`intcos^2 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`

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