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

• Mar 25th 2012, 08:51 AM
bignaughtydog
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
• Mar 25th 2012, 09:03 AM
Plato
Re: integral (cos(x))/1+4sin^(2)x
Look at this.
Click the show steps button.
• Mar 25th 2012, 09:04 AM
Ridley
Re: integral (cos(x))/1+4sin^(2)x
$\displaystyle u = \sin{x} \implies du = \cos{x}dx$

That transforms the integral into $\displaystyle \int\frac{1}{1+4u^2}du$
• Mar 25th 2012, 09:46 AM
bignaughtydog
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......
• Mar 25th 2012, 09:47 AM
bignaughtydog
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.
• Mar 25th 2012, 10:32 AM
Ridley
Re: integral (cos(x))/1+4sin^(2)x
You can make another variable change if you want to.

$\displaystyle y = 2u$ gives us $\displaystyle dy = 2du$ and the integral

$\displaystyle \frac{1}{2}\int\frac{1}{1+y^2}dy=\frac{1}{2} \arctan{(y)}+C=$
$\displaystyle =\frac{1}{2}\arctan{(2u)}+C=\frac{1}{2} \arctan {(2\sin{x})}+C$
• Mar 26th 2012, 11:25 AM
bignaughtydog
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
• Mar 26th 2012, 11:26 AM
bignaughtydog
Re: integral (cos(x))/1+4sin^(2)x
Could you show me where the cos x disapears to please???Thanks
• Mar 26th 2012, 12:06 PM
Ridley
Re: integral (cos(x))/1+4sin^(2)x
That terms disappears when you do this variable change: $\displaystyle u = \sin x \implies du = \cos x dx \implies dx = \frac{du}{\cos x}$

$\displaystyle \int \frac{\cos x}{1 + 4\sin ^2{x}}dx=\int \frac{\cos x}{1 + 4u^2}\cdot \frac{du}{\cos x}=\int\frac{1}{1+4u^2}du$
• Mar 27th 2012, 11:06 AM
bignaughtydog
Re: integral (cos(x))/1+4sin^(2)x
thanks for your pointers.the arrangement for dx made things clear.youre a star.thanks again.