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    how to integrate :dx/(cosx-cos5x)...thx
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    (Original post by liviabruma)
    how to integrate :dx/(cosx-cos5x)...thx
    Is that \displaystyle\int \dfrac{dx}{\cos(x) - \cos(5x)}?
    Because it's really complicated
    Wolfram alpha tends to overcomplicate things, so I simplified as far as I could and fed it in and the answer looks horrible
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    it's surely got to be

    \displaystyle \int\frac{dx}{cosx-cos^5x}
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    it depends on if the 5 is a power or an integer?
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    If you could LaTeX your maths, that would be good - I assume you mean \int \frac{1}{\cos(x)-\cos(5x)} dx. I would strongly recommend avoiding doing that integral - it's truly disgusting, and the answer is \displaystyle  \int \frac{1}{\cos (x)-\cos (5 x)} \, dx=\frac{1}{36} \left(-3 \csc (x)+8 \sqrt{3} 

\tanh ^{-1}\left(\frac{\tan   \left(\frac{x}{2}\right)}{\sqrt{  3}}\right)+8 \sqrt{3} \tanh {-1}\left(\sqrt{3} \tan \left(\frac{x}{2}\right)\right)+  9 \log \left(\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)\right)-9 \log \left(\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)\right) \right)
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    What's with the neg? I was trying to clarify what the OP was asking.
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    (Original post by Smaug123)
    If you could LaTeX your maths, that would be good - I assume you mean \int \frac{1}{\cos(x)-\cos(5x)} dx. I would strongly recommend avoiding doing that integral - it's truly disgusting, and the answer is \displaystyle  \int \frac{1}{\cos (x)-\cos (5 x)} \, dx=\frac{1}{36} \left(-3 \csc (x)+8 \sqrt{3} 

\tanh ^{-1}\left(\frac{\tan   \left(\frac{x}{2}\right)}{\sqrt{  3}}\right)+8 \sqrt{3} \tanh {-1}\left(\sqrt{3} \tan \left(\frac{x}{2}\right)\right)+  9 \log \left(\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)\right)-9 \log \left(\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)\right) \right)
    And you latex'd that instead of doing?
    (Original post by joostan)
    What's with the neg? I was trying to clarify what the OP was asking.
    Me too...
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    (Original post by I am Ace)
    And you latex'd that instead of doing…
    Nah, I just shoved it into WolframAlpha in Mathematica, then used the rather handy function TeXForm on the resulting output. All of twenty seconds, and it was mostly waiting for the internet to hurry up.
    Having said that, I'm meant to be doing past papers, and this is good procrastination :P
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    (Original post by liviabruma)
    how to integrate :dx/(cosx-cos5x)...thx
    (Original post by joostan)
    Is that \displaystyle\int \dfrac{dx}{\cos(x) - \cos(5x)}?
    Because it's really complicated
    Wolfram alpha tends to overcomplicate things, so I simplified as far as I could and fed it in and the answer looks horrible
    Use complex numbers to expand \cos5x and then use the t substitution (x=2\arctan{t}).
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    (Original post by Jkn)
    Use complex numbers to expand \cos5x and then use the t substitution (x=2\arctan{t}).
    I just realised how many integrals of this type can be solved by complex number expansions!
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    (Original post by I am Ace)
    I just realised how many integrals of this type can be solved by complex number expansions!
    Mwuhahahaa

    Now you all must try this: \int_{-\infty}^{\infty}\left(\frac{sinx  }{x}\right)^ndx
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    The Tangent half angle sub can probably be used too.
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    (Original post by Jkn)
    Mwuhahahaa

    Now you all must try this: \int_{-\infty}^{\infty}\left(\frac{sinx  }{x}\right)^ndx
    Someone's going about just negging everyone who writes an equation...
    maybe he's a latex-ophobe
    Impossebleh,
    I'd give it a go (rough idea what I might do using that complex formula) but to be honest,

    AIN'T NOBODY GAT TIME FO DAT
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    (Original post by I am Ace)
    Someone's going about just negging everyone who writes an equation...
    maybe he's a latex-ophobe
    Impossebleh,
    I'd give it a go (rough idea what I might do using that complex formula) but to be honest,

    AIN'T NOBODY GAT TIME FO DAT
    Yeah, I was thinking that :lol: Odd! Let's balance it out and start giving everything on here the thumbs up! :lol:

    It's ****ing brutal! I can do it when n is small but it's so hard to do it for all n (I think it's possible though!) The solution of the first few values of n was posted by LotF on TpiT (using Laplace Tranforms... which are awesome and surprisingly easy once you learn them!)

    Also, whenever n is a natural number, it seems like the integral is a always in the form \frac{p\pi}{q} for positive integers p,q.
 
 
 
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