ATAR Notes: Forum

VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: Tea.bag on April 16, 2009, 08:23:55 pm

Title: integrate
Post by: Tea.bag on April 16, 2009, 08:23:55 pm: how do u integrate

$\int \frac {x^2}{\sqrt{x^2+1}}.dx$
Title: Re: integrate
Post by: vibs23 on April 16, 2009, 08:39:36 pm: use the process of substitution;

turn u = x^2 +1

And follow on from there
Title: Re: integrate
Post by: vibs23 on April 16, 2009, 08:42:29 pm: i cant post whole solution because i cant figure out how you type all the stuff in, but it shud be easy from there on....
Title: Re: integrate
Post by: shinny on April 16, 2009, 08:46:17 pm: I think everyone's missing the typo'd square root. I imagine it's meant to cover the entire denominator so that complicates things no?

EDIT: Pfft you deleted your post =P
Title: Re: integrate
Post by: Flaming_Arrow on April 16, 2009, 08:52:50 pm: i think he means

$\int \frac {x^2}{\sqrt{x^2+1}}.dx$
Title: Re: integrate
Post by: vibs23 on April 16, 2009, 08:55:22 pm: yeh i couldnt figure out what marshmallow meant, just turn everything inside the square root = u.

So the u equation shud look like

U-1/(U^(1/2))

Can someone tell me how you guys who do those special typing?
Title: Re: integrate
Post by: Tea.bag on April 16, 2009, 08:57:25 pm: yea..i know all that but what about the

dx = du/2x
Title: Re: integrate
Post by: Tea.bag on April 16, 2009, 08:59:33 pm: im guessin its something to do with hyperbolic functions?
Title: Re: integrate
Post by: enwiabe on April 16, 2009, 09:09:50 pm: yeah teabag

do it in parts

i was discussing it with mao on MSN, he knows latex so he'll be posting the soln
Title: Re: integrate
Post by: Mao on April 16, 2009, 09:29:41 pm: Let $x=\sinh \theta$

$\implies dx = \cosh \theta\; d\theta$

$\implies \theta = \sinh^{-1} x$

$\implies \cosh \theta = \sqrt{1 + \sinh^2 \theta} = \sqrt{1+x^2}$

$I = \int \frac{\sinh^2 \theta}{\cosh \theta}\cosh \theta\; d\theta = \int \sinh^2 \theta\; d\theta = \frac{1}{2} \int \cosh 2\theta - 1\; d\theta = \frac{1}{2}\left(\frac{1}{2}\sinh 2\theta - \theta \right) + C = \frac{1}{2}\left( \sinh \theta \cosh \theta - \theta \right) + C$

$\implies \int \frac{x^2}{\sqrt{1+x^2}}\; dx = \frac{x}{2}\left( \sqrt{1 + x^2} - \sinh^{-1}x\right) + C = \frac{x}{2}\sqrt{1+x^2} - \frac{1}{2}\log_e \left( x + \sqrt{x^2 + 1}\right) + C$

apologies for the delay, really rusty on this stuff, messed up negatives and identities everywhere.