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

[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 a lot of ugly calculations. and that you could just integrate by parts directly. So if you could explain that, that would nice.

 

[LCKurtz]

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

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.

 

[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)

 

[randoreds]

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.

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