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Math Help - Integral of sqrt(1+cos x)

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

    Integral of sqrt(1+cos x)

    The radical has stumped me on this. Can anybody help?
    Last edited by mr fantastic; June 3rd 2010 at 03:53 PM. Reason: Copied post title into main body of post.
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  2. #2
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    Try substituting x = cos^{-1}(u) and it will reduce to a known rational integral.
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  3. #3
    Super Member General's Avatar
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    Or:

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

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

    Now, substitute u=\frac{x}{2} ..
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  4. #4
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    THE BEST way :


    we will multiply and devide \sqrt{1-cosx} into the integral .

    \int \sqrt{1+cosx} * \frac{\sqrt{1-cosx}}{\sqrt{1-cosx}} dx<br />

    and now from twofold we have :

    \int \frac{|sinx|}{\sqrt{1-cosx}} dx

    and then u = 1-cosx and du = sinx and :

    \int \frac{du}{\sqrt{u}} = 2\sqrt{u} = 2\sqrt{1-cosx} + C


    We have indeterminate Integral . so for |sinx| we can suppose that

    0<x<\pi .

    and for \pi<x<2\pi we should multiply -1 to answer .
    Last edited by mr fantastic; June 3rd 2010 at 03:58 PM. Reason: Merged posts, minor consequential editing.
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  5. #5
    MHF Contributor Unknown008's Avatar
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    Quote Originally Posted by General View Post
    Or:

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

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

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

    Do we really need to use a substitution?

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

    as a reverse of \frac{d(sin(ax))}{dx} = a cos(ax)
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  6. #6
    Super Member General's Avatar
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    Quote Originally Posted by Unknown008 View Post
    Do we really need to use a substitution?

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

    as a reverse of \frac{d(sin(ax))}{dx} = a cos(ax)
    I know that formula ..
    But the substitution will give the OP a standard integral, namely, \int cos(u) \, du
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  7. #7
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    Quote Originally Posted by General View Post
    Or:

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

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

    Now, substitute u=\frac{x}{2} ..
    Please be careful.

    \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

    \sqrt{2} \int \left| \cos \left(\frac{x}{2}\right) \right| \, dx
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  8. #8
    Super Member General's Avatar
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    Quote Originally Posted by drumist View Post
    Please be careful.

    \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

    \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
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  9. #9
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    Quote Originally Posted by parkhid View Post
    THE BEST way :


    we will multiply and devide \sqrt{1-cosx} into the integral .

    \int \sqrt{1+cosx} * \frac{\sqrt{1-cosx}}{\sqrt{1-cosx}} dx<br />

    and now from twofold we have :

    \int \frac{|sinx|}{\sqrt{1-cosx}} dx

    and then u = 1-cosx and du = sinx and :

    \int \frac{du}{\sqrt{u}} = 2\sqrt{u} = 2\sqrt{1-cosx} + C


    We have indeterminate Integral . so for |sinx| we can suppose that

    0<x<\pi .

    and for \pi<x<2\pi we should multiply -1 to answer .
    Thanks Parkhid,
    I like your method - rationalize the numerator.
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  10. #10
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    Quote Originally Posted by parkhid View Post
    THE BEST way :


    we will multiply and devide \sqrt{1-cosx} into the integral .

    \int \sqrt{1+cosx} * \frac{\sqrt{1-cosx}}{\sqrt{1-cosx}} dx<br />

    and now from twofold we have :

    \int \frac{|sinx|}{\sqrt{1-cosx}} dx

    and then u = 1-cosx and du = sinx and :

    \int \frac{du}{\sqrt{u}} = 2\sqrt{u} = 2\sqrt{1-cosx} + C


    We have indeterminate Integral . so for |sinx| we can suppose that

    0<x<\pi .

    and for \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 u substitution...
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  11. #11
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    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?
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  12. #12
    MHF Contributor Unknown008's Avatar
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    You know the identity:

    cos(2A) = 2cos^2(A) - 1

    In the form you asked, it's like this:

    2cos^2(A) = cos(2A) + 1

    Now, here your 2A = x. So, A = x/2.

    Replace that in the identity:

    2cos^2(\frac{x}{2}) = cos(x) + 1

    There you are
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