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    Can someone help me integrate (cot x) / (rt(1-cos 2x)) please.

    I think it's possible to rewrite it as (1/ rt2) cosec x cot x or something very similar, but have no idea how to get there, or perform the subsequent integration.

    Any help greatly received.
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    I'm guessing rt is a constant so just take that outside of the integral

    double angle formula - cos 2x = 1-2sin^2(x)
    substitution, let u=sinx => du/cosx = dx

    Work that through and you should get a result
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    so getting from the first part to the second part, it is just algebraic manipulation, when there is a cos2x, what comes to mind?

    and to integrate it, it is a standard integral (if you remember what the derivative of the cosec, sec, and cot is)
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    (Original post by Ewan)
    I'm guessing rt is a constant so just take that outside of the integral

    double angle formula - cos 2x = 1-2sin^2(x)
    substitution, let u=sinx => du/cosx = dx

    Work that through and you should get a result
    rt is short for "square root".

    Thanks for the other points, I'll have another go!
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    (Original post by baffled_mathman)
    rt is short for "square root".

    Thanks for the other points, I'll have another go!
    Ah fair enough, don't need to edit my post in that case xD (you'll see why)
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    (Original post by 1729)
    so getting from the first part to the second part, it is just algebraic manipulation, when there is a cos2x, what comes to mind?

    and to integrate it, it is a standard integral (if you remember what the derivative of the cosec, sec, and cot is)

    I could rewrite is as cot x / (rt (1-2cos^2x-1) which gives cot x / -2 cos x.

    I'm not sure where to take it from the though. I can recall cot, sec and cosec derivatives, but they aren't helping yet!

    Thanks for your help.
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    writing cos2x as in what you did is one way, but is there another way of writing cos2x?
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    oh and btw, 1 - cos2x is not 1-2cos^2x-1



    (Original post by baffled_mathman)
    I could rewrite is as cot x / (rt (1-2cos^2x-1) which gives cot x / -2 cos x.

    I'm not sure where to take it from the though. I can recall cot, sec and cosec derivatives, but they aren't helping yet!

    Thanks for your help.
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    (Original post by baffled_mathman)
    I could rewrite is as cot x / (rt (1-2cos^2x-1) which gives cot x / -2 cos x.

    I'm not sure where to take it from the though. I can recall cot, sec and cosec derivatives, but they aren't helping yet!

    Thanks for your help.
    No you can't, check your minus signs.
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    (Original post by 1729)
    oh and btw, 1 - cos2x is not 1-2cos^2x-1
    (Original post by Ewan)
    No you can't, check your minus signs.
    Thanks, no idea what I was thinking there!

    (Original post by 1729)
    writing cos2x as in what you did is one way, but is there another way of writing cos2x?
    Thanks! I didn't think to use the cos 2x = 1-2sin^2 form. It then becomes cot x / (rt 2 sin x). Then take out 1/rt2 leaving cosec x cot x. The integral of which is -cosec x. So the solution is (-1/rt2 ) cosec x + c.

    I appreciate it people.
 
 
 
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