful.
$\displaystyle \sqrt{2 \cos^2 \left(\frac{x}{2}\right)} = \sqrt{2} \left| \cos \left(\frac{x}{2}\right) \right| \ne \sqrt{2} \cos \left(\frac{x}{2}\right)$
The correct integral is
$\displaystyle \sqrt{2} \int \left| \cos \left(\frac{x}{2}\right) \right| \, dx$
I know it should be the absloute value ..
and we should consider the both cases when its poisitve and when its negative ..
The general case is consider the positive one ..
I think you know what I mean ..
The problem here is that I do not know how to say what I mean in English
You know the identity:
$\displaystyle cos(2A) = 2cos^2(A) - 1$
In the form you asked, it's like this:
$\displaystyle 2cos^2(A) = cos(2A) + 1$
Now, here your 2A = x. So, A = x/2.
Replace that in the identity:
$\displaystyle 2cos^2(\frac{x}{2}) = cos(x) + 1$
There you are