How do you evaluate the integral $int cos^2(lnx)$ ?

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Answer 1 · 2017-01-29T01:04:10

$1/10xcos(2lnx)+1/5xsin(2lnx)+1/2x+C$

$I=intcos^2(lnx)color(red)(dx)$

First, let $t=lnx$ . This implies that $x=e^t$ and that $dt=1/xdx$ . Then:

$I=intcos^2(lnx)(x)(1/xdx)$

$I=intcos^2(t)e^tdt$

From the identity $cos(2t)=2cos^2(t)-1$ , we write that $cos^2(t)=1/2cos(2t)+1/2$ . Then:

$I=int(1/2cos(2t)+1/2)e^tdt$

$I=1/2inte^tcos(2t)dt+1/2inte^tdt$

$I=1/2inte^tcos(2t)dt+1/2e^t$

Let $J=inte^tcos(2t)dt$ . This will be tackled in isolation. We will use integration by parts to solve it. 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 $u$ and $dv$ :

${(u=2sin(2t),=>,du=4cos(2t)),(dv=e^tdt,=>,v=e^t):}$

So:

$J=e^tcos(2t)+2e^tsin(2t)-4inte^tcos(2t)dt$

Notice that $J=inte^tcos(2t)dt$ has reappeared in the expression. We can now do some "integral algebra" to solve for $J$ , the integral.

$J=e^tcos(2t)+2e^tsin(2t)-4J$

$5J=e^tcos(2t)+2e^tsin(2t)$

$J=1/5e^tcos(2t)+2/5e^tsin(2t)$

Returning to $I$ :

$I=1/2J+1/2e^t$

$I=1/10e^tcos(2t)+1/5e^tsin(2t)+1/2e^t$

Our original substitution was $t=lnx$ , also implying that $x=e^t$ , so:

$I=1/10xcos(2lnx)+1/5xsin(2lnx)+1/2x+C$

How do you evaluate the integral int cos^2(lnx)int cos^2(lnx)?