Recall the half-angle identity for cosine:
cos²(x) = 1/2 (1 + cos(2x))
Then we can rewrite the integrand as
cos³(4x) = cos(4x) cos²(4x) = 1/2 cos(4x) (1 + cos(8x))
So we have
[tex]\displaystyle \int \cos^3(4x) \, dx = \frac12 \int (\cos(4x) + \cos(4x)\cos(8x)) \, dx[/tex]
Next, recall the cosine product identity,
cos(a) cos(b) = 1/2 (cos(a - b) + cos(a + b))
so that the integral is equivalent to
[tex]\displaystyle \int \cos^3(4x) \, dx = \frac12 \int \cos(4x) \, dx + \frac14 \int (\cos(4x - 8x) + \cos(4x + 8x)) \, dx[/tex]
[tex]\displaystyle \int \cos^3(4x) \, dx = \frac34 \int \cos(4x) \, dx + \frac14 \int \cos(12x) \, dx[/tex]
Computing the rest is trivial:
[tex]\displaystyle \int \cos^3(4x) \, dx = \boxed{\frac3{16} \sin(4x) + \frac1{48} \sin(4x) + C}[/tex]
