`intcos^2x\ dx`

Let `u=2x`, then `du=2\ dx`

This gives:

`{: (intcos^2u(du)/2,=1/2intcos^2 u\ du),(,=1/2int(1+cos\ 2u)/2du),(,=1/4int1+cos\ 2u\ du),(,=1/4[u+(sin\ 2u)/2]+K),(,=1/4[2x+(sin\ 2(2x))/2]+K),(,=x/2+(sin\ 4x)/8+K) :}`