Printable View

- Jan 27th 2009, 03:49 PMsobadinevaluate definite integral
- Jan 27th 2009, 06:05 PMKrizalid
Put $\displaystyle u={{x}^{2}}+\sqrt{{{x}^{4}}-3}.$

- Jan 27th 2009, 06:08 PMJhevon
alternatively, you could use a trig substitution. let $\displaystyle x^2 = \sqrt{3} \sec \theta$

- Jan 27th 2009, 06:17 PMsobadinumm
im still confused, how would i use this http://www.mathhelpforum.com/math-he...48c6e6bb-1.gif) or http://www.mathhelpforum.com/math-he...b444a9c7-1.gif

hey Jhevon you a Jamaican man...bless i from st. kitts - Jan 27th 2009, 06:28 PMJhevon
- Jan 27th 2009, 06:42 PMsobadin
yes just not at this level of difficulty

- Jan 27th 2009, 06:47 PMJhevon
the important thing is to be able to recognize when you can use it. the method is essentially the same in all cases.

for my method:

$\displaystyle x^2 = \sqrt{3} \sec t$

$\displaystyle \Rightarrow 2x~dx = \sqrt{3} \sec t \tan t ~dt$

$\displaystyle \Rightarrow x~dx = \frac {\sqrt{3}}2 \sec t \tan t ~dt$

so our integral becomes:

$\displaystyle \frac {\sqrt{3}}2 \int \frac {\sec t \tan t}{\sqrt{3 \sec^2 t - 3}}~dt$

this simplifies to

$\displaystyle \frac 12 \int \sec t~dt$

now can you finish up? don't forget this is a definite integral, so you have to deal with the limits at some point - Jan 27th 2009, 08:00 PMbecker89
hope its clear!

- Jan 27th 2009, 08:11 PMKrizalid
Don't complicate to much, note that,

$\displaystyle u={{x}^{2}}+\sqrt{{{x}^{4}}-3}\implies du=\left( 2x+\frac{2{{x}^{3}}}{\sqrt{{{x}^{4}}-3}} \right)\,dx\implies \frac{du}{2u}=\frac{x}{\sqrt{{{x}^{4}}-3}}\,dx.$ - Jan 27th 2009, 08:20 PMbecker89
you`re right my friend,

but luckily all roads lead to Rome.

i`m not too much familiar with the subject,which explains the length :]