What is the best way to integrate this?

  • #1
LinearAlgebra
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0
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.
 

Answers and Replies

  • #2
quasar987
Science Advisor
Homework Helper
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What are you choosing as u and dv?
 
  • #3
d_leet
1,077
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Try a u-substitution. The most obvious being u=sin(x)
 
  • #4
LinearAlgebra
22
0
Okay so here's 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)


??
 
  • #5
LinearAlgebra
22
0
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??
 
  • #6
d_leet
1,077
1
Okay so here's 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.
 
  • #7
d_leet
1,077
1
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?
 
  • #8
LinearAlgebra
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0
No it absolutely didnt say intgrate by parts...i'm in a higher level class and haven't 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 that's what it said to do. If you have a simpler way, I'm sure that's what we should do.
 
  • #9
LinearAlgebra
22
0
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).
 
  • #10
d_leet
1,077
1
No it absolutely didnt say intgrate by parts...i'm in a higher level class and haven't 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 that's what it said to do. If you have a simpler way, I'm sure that's what we should do.

I don't even believe a freshman level book would say to do integration by parts for this problem.
 
  • #11
LinearAlgebra
22
0
um. okay. well I'm not lying?? I mean i don't know how that it supposed to be constructive. do you have the name of the method, the easier way, that you would use?
 
  • #12
d_leet
1,077
1
um. okay. well I'm not lying?? I mean i don't 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.
 
  • #13
LinearAlgebra
22
0
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...
 
  • #14
d_leet
1,077
1
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.
 
  • #15
Curious3141
Homework Helper
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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.
 
  • #16
Link-
100
0
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.

-Link
 
  • #17
Link-
100
0
Okay so here's 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.

-link
 
  • #18
ssd
268
6
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


-Link
The original problem does not contain sin(2t), it has sin^2t.
As d_leet has mentioned, it appears, that the simplest way is to put u=sint.
 
Last edited:
  • #19
Link-
100
0
The original problem does not contain sin(2t), it has sin^2t.
As d_leet has mentioned, it appears, that the simplest way is to put u=sint.

OHHH, I didn't notice, the that's the way, you are right.
 
  • #20
D H
Staff Emeritus
Science Advisor
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Okay so here's 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 one is quite easy with integration by parts. It's no harder (and maybe even easier) than u-substitution.

[tex]\int \sin^2t \cos t \; dt[/tex]

Applying integration by parts with
[tex]u = \sin^2 t, dv = \cos t\;dt \text{\ for which\ } du = 2 \sin t \cos t\; dt, v = \sin t[/tex]
[tex]\int \sin^2t \cos t \; dt = \sin^3 t - 2 \int \sin^2t \cos t \; dt[/tex]
The integral on the left- and right-hand sides are the same integral. Thus
[tex]3 \int \sin^2t \cos t \; dt = \sin^3 t [/tex]
or
[tex]\int \sin^2t \cos t \; dt = \frac{\sin^3 t}{3}[/tex]
 
Last edited:
  • #21
MrSparky
8
0


sub u= sin t
du/dt = cos t
-250 sin^2 t . du/dt . dt = -250 u^2 du
and then = (-250 u ^3)/3
Substitute 'u' back in and u get (-250 sin^3 t)/3
then the result is (-250 sin^3 t)/3 + c since there are no boundaries
 
Last edited:
  • #22
acsi
1
0


This one is quite easy with integration by parts. It's no harder (and maybe even easier) than u-substitution.

[tex]\int \sin^2t \cos t \; dt[/tex]

Applying integration by parts with
[tex]u = \sin^2 t, dv = \cos t\;dt \text{\ for which\ } du = 2 \sin t \cos t\; dt, v = \sin t[/tex]
[tex]\int \sin^2t \cos t \; dt = \sin^3 t - 2 \int \sin^2t \cos t \; dt[/tex]
The integral on the left- and right-hand sides are the same integral. Thus
[tex]3 \int \sin^2t \cos t \; dt = \sin^3 t [/tex]
or
[tex]\int \sin^2t \cos t \; dt = \frac{\sin^3 t}{3}[/tex]

finally :smile: , I just spent 20 minutes doing this integral.
 
  • #23
poutsos.A
102
1


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.



[tex]\int-250.cost\sin^2tdt = \int-250\sin^2td(sint)= -250 \frac{\ sin^3t}_{ 3}[/tex]
 

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