# Integrate cos(lnx)dx

1. Jan 25, 2011

### Cudi1

1. The problem statement, all variables and given/known data
let u=lnx
du=1/x*dx
dv=cosdx
v=-sin
2. Relevant equations
Now im confused as im getting nowhere with this substiution, i learned the LIPTE rule but its quite confusing, i have a function within a function

3. The attempt at a solution

2. Jan 25, 2011

### Cudi1

for integration by parts to work, i would need two differentiable functions, but the cosdx is not differentiable would it need to be cosxdx for it to be differentiated?

3. Jan 25, 2011

### Dickfore

Is the integral:

$$\int{\cos{(\ln{(x)})} \, dx}$$

If it is, make the substitution:

$$t = \ln{(x)} \Rightarrow x = e^{t}$$

and substitute everywhere. The integral that you will get can be integrated by using integration by parts twice, or, if you know complex numbers, by representing the trigonometric function through the complex exponential.

4. Jan 25, 2011

### Cudi1

it is just cos(lnx)dx

5. Jan 25, 2011

### Dickfore

then proceed as I told you.

6. Jan 25, 2011

### Cudi1

ok im getting an integral of the form : coste^tdt. is this correct?

7. Jan 25, 2011

### Dickfore

yes. proceed by integration by parts or using complex exponentials.

8. Jan 25, 2011

### Cudi1

ok thank you for the help, quick question why do we have to let t=lnx?

9. Jan 25, 2011

### Dickfore

We have a composite function $\cos{(\ln{x})}$ of two elementary functions (trigonometric and logarithmic). As a combination, it does not have an immediate table integral. But, the method of substitution, which is nothing but inverting the chain rule for derivatives of composite functions, works exactly for such compound functions.

10. Jan 25, 2011

### Cudi1

thank you, lastly i end up with 1/2(e^tcost-e^tsint), do i sub back , so that x=e^t and t=lnx
which leaves me with : 1/2x(cos(lnx)-sin(lnx)+c?

11. Jan 25, 2011

### Dickfore

I think the sign in front of the sine should be + and the result should be:

$$\frac{1}{2} x \left[\cos{\left(\ln{(x)}\right)} + \sin{\left(\ln{(x)}\right)}\right]+ C$$

12. Jan 25, 2011

### Cudi1

yes, made a slight mistake thank you very much for the help, i tried doing it another way by letting u=cos(lnx) and dv=dx and i arrived at the same answer

13. Jan 25, 2011

### Dickfore

yes, the x's will cancel.