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Integrate cos(lnx)dx 
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#1
Jan2511, 07:12 PM

P: 100

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
Jan2511, 07:16 PM

P: 100

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
Jan2511, 07:21 PM

P: 3,014

Is the integral:
[tex] \int{\cos{(\ln{(x)})} \, dx} [/tex] If it is, make the substitution: [tex] t = \ln{(x)} \Rightarrow x = e^{t} [/tex] 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
Jan2511, 07:23 PM

P: 100

Integrate cos(lnx)dx
it is just cos(lnx)dx



#5
Jan2511, 07:27 PM

P: 3,014

then proceed as I told you.



#6
Jan2511, 07:37 PM

P: 100

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



#7
Jan2511, 07:37 PM

P: 3,014

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



#8
Jan2511, 07:39 PM

P: 100

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



#9
Jan2511, 07:41 PM

P: 3,014

We have a composite function [itex]\cos{(\ln{x})}[/itex] 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
Jan2511, 08:31 PM

P: 100

thank you, lastly i end up with 1/2(e^tcoste^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
Jan2511, 08:40 PM

P: 3,014

I think the sign in front of the sine should be + and the result should be:
[tex] \frac{1}{2} x \left[\cos{\left(\ln{(x)}\right)} + \sin{\left(\ln{(x)}\right)}\right]+ C [/tex] 


#12
Jan2511, 08:42 PM

P: 100

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
Jan2511, 08:43 PM

P: 3,014

yes, the x's will cancel.



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