How do you evaluate the integral of #int (cos(lnx))/x dx#?

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Answer 1 · 2018-04-15T01:16:52

#intcos(lnx)/xdx=sin(lnx)+C#

A substitution solves this.

#u=lnx#

#du=dx/x#

Then, we have

#intcosudu=sinu+C#

Rewriting in terms of #x# yields

#intcos(lnx)/xdx=sin(lnx)+C#