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Math Help - An integral with trigonometric functions

  1. #1
    Junior Member gusztav's Avatar
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    An integral with trigonometric functions

    Hello,

    I'm trying to solve the following integral:

    \int \frac{\cos x}{\sqrt{\sin 2x}} dx



    Here's what I've tried:

    \int \frac{\cos x}{\sqrt{\sin 2x}} dx=

    \int \frac{\cos x}{\sqrt{2 \sin x  \cos x}} dx=

    \frac{1}{\sqrt{2}} \int \frac{\sqrt{\cos x}}{\sqrt{\sin x}} dx=

    universal substitution; t=\tan \frac{x}{2}; \sin x=\frac{2t}{1+t^2}; \cos x=\frac{1-t^2}{1+t^2}; dx=\frac{2dt}{1+t^2}

    ...and then I get

    \int \sqrt{\frac{1-t^2}{t}}  \cdot \frac{dt}{1+t^2}

    Can you give me a clue how to tackle this integral?

    Thanks!
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    This May be helpful

    I don't think this has an antiderivatitive, but It does look alot like the Beta function or the Incomplete Beta function.

    Beta function - Wikipedia, the free encyclopedia

    Do you have limits of integration or do you need an antiderivative?
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  3. #3
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    Quote Originally Posted by gusztav View Post
    Hello,

    I'm trying to solve the following integral:

    \int \frac{\cos x}{\sqrt{\sin 2x}} dx


    Can you give me a clue how to tackle this integral?

    Thanks!
    Here's what I'd do: Yeah, I'd try some things first but if I couldn't figure it out, I'd input it into Mathematica so that I could at least figure out how to arrive at the answer:

    \int \frac{\cos(x)}{\sqrt{\sin(2x)}}dx=-\frac{cos(x)\sqrt{\sin(2x)}}{3 \sqrt{\sin(x)}}\sum_{k=0}^{\infty}\frac{(3/4)_k (3/4)_k}{(7/4)_k}\frac{\cos^{2k}(x)}{k!}

    At this point I'd ask myself, why is Mathematica reporting this? Can it be that there is no finite antiderivative? Mathematica is sometimes wrong of course and frequently expresses answers in "uncommon" ways. If there is no finite answer, then I'd try and figure out how did Mathematica come up with that answer (which is in terms of a hypergeometric series)?
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  4. #4
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    Quote Originally Posted by shawsend View Post
    Here's what I'd do: Yeah, I'd try some things first but if I couldn't figure it out, I'd input it into Mathematica so that I could at least figure out how to arrive at the answer:

    \int \frac{\cos(x)}{\sqrt{\sin(2x)}}dx=-\frac{cos(x)\sqrt{\sin(2x)}}{3 \sqrt{\sin(x)}}\sum_{k=0}^{\infty}\frac{(3/4)_k (3/4)_k}{(7/4)_k}\frac{\cos^{2k}(x)}{k!}

    At this point I'd ask myself, why is Mathematica reporting this? Can it be that there is no finite antiderivative? Mathematica is sometimes wrong of course and frequently expresses answers in "uncommon" ways. If there is no finite answer, then I'd try and figure out how did Mathematica come up with that answer (which is in terms of a hypergeometric series)?
    What I find amazing is that I can differentiate that expression in Mathematica and it returns the integrand. How is it doing that? I just don't understand big business.
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  5. #5
    Senior Member pankaj's Avatar
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    \frac{1}{2}\int \frac{2\cos x}{\sqrt{\sin 2x}}dx

    =\frac{1}{2}\int\frac{\cos x+\sin x+\cos x-\sin x}{\sqrt{\sin 2x}}dx

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

    Now,put t=\sin x-\cos x and u=\sin x+\cos x

    =\frac{1}{2}\int\frac{dt}{\sqrt{1-t^2}}+\frac{1}{2}\int\frac{du}{\sqrt{u^2-1}}

    =\frac{1}{2}sin^{-1}t+\frac{1}{2}ln|u+\sqrt{u^2-1}|+c

    =\frac{1}{2}sin^{-1}(\sin x-\cos x)+\frac{1}{2}ln|(\sin x+\cos x)+\sqrt{\sin 2x}|+c
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  6. #6
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    Can {\sqrt {2\tan x}} +c be the solution?

    \frac{1}{\sqrt{2}} \int \frac{\sqrt{\cos x}}{\sqrt{\sin x}} dx

    \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt {\tan x}}

    \frac{1}{\sqrt{2}} \int ({\tan x})^{- \frac{1}{2}}

    \frac{2(\tan x)^{\frac {1}{2}}}{\sqrt{2}}
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  7. #7
    Math Engineering Student
    Krizalid's Avatar
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    that's wrong, check the basics of integration.
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  8. #8
    Super Member Random Variable's Avatar
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    Quote Originally Posted by chengbin View Post
    Can {\sqrt {2\tan x}} +c be the solution?

    \frac{1}{\sqrt{2}} \int \frac{\sqrt{\cos x}}{\sqrt{\sin x}} dx

    \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt {\tan x}}

    \frac{1}{\sqrt{2}} \int ({\tan x})^{- \frac{1}{2}}

    \frac{2(\tan x)^{\frac {1}{2}}}{\sqrt{2}}
    You could make the substiution  u=tan(x) , but then you have

     \frac {1}{\sqrt{2}} \int u^{- \frac {1}{2}}cos^{2}(x)du

    which isn't very helpful
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