# Homework Help: Integration by parts formula

1. Sep 14, 2004

### Whatupdoc

i will use "\int" as integral signs, cause latex seems to be down.

uv - \int v*du

\int 8x^2cos(2x)*dx

u = 8x^2
du = 16x*dx
dx = 1/16

dv = cos(2x)
v = 1/2sin(2x)

plug in what i found for the formula

8x^2*1/2*sin(2x) - \int 1/2*sin(2x)*16x

take out the 1/2, because it's a number.

4x^2*sin(2x) - 1/2 \int sin(2x) *16x

k this is the part where im stuck on. do i use integration by parts on the right side agian? my attempt of u-du:
u=16x
du = 16*dx
dx = 1/16
sin(1/8*u) * u

hmm... looks wrong

2. Sep 14, 2004

### NateTG

I find the equation
dx=1/16
to be rather odd.

Also, the lack of a 'dx' at the end of the expression:
8x^2*1/2*sin(2x) - \int 1/2*sin(2x)*16x
seems a bit disconcerting.

That said, what's wrong with dv=sin(2x) in the second parts?

3. Sep 14, 2004

### Whatupdoc

"That said, what's wrong with dv=sin(2x) in the second parts?"

hmm it's suppose to be v=sin(2x)*dx right? not dv. hmm i dont see anything wrong with that part, unless im missing something.

"I find the equation
dx=1/16
to be rather odd."

yea it's suppose to be dx=1/16*du, it's suppose to balance du right?

how would i solve the integral on the right side? \int 1/2*sin(2x)*16x

another parts sub?

4. Sep 14, 2004

### NateTG

How about dx=1/(16 x) * du so that it agress with the equation above it?

...

\int sin(2x) * 16x dx
is the same as
\int 16x * sin(2x) dx
which should look somewhat familiar. (yes, parts should work.)

5. Sep 14, 2004

### Whatupdoc

ok i finished the problem, but i cant seem to get it correct.

4x^2*sin(2x) - 1/2 \int sin(2x) *16x*dx

that is what i got so far form the previous post

ok using parts agian...
uv - \int v*du

u = 16x
du = 16*dx
dx = 1/16 *du
dv = sin(2x)
v = -1/2*cos(2x)

ok time to fill in what i know...
16x*-1/2*cos(2x) - \int -1/2*cos(2x)*16*dx
take out the -1/2 and 16 cause it's constant
-8x*cos(2x) +8 \int cos(2x)*dx
find the anti-derv.
-8x*cos(2x) + 4 sin(2x)*dx
put everything together...
4x^2*sin(2x) - 1/2(-8x*cos(2x) + 4 sin(2x)*dx)
4x^2*sin(2x) +4x*cos(2x) -2sin(2x)*dx
ok that's my final answer and it's correct, but the DX is still there... hmm i thought i was suppose to add in 1/16 for dx,cause that's the value that i found

Last edited: Sep 15, 2004
6. Sep 15, 2004

### NateTG

You shouldn't be getting any differentials at the end of the process.
Still no Latex :)

The last integration should be something like:
\int -cos(2x) dx = -1/2 sin(2x)

I'm guessing you did:
\int -cos(2x) dx -> -1/2 sin(2x) dx

7. Sep 15, 2004

### Whatupdoc

yea, but why did the dx get removed?

8. Sep 16, 2004

### NateTG

In an integral, the 'dx' indicates the variable of integration, and where the expression 'inside' the integral ends.

This isn't necessarily the most satisfying answer, but really, the 'dx' gets removed when integration happens because that's the way the notation works. There's probably a historic explanation, and there may well also be an explanation based on differentials, but I haven't seen either. If you like, you could think of the \int as a sort of opening parenthesis and the 'dx' as a sort of closing parenthesis.

For now, it sufficient to know that every time that integration takes place using the usual notation, a differential (e.g. 'dx', 'dt', or 'dθ') gets removed. Integration by parts is a bit confusing in this regard because it's really a shorthand for something like the this:

/int u(x)v'(x) dx =
/int u(x)v'(x) + 0 dx =
/int u(x)v'(x) + (u'(x)v(x) - u'(x)v(x)) dx =
/int (u(x)v'(x)+u'(x)v(x)) - u'(x)v(x) dx =
/int (u(x)v'(x)+u'(x)v(x)) dx - \int u'(x)v(x) dx =
u(x)v(x) + C - \int u'(x)v(x) dx

Although I advise you against it, and it is not standard, you could develop your own notation for integration. On a computer, for example, an integral might look like "integral(f(x),x,a,b)" rather than using the symbols that you're familiar with.