How do you evaluate the integral #int cos^2(lnx)#?
1 Answer
Explanation:
#I=intcos^2(lnx)color(red)(dx)#
First, let
#I=intcos^2(lnx)(x)(1/xdx)#
#I=intcos^2(t)e^tdt#
From the identity
#I=int(1/2cos(2t)+1/2)e^tdt#
#I=1/2inte^tcos(2t)dt+1/2inte^tdt#
#I=1/2inte^tcos(2t)dt+1/2e^t#
Let
#{(u=cos(2t),=>,du=-2sin(2t)dt),(dv=e^tdt,=>,v=e^t):}#
Then by the IBP formula:
#J=e^tcos(2t)+int2e^tsin(2t)dt#
Reapplying IBP to the remaining integral with new
#{(u=2sin(2t),=>,du=4cos(2t)),(dv=e^tdt,=>,v=e^t):}#
So:
#J=e^tcos(2t)+2e^tsin(2t)-4inte^tcos(2t)dt#
Notice that
#J=e^tcos(2t)+2e^tsin(2t)-4J#
#5J=e^tcos(2t)+2e^tsin(2t)#
#J=1/5e^tcos(2t)+2/5e^tsin(2t)#
Returning to
#I=1/2J+1/2e^t#
#I=1/10e^tcos(2t)+1/5e^tsin(2t)+1/2e^t#
Our original substitution was
#I=1/10xcos(2lnx)+1/5xsin(2lnx)+1/2x+C#