integrate cos(lnx)dx


by Cudi1
Tags: coslnxdx, integrate
Cudi1
Cudi1 is offline
#1
Jan25-11, 07:12 PM
P: 102
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
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Cudi1
Cudi1 is offline
#2
Jan25-11, 07:16 PM
P: 102
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?
Dickfore
Dickfore is offline
#3
Jan25-11, 07:21 PM
P: 3,015
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.

Cudi1
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#4
Jan25-11, 07:23 PM
P: 102

integrate cos(lnx)dx


it is just cos(lnx)dx
Dickfore
Dickfore is offline
#5
Jan25-11, 07:27 PM
P: 3,015
then proceed as I told you.
Cudi1
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#6
Jan25-11, 07:37 PM
P: 102
ok im getting an integral of the form : coste^tdt. is this correct?
Dickfore
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#7
Jan25-11, 07:37 PM
P: 3,015
yes. proceed by integration by parts or using complex exponentials.
Cudi1
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#8
Jan25-11, 07:39 PM
P: 102
ok thank you for the help, quick question why do we have to let t=lnx?
Dickfore
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#9
Jan25-11, 07:41 PM
P: 3,015
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.
Cudi1
Cudi1 is offline
#10
Jan25-11, 08:31 PM
P: 102
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?
Dickfore
Dickfore is offline
#11
Jan25-11, 08:40 PM
P: 3,015
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]
Cudi1
Cudi1 is offline
#12
Jan25-11, 08:42 PM
P: 102
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
Dickfore
Dickfore is offline
#13
Jan25-11, 08:43 PM
P: 3,015
yes, the x's will cancel.


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