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
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: [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.
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.
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: [tex] \frac{1}{2} x \left[\cos{\left(\ln{(x)}\right)} + \sin{\left(\ln{(x)}\right)}\right]+ C [/tex]
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