# Thread: Integral of sqrt(1+cos x)

1. ## Integral of sqrt(1+cos x)

Integral of sqrt(1+cos x)

The radical has stumped me on this. Can anybody help?

2. Try substituting $\displaystyle x = cos^{-1}(u)$ and it will reduce to a known rational integral.

3. Or:

$\displaystyle \int \, \sqrt{1+cos(x)} \, dx=\int \, \sqrt{2 \, cos^2\left(\frac{x}{2}\right) }$

$\displaystyle =\sqrt{2} \, \int \, cos\left(\frac{x}{2}\right) \, dx$

Now, substitute $\displaystyle u=\frac{x}{2}$ ..

4. 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 .

5. Originally Posted by General
Or:

$\displaystyle \int \, \sqrt{1+cos(x)} \, dx=\int \, \sqrt{2 \, cos^2\left(\frac{x}{2}\right) }$

$\displaystyle =\sqrt{2} \, \int \, cos\left(\frac{x}{2}\right) \, dx$

Now, substitute $\displaystyle u=\frac{x}{2}$ ..

Do we really need to use a substitution?

$\displaystyle \int cos(ax) dx = \frac{1}{a} sin(ax) + c$

as a reverse of $\displaystyle \frac{d(sin(ax))}{dx} = a cos(ax)$

6. Originally Posted by Unknown008
Do we really need to use a substitution?

$\displaystyle \int cos(ax) dx = \frac{1}{a} sin(ax) + c$

as a reverse of $\displaystyle \frac{d(sin(ax))}{dx} = a cos(ax)$
I know that formula ..
But the substitution will give the OP a standard integral, namely, $\displaystyle \int cos(u) \, du$

7. Originally Posted by General
Or:

$\displaystyle \int \, \sqrt{1+cos(x)} \, dx=\int \, \sqrt{2 \, cos^2\left(\frac{x}{2}\right) }$

$\displaystyle =\sqrt{2} \, \int \, cos\left(\frac{x}{2}\right) \, dx$

Now, substitute $\displaystyle u=\frac{x}{2}$ ..

$\displaystyle \sqrt{2 \cos^2 \left(\frac{x}{2}\right)} = \sqrt{2} \left| \cos \left(\frac{x}{2}\right) \right| \ne \sqrt{2} \cos \left(\frac{x}{2}\right)$

The correct integral is

$\displaystyle \sqrt{2} \int \left| \cos \left(\frac{x}{2}\right) \right| \, dx$

8. Originally Posted by drumist

$\displaystyle \sqrt{2 \cos^2 \left(\frac{x}{2}\right)} = \sqrt{2} \left| \cos \left(\frac{x}{2}\right) \right| \ne \sqrt{2} \cos \left(\frac{x}{2}\right)$

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

9. Originally Posted by parkhid
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 .
Thanks Parkhid,
I like your method - rationalize the numerator.

10. Originally Posted by parkhid
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 .
I don't see why this is the "best" way...

I think converting to a single cosine value via the half angle identity is an easier and "better" integral than getting it to a form that requires a $\displaystyle u$ substitution...

11. Yes - I like the half angle identity also. One question: from cos^2 x = 1/2(1+cos 2x) I get 2cos^2 x = 1+cos 2x. How do you reduce 1+cos 2x to simply 1+cos x. Is it just a matter of dividing each angle by 2?

12. 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

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