• We’re currently investigating an issue related to the forum theme and styling that is impacting page layout and visual formatting. The problem has been identified, and we are actively working on a resolution. There is no impact to user data or functionality, this is strictly a front-end display issue. We’ll post an update once the fix has been deployed. Thanks for your patience while we get this sorted.

Quick integration help - math

RESmonkey

Diamond Member
How can I integrate this:

1 / sqr.root(4-x^2)

or just

(4-x^2)^(-.5)

It looks like an arcsine thing, but where did the four come from?
 
sqrt(a^2-x^2) where x = a sin(theta),
from the identity 1 - sin^2(theta) = cos^2(theta)

so the 4 in your equation would = a^2, in which a = 2?...
(making an assumption, might be wrong)

Don't think it'd be much help though...
 
Never mind...I figured it out:

If anyone wanted to know:

4-x^2
4(1-.25x^2)
rip the four outta of the sqrroot:
2(1-.25x^2)^.5

Rip 1/2 outta of the integral. use u substitution from there on out.

 
Originally posted by: Ricemarine
sqrt(a^2-x^2) where x = a sin(theta),
from the identity 1 - sin^2(theta) = cos^2(theta)

so the 4 in your equation would = a^2, in which a = 2?...
(making an assumption, might be wrong)

Don't think it'd be much help though...

Hmm...that kind of does parallel what I did. It might even be the same thing, dunno. :0
 
***This is using my previous post***

( 1 / sqrt(4-x^2) )dx

x = 2sin(theta)
dx = 2cos(theta)

(1 / sqrt(4 - 4sin^2(theta))dx

since 1 - sin^2(theta) = cos^2(theta)

( 1/ sqrt(4-4sin^2(theta))dx = (1/sqrt(4cos^2(theta))dx

which then becomes

dx / 2cos(theta)

and since dx = 2cos(theta)

plug the thing into the integral and use U substitution from there...


Umm, might wanna double check if I'm wrong. My head's kinda iffy today.
 
Originally posted by: RESmonkey
How can I integrate this:

1 / sqr.root(4-x^2)

or just

(4-x^2)^(-.5)

It looks like an arcsine thing, but where did the four come from?

(4 - x^2)^(-.5) = [4 * (1 - x^2 / 4)]^-.5 = .5 * (1 - x^2 / 4)^-.5

so your integral takes the form of 1 / sqrt(1 - u^2) where u = x/2 and du = .5dx

so u look up the integral in a table or whatever and u get asin(x/2) + c

edit: you were on the right track but u need to remember that u when u can break down equations into simpler pieces, you should do it.
 
Originally posted by: Fenixgoon
looks good so far

edit: ricemarine, wtf are you doing his hw?

No, I'm just copying the example, writing step by step of what I try and do to solve it with the help of my calculus book, and posting it on a forum, specifically the topic which originally had the example...

Plus, he already solved the problem before I posted that. 🙂
 
Yikes. My grade for last q was a B+ ....an 89.something percent. Grr...I've never gotten anything but A's in math/science 🙁
I gotta average out an A for the semester.
 
Back
Top