Question

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

Answer 1

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

Explanation:

`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`