# Problem with integration by parts

1. Apr 20, 2010

### bobey

1. The problem statement, all variables and given/known data
the question :

integrate the following :

integration of d(y/x) = integration of(c cos x/x^2) dx , where c is a constant

2. Relevant equations

integration of d(y/x) = integration of(c cos x/x^2) dx
y/x = c integration of (c cos x/x^2) dx (*)
= c(x^-2 sin x -integration(sin x (-2x^-3))dx

(*) i let u = x^-2
du = -2x^-3

dv= cos x dx
v = sin x

and by integration by parts, i got (*)

but the integration on the RHS seems to complex which contradicts with the principle of integration by parts, which makes the integration simpler... i think i made some mistake some where... can anyone highlight it to me... plz...

3. The attempt at a solution

2. Apr 21, 2010

### Susanne217

Do you own the Edwards and Penny Calculus Bible? In there you are presented with several integrals which can't be solved using standard methods. This is one them...

I learned to find a solution in Calculus. But if you can't find in your book. Please report back to me and I will guide you :)

Remember you want evaluate $$\int c \cdot \frac{cos(x)}{x^2} dx$$ which is a series also. Did you know that?

Find a series for $$f(x) = \frac{cos(x)}{x^2}$$ and then report back to me :)

Last edited: Apr 21, 2010
3. Apr 22, 2010

### SgtSniper90

If you change your u=cos(x) and v'=x^-2 you can integrate it easier but you get to a point where you get the integral of sin(x)/x. Which if you dont know is Si(x). If you want to keep going you can use the taylor polynomial of sin(x) which is x-(x^3)/3!+(x^5)/5!-(x^7)/7!. It really keeps going forever but this will be really really close integrate this and you should have an answer. But depending on where you are in Calculus you should just leave it:

-cos(x)/x - Si(x)

4. Apr 22, 2010

### Susanne217

I'm not trying argue but isn't the Si function post Calculus?

5. Apr 22, 2010

### SgtSniper90

I think so but there really is no other answer at your level of calc (not to insult you). Even if you used the taylor polynomial you can never get it to fit the entire graph, you can get really really close but never perfect. I havent tried but i dont think that a series would help here anyways