# Integrate cos(lnx)dx

let u=lnx
du=1/x*dx
dv=cosdx
v=-sin

## Homework 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

## The Attempt at a Solution

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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?

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.

it is just cos(lnx)dx

then proceed as I told you.

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

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

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

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.

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?

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

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

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

yes, the x's will cancel.