# What is the difference b/w cos(lnx) and cosxlnx? integration by parts!

1. Oct 2, 2012

### randoreds

Ok I have to integrate -->∫cos(lnx) dx. could I use cos =U, -sinx=du, dv=lnxdx, v = 1/x

I know the difference technically, but in this situation it is kinda weird.
because the formula f(x)g(x)= uv-∫vdu. I thinking if they were number like 9(3) it would equal 27 so f(g) = f times G? but then that would mean ∫cos(lnx)dx = ∫cosxlnxdx . Which I don't think is right.

last guess: could you do cos(lnx)= U, -sinx(lnx) times (1/2) dx = du, DV = lnxdx, v =1/x

So I'm confused.

I asked by teacher, because my first taught was to do U-substituion then integrate by parts. She said that would work, but it would end up being alot of ugly calculations. and that you could just integrate by parts directly. So if you could explain that, that would nice.

Last edited: Oct 2, 2012
2. Oct 2, 2012

### LCKurtz

What do you mean by cos = u? Do you mean u = cos(ln(x))? Then du isn't sin(x). And cos(ln(x)) is not the product of cos and ln(x) so your integrand is not udv. A long winded way of saying what you have written is nonsense.

Try u = cos(ln(x)) and dv = 1 dx and see where that goes. You may have to do parts more than once.

3. Oct 2, 2012

### dextercioby

I think you can use a trick:

$$I = \int \cos \ln x ~ dx = \displaystyle{\mathcal{Re}} \left(\int e^{i\ln x} ~dx\right)$$

4. Oct 2, 2012

### randoreds

I mean't cos(x) then du = -sintheta. But that was totally wrong! you had the right substitution. I tried u = cos(ln(x)) and dv =dx. It worked out quite nicely. I only had to integrate it twice! Thanks