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Calculus help!!

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funk.
Posts: 2679 - Karma: 33
Integrate: (1-t^3)^-(1/2) dt

I'm stumped.  Thanks.  Plus karma
Dec 19 2008 1:23AM
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PKalow
Posts: 254 - Karma: 11
So that would be -√1-t³ dt?

I'd need to know what integer it's for...
Dec 19 2008 1:42AM
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wang_bong
Posts: 766 - Karma: 13
http://integrals.wolfram.com/index.jsp?expr=1%2Fsqrt[(1-x^3)]&random=false
Dec 19 2008 1:47AM
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killa_cam
Posts: 13143 - Karma: 15
looks hard

imma try it
Dec 19 2008 1:48AM
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funk.
Posts: 2679 - Karma: 33
no.  1/(1-t^3)^1/2
Dec 19 2008 1:55AM
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funk.
Posts: 2679 - Karma: 33
ugly.  damn.
Dec 19 2008 1:56AM
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killa_cam
Posts: 13143 - Karma: 15
the answer is.

this integral cannot be expressed in terms of elementary functions.

the integral exists, but it's very complicated and can't be expressed simply as a function of t
Dec 19 2008 1:59AM
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funk.
Posts: 2679 - Karma: 33
^shitty.  yeah the original problem was find F'(x) if f(x) = integral[x^2,0] of (1-t^3)^(-1/2) dt

OR

d/dx {integral[x^2,0] of (1-t^3)^(-1/2) dt}

which is fairly straight forward and simple.  I started it out by trying to find the integral and was like whaat!?  so I thought I'd throw it at you guys to see if that part was possible or not.  Thanks for the help!
Dec 19 2008 2:05AM
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whattheshit
Posts: 6562 - Karma: 107
can't help you.  but this is funny.

Click for larger image



Dec 19 2008 2:14AM
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PKalow
Posts: 254 - Karma: 11
Here's what I got so far on the integral part of it, I don't want to simplify it any more

Click for larger image
Dec 19 2008 2:21AM
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[MEGAMO]
Posts: 1222 - Karma: 18
probably needs a u-sub or something. it actually looks like a trig sub, but i left my list of trig subs at school
Dec 19 2008 2:35AM
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CFD
Posts: 1009 - Karma: 197
second line is wrong i think.

its so hard to do calc on the computer. i need to write it down on paper for it to make any sense in my head.
Dec 19 2008 8:00AM
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49566738265
Posts: 0 - Karma: 7,704
No, that's not how you integrate. 

You need to integrate by parts to solve this problem.
Dec 19 2008 11:55AM
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mel220
Posts: 31 - Karma: 20

yeah you need to do u-sub, its like 1/a square root of blah blah right? so what ever is inside the square root is u and then du is teh derivative of that, and then you integrate for you and plug u back in

sorry its kinda hard to do calculus on the internet, but hope it helps

Dec 19 2008 12:20PM
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COskier
Posts: 1066 - Karma: 131
do a u substitution of like t=u^(2/3).
then its just like the derivative of inverse sin i believe.
then take the integral so its inverse sin.
Don't forget to change your limits of integration when you do your substitution.

At least that what I'd do.
Dec 19 2008 3:48PM
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COskier
Posts: 1066 - Karma: 131
You just gots to have your identities memorized dawg.
Dec 19 2008 3:49PM
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COskier
Posts: 1066 - Karma: 131
Fuck I'm an idiot, I think if you do that you'll have a random u^(/1/3) hanging out n top from changing dt to du. So, I'm done.
Dec 19 2008 3:53PM
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KLovelace
Posts: 2118 - Karma: 289
oh my god fuck this shit!!


my brain hurts just from scan reading this thread
Dec 19 2008 3:59PM
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COskier
Posts: 1066 - Karma: 131
What if you use the thing that d/dx(int(a,g(x)) of f(t)dt=g(g(x))g'(x)

then it equals int of 2x/(1-x^6)^(1/2) if you treat your original problem as the integral of the left side of the equation right?

I think thats right and that last part is an easy integral.

Probably wrong again though
Dec 19 2008 4:04PM
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OrangGrant
Posts: 83 - Karma: 13
im pretty sure its a trig substitution. so that whole function would go to cos(t) or something to that nature. not sure though my calc book is in newyork lol.
Dec 19 2008 4:46PM
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doublexl
Posts: 8407 - Karma: 27
math is so gay
Dec 19 2008 4:53PM
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Jwenz
Posts: 8893 - Karma: 7,634
Okay so you are looking for F' and you have F...you just plug in the x^2 into the t value. that gives you the derivitive. depending on the book that is the Fundamental therom of calculus 1 or 2...
Dec 19 2008 6:25PM
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CFD
Posts: 1009 - Karma: 197
i can't see any suitable parts
Dec 19 2008 8:02PM
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kepler.
Posts: 8761 - Karma: 139
Where's GLrunner when you need him!?!
Dec 19 2008 8:38PM
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quixotic
Posts: 6937 - Karma: 2,978
I'm so glad my last day of math in my life was on Monday...... WEEEEEEEE!!!
Dec 19 2008 9:03PM
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jib_kid_01
Posts: 248 - Karma: 10
If that is the question i don't think you have to integrate at all. You use one of the fundamental theorems of calculus to find that the derivative is (1-t^3)^-(1/2) then you plug in x^2 and use the chain rule to find your solution
Dec 19 2008 9:41PM
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jib_kid_01
Posts: 248 - Karma: 10
Oh opps just look at PKalow's answer thats right.
Dec 19 2008 9:48PM
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An10a_Eng
Posts: 2505 - Karma: 26
no. this is wrong. you are trying to use the second fundamental theorem of calculus which can not be applied to this integral. that is for finding derivatives of integrals where the upper bound is a function.
the problem is integrate (indefinately) (1-t^3)^-1/2....if your in calc 2 you can use binomial expansion where m = -1/2. you can take the binomial expansion for (1-t)^-1/2 and sub t^3 in for t as your last step
Dec 20 2008 12:01AM
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An10a_Eng
Posts: 2505 - Karma: 26
and its not a trig sub, not a u sub, and not inverse sin
Dec 20 2008 12:05AM
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An10a_Eng
Posts: 2505 - Karma: 26
i didnt see this post. my first post is wrong. the second fundamental theorem is how you do it if your lookin for the derivative
Dec 20 2008 12:09AM
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Bepo
Posts: 8195 - Karma: 17
shit, this just reminded me i haven't done calc in a year and i'm gonna need it for a class next semester. fuck
Dec 20 2008 12:10AM
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funk.
Posts: 2679 - Karma: 33
so to let all of you knowww:

substitute x^2 for t
take the derivative
multiply by the derivative of the upper limit

That's what I learned in school today mommy
Dec 20 2008 3:23AM
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killa_cam
Posts: 13143 - Karma: 15
sorry but that is so wrong
Dec 25 2008 1:30AM
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St.Steezy
Posts: 4228 - Karma: 20
wow.. that integral actually took sometime for my computer to create a curve for.. definitely a tricky one.
Dec 25 2008 3:00AM
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freeskier824
Posts: 813 - Karma: 11
yeah i used to know how to do this... then i got to the real world
Dec 25 2008 10:58AM
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STwoops
Posts: 2941 - Karma: 5,126
this thread is so sick

Dec 29 2008 2:53AM
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