I am not sure where I am going wrong but this problem gets more complicated

$$\int \cos^4 x\;dx
$$

$u =\cos^3 x$

$du = -3\cos^2 x \sin x \;dx$

$dv = \cos x\;dx$

$v = \sin x$

$$\sin x \cos^3 x + 3\int \sin^2x \cos^2 x \;dx$$

Now I have two squared trig functions, I can take either in the form of 1 + squared trig function but either way I get a power of 4 and nothing has gotten more simple. What is wrong?