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Math Help - integral (cos(x))/1+4sin^(2)x

  1. #1
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    integral (cos(x))/1+4sin^(2)x

    I would be grateful if someone would have a look at the working on this link.I cant paste picture here or save in correct format and upload..
    http://www.wolframalpha.com/input/po...&s=41&button=1

    could someone please talk me thru this problem???
    I understand to use u and du but am unsure where the values 1/2 and the number 2 in 2u come from.In my book it says to use a compensatory value.Im guessin here that 2 compensates 1/2 or is it the other way round? I will be using arctan rather than tan^-1.Any help will be great.Thanks
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  2. #2
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    Re: integral (cos(x))/1+4sin^(2)x

    Look at this.
    Click the show steps button.
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    Re: integral (cos(x))/1+4sin^(2)x

    u = \sin{x} \implies du = \cos{x}dx

    That transforms the integral into \int\frac{1}{1+4u^2}du
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    Re: integral (cos(x))/1+4sin^(2)x

    Thanks Plato.Yes thats where i tried initially but im still unsure about the numbers 2 and 1/2.Im a little lost with it......
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    Re: integral (cos(x))/1+4sin^(2)x

    Thanks Ridley.Im a little unsure about using the numbers 2 and 1/2.Plus i cant see du in the solution.Many thanks.
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    Re: integral (cos(x))/1+4sin^(2)x

    You can make another variable change if you want to.

    y = 2u gives us dy = 2du and the integral

    \frac{1}{2}\int\frac{1}{1+y^2}dy=\frac{1}{2} \arctan{(y)}+C=
    =\frac{1}{2}\arctan{(2u)}+C=\frac{1}{2} \arctan {(2\sin{x})}+C
    Last edited by Ridley; March 25th 2012 at 11:36 AM.
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    Re: integral (cos(x))/1+4sin^(2)x

    but where does the cos x dissapear to in the equasion??? I must be having a really dumb day.
    Thanks
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    Re: integral (cos(x))/1+4sin^(2)x

    Could you show me where the cos x disapears to please???Thanks
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    Re: integral (cos(x))/1+4sin^(2)x

    That terms disappears when you do this variable change: u = \sin x \implies du = \cos x dx \implies dx = \frac{du}{\cos x}

    \int \frac{\cos x}{1 + 4\sin ^2{x}}dx=\int \frac{\cos x}{1 + 4u^2}\cdot \frac{du}{\cos x}=\int\frac{1}{1+4u^2}du
    Last edited by Ridley; March 26th 2012 at 01:10 PM.
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    Re: integral (cos(x))/1+4sin^(2)x

    thanks for your pointers.the arrangement for dx made things clear.youre a star.thanks again.
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