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

Re: integral (cos(x))/1+4sin^(2)x

Look at this.

Click the show steps button.

Re: integral (cos(x))/1+4sin^(2)x

That transforms the integral into

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......

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.

Re: integral (cos(x))/1+4sin^(2)x

You can make another variable change if you want to.

gives us and the integral

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

Re: integral (cos(x))/1+4sin^(2)x

Could you show me where the cos x disapears to please???Thanks

Re: integral (cos(x))/1+4sin^(2)x

That terms disappears when you do this variable change:

Re: integral (cos(x))/1+4sin^(2)x

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