integration (1 Viewer)

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i got stuck in this problem

integrate:
x²sin³(x) dx

and err..

integrate:
sin(log(x)) dx

big wtf for me

 

[George W. Bush]

are there any limits on those integrals?

 

[...]

nope

 

[:: ck ::]

ibp ?

 

[...]

wat?!?!?!

 

[redruM]

integration by parts

 

[George W. Bush]

integration by parts

 

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Originally posted by redruM
integration by parts

haha

err...can u start it off for me?

 

[redruM]

ehh...i dont know?

its should be in any 4 unit textbook. try googling it. i have forgotten it...maybe someone else remembers/knows..

 

[CM_Tutor]

The first one is nasty - I don't see an easy solution at this moment, but I'll give it some thought.

For the second one, start with a substitution of u = ln x. It should transform to int e^usin u du, which can be done by integrating by parts twice, then simplifying. This starts as:

int e^usin u du = int sin u de^u = e^usin u - int e^u dsin u = e^usin u - int e^ucos u du = ...

I get the answer to be:

int sin(ln x) dx = (x / 2) * [sin(ln x) - cos(ln x)] + C, for some constant C

 

[martin]

Warning: What I am about to do is very messy: hopefully there's a way to do this much more easily. But I can't see it.

integrate: x^2sin^3(x) dx

Basically we want to use integration by parts
Integral(uv') = uv - Integral(u'v)
with u=x^2 so we reduce the power of x^2 down via IBP twice.

But the problem is that we can't integrate [sin(x)]^3 so we need to write in in terms of sinx, sin(3x) that kind of stuff.

There are a few ways of doing this but probably the easiest in 4U is to use DeMoivres and binomial theorem

(cosx + isinx)^3
= cos(3x) + isin(3x)
and by binomial therem with c=cosx, s=sinx
= c^3 + 3(c^2)is + 3c(i^2)(s^2) + (i^3)s^3
so equating imaginary parts
sin(3x) = 3(c^2)s - s^3
= 3(1-s^2)s - s^3
=3s-4s^3

so 4(sinx)^3 = 3sinx - sin(3x)
(sinx)^3 = 3/4sinx - 1/4sin(3x)

so Int(x^2 (sinx)^3)
= 3/4Int(x^2*sinx) - 1/4Int(x^2*sin3x)

I'll do one step of the first integral for you

Int(x^2*sinx)
=x^2*-cosx - Int(2x*-cosx)
= -x^2 cosx + 2Int(x cosx)

So you just need to use IBP again on this integral and then twice on the other integral

My final answer is
Int(x^2*(cosx)^3)
= 1/12*x^2*cos3x - 3/4*x^2*cosx + 2/9*x*sin3x + 3/2*x*sinx +2/27*cos3x + 3/2*sinx

 

[martin]

I just went to the integrator (integrals.wolfram.com) and put in x^2*(Sin[x])^3. The answer it gave back is:

-3/4*(-2+x^2)cos[x] + 1/108*(-2+9x^2) + 3/2*x*sinx - 1/18*x*sin(3x)

This isn't exactly what I had but it seems to be of the same form. I obviously made a few mistakes somewhere but I'm not going to bother finding them.

 

[Jase]

what the..
well i can honestly say im just too lazy to simplify my answer...
i used parts many times and got very confused with my positives and negatives.

there is also something i have just become very confused about, concerning the post by martin.

Quote "But the problem is that we can't integrate [sin(x)]^3 so we need to write in in terms of sinx, sin(3x) that kind of stuff."

dont you just simply substitute u = cos(x) -> 3 unit integration?
or is there something im missing.

 

[martin]

Quote "But the problem is that we can't integrate [sin(x)]^3 so we need to write in in terms of sinx, sin(3x) that kind of stuff."

dont you just simply substitute u = cos(x) -> 3 unit integration?

Yeah, maybe I should of said "can't integrate easily". (In fact I didn't think of doing it this way)

But it would be possible to do it this way

Int(sin^3(x))
=Int((1-cos^2(x)) sinx)
= -cosx + 1/3 cos^3(x)

then
Int (x^2 sin^3(x))
=x^2(1/3 cos^3(x) - cosx) - Int(2x(1/3 cos^3(x) - cosx))

This integral could be done in two seperate sections, both of them needing integration by parts. So I don't think it would be any easier.

try both of them if you really want to