# Math Help - evaluate definite integral

1. ## evaluate definite integral

2. Put $u={{x}^{2}}+\sqrt{{{x}^{4}}-3}.$

3. alternatively, you could use a trig substitution. let $x^2 = \sqrt{3} \sec \theta$

4. ## umm

im still confused, how would i use this ) or

hey Jhevon you a Jamaican man...bless i from st. kitts

5. Originally Posted by sobadin
im still confused, how would i use this ) or

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?

6. yes just not at this level of difficulty

7. Originally Posted by sobadin
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:

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

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

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

so our integral becomes:

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

this simplifies to

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

now can you finish up? don't forget this is a definite integral, so you have to deal with the limits at some point

8. hope its clear!

9. Don't complicate to much, note that,

$u={{x}^{2}}+\sqrt{{{x}^{4}}-3}\implies du=\left( 2x+\frac{2{{x}^{3}}}{\sqrt{{{x}^{4}}-3}} \right)\,dx\implies \frac{du}{2u}=\frac{x}{\sqrt{{{x}^{4}}-3}}\,dx.$

10. youre right my friend,
but luckily all roads lead to Rome.
im not too much familiar with the subject,which explains the length :]