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What is the best way to integrate this??

 
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Feb10-07, 07:46 PM   #1
 

What is the best way to integrate this??


What is the best way to integrate -250*cost*sin^2t dt ??

I'm looking in my old calc book and it says to integrate by parts-is this the only way to do it? When I try this it makes the integral even more complicated than it was to begin with.
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Feb10-07, 08:04 PM   #2
 
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What are you choosing as u and dv?
Feb10-07, 08:07 PM   #3
 
Try a u-substitution. The most obvious being u=sin(x)
Feb10-07, 08:08 PM   #4
 

What is the best way to integrate this??


Okay so heres how I did it:

The general rule is Integral of udv = u*v - int(v*du)

So i said that v = sin^2t dv/dt = 2sintcost
du=costdt --> u=sint

Integral = sin^3t - int (sin^2t*costdt)


??????????????????????????
Feb10-07, 08:09 PM   #5
 
Quote by d_leet View Post
Try a u-substitution. The most obvious being u=sin(x)
But then what do you do with cost ?? how do you put that in terms of u??
Feb10-07, 08:10 PM   #6
 
Quote by LinearAlgebra View Post
Okay so heres how I did it:

The general rule is Integral of udv = u*v - int(v*du)

So i said that v = sin^2t dv/dt = 2sintcost
du=costdt --> u=sint

Integral = sin^3t - int (sin^2t*costdt)


??????????????????????????

This is really messy, are you absolutely sure the book said to use integration by parts because if the problem is exactly as you've posted it then there is a much simpler way.
Feb10-07, 08:11 PM   #7
 
Quote by LinearAlgebra View Post
But then what do you do with cost ?? how do you put that in terms of u??
Well how do you normally do a u-substitution? What is du?
Feb10-07, 08:14 PM   #8
 
No it absolutely didnt say intgrate by parts...i'm in a higher level class and havent touched calc in 5 years and so i forget how to do this basic operation. I looked in my old freshman yr calc book and thats what it said to do. If you have a simpler way, i'm sure thats what we should do.
Feb10-07, 08:17 PM   #9
 
We have to do line integrals and as a part of the g(x,y)dx, dy, ds..this was one of the problems i was having (actually performing the integral).
Feb10-07, 08:17 PM   #10
 
Quote by LinearAlgebra View Post
No it absolutely didnt say intgrate by parts...i'm in a higher level class and havent touched calc in 5 years and so i forget how to do this basic operation. I looked in my old freshman yr calc book and thats what it said to do. If you have a simpler way, i'm sure thats what we should do.
I don't even believe a freshman level book would say to do integration by parts for this problem.
Feb10-07, 08:18 PM   #11
 
um. okay. well i'm not lying?? I mean i dont know how that it supposed to be constructive. do you have the name of the method, the easier way, that you would use?
Feb10-07, 08:19 PM   #12
 
Quote by LinearAlgebra View Post
um. okay. well i'm not lying?? I mean i dont know how that it supposed to be constructive. do you have the name of the method, the easier way, that you would use?
Yes I do, and I've mentioned it twice already.
Feb10-07, 08:21 PM   #13
 
Ooooh. got it, thanks. so when should i use the u substitution method vs integration by parts? the u sub worked out perfectly in this case...
Feb10-07, 08:22 PM   #14
 
Quote by LinearAlgebra View Post
Ooooh. got it, thanks. so when should i use the u substitution method vs integration by parts? the u sub worked out perfectly in this case...
Parts often seems to be reserved for a last resort situation when nothing else seems to work, so it's probably best to try and use a u-substitution or other method before resorting to parts.
Feb10-07, 09:05 PM   #15
 
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Another way to do it would be to use double angle formula (for cos) and then factor formulae to simplify, then just integrate. Not difficult either.
Feb11-07, 11:30 AM   #16
 
You could use the double angle of sin. (sin(2t)=2sintcost) and substitut it .

Would be somethink like this: -250 S cos t sin 2t dt
-250 S cos t 2sint cost dt
-500 S (cos t)^2 sint dt
Then u=cost du=-sint dt, then 500 S u^2 dt
500 (u^3)/3
substituting back u= cost 500 (cost)^3* 1 / 3


Wala, there is, no integration by parts is useless if you have the du on the integral. Just the double angle formula.

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Feb11-07, 11:34 AM   #17
 
Quote by LinearAlgebra View Post
Okay so heres how I did it:

The general rule is Integral of udv = u*v - int(v*du)

So i said that v = sin^2t dv/dt = 2sintcost
du=costdt --> u=sint

Integral = sin^3t - int (sin^2t*costdt)


??????????????????????????
With all respect,
You are the rule right but the dv/dt= -2 cos 2t, you just use the double angle formula.

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