$I_1 =\int \cos^4 3x dx =\dfrac{1}{4} \int (1+\cos 6x)^2 dx =\dfrac{1}{4}\int (1 +2\cos 6x + \cos^2 6x)dx$
$=\dfrac{1}{4}x + \dfrac{1}{12}\sin 6x +\dfrac{1}{4}\int \cos^2 6x dx =\dfrac{1}{4}x + \dfrac{1}{12}\sin 6x+\dfrac{1}{8}\int (1+\cos 12x)dx$
$=\dfrac{3}{8}x + \dfrac{1}{12}\sin 6x+\dfrac{1}{96}\sin 12x + C$