# What is the best way to integrate this?

1. Feb 10, 2007

### LinearAlgebra

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.

2. Feb 10, 2007

### quasar987

What are you choosing as u and dv?

3. Feb 10, 2007

### d_leet

Try a u-substitution. The most obvious being u=sin(x)

4. Feb 10, 2007

### LinearAlgebra

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)

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

5. Feb 10, 2007

### LinearAlgebra

But then what do you do with cost ?? how do you put that in terms of u??

6. Feb 10, 2007

### d_leet

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. Feb 10, 2007

### d_leet

Well how do you normally do a u-substitution? What is du?

8. Feb 10, 2007

### LinearAlgebra

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.

9. Feb 10, 2007

### LinearAlgebra

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. Feb 10, 2007

### d_leet

I don't even believe a freshman level book would say to do integration by parts for this problem.

11. Feb 10, 2007

### LinearAlgebra

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?

12. Feb 10, 2007

### d_leet

Yes I do, and I've mentioned it twice already.

13. Feb 10, 2007

### LinearAlgebra

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. Feb 10, 2007

### d_leet

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. Feb 10, 2007

### Curious3141

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. Feb 11, 2007

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.

17. Feb 11, 2007

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

18. Feb 12, 2007

### ssd

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: Feb 12, 2007
19. Feb 14, 2007

OHHH, I didn't notice, the that's the way, you are right.

20. Feb 14, 2007

### D H

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

$$\int \sin^2t \cos t \; dt$$

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

Last edited: Feb 14, 2007