# What is the best way to integrate this?

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.

## Answers and Replies

Science Advisor
Homework Helper
Gold Member
What are you choosing as u and dv?

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

LinearAlgebra
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)

??

LinearAlgebra
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??

d_leet
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.

d_leet
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?

LinearAlgebra
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.

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).

d_leet
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.

LinearAlgebra
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?

d_leet
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.

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...

d_leet
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.

Homework Helper
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.

Link-
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

Link-
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

ssd
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:
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.

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

Staff Emeritus
Science Advisor
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.

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

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:
MrSparky

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:
acsi

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
$$\int \sin^2t \cos t \; dt = \frac{\sin^3 t}{3}$$

finally , I just spent 20 minutes doing this integral.

poutsos.A

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.

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