# evaluate definite integral

• Jan 27th 2009, 03:49 PM
evaluate definite integral
• Jan 27th 2009, 06:05 PM
Krizalid
Put $\displaystyle u={{x}^{2}}+\sqrt{{{x}^{4}}-3}.$
• Jan 27th 2009, 06:08 PM
Jhevon
alternatively, you could use a trig substitution. let $\displaystyle x^2 = \sqrt{3} \sec \theta$
• Jan 27th 2009, 06:17 PM
umm
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 PM
Jhevon
Quote:

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

lets start with a basic question. do you know how to do integration by substitution and integration by trigonometric substitution?
• Jan 27th 2009, 06:42 PM
yes just not at this level of difficulty
• Jan 27th 2009, 06:47 PM
Jhevon
Quote:

yes just not at this level of difficulty

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 PM
becker89
hope its clear!
• Jan 27th 2009, 08:11 PM
Krizalid
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 PM
becker89
youre right my friend,