Apply the double-angle formula twice

$\cos^4x=(\cos^2x)^2=[(1+\cos2x)/2]^2=(1+2\cos2x+\cos^2(2x))/4$. Apply double angle again to $\cos^2(2x)=(1+\cos4x)/2$. This gives $(1+2\cos2x)/4+(1+\cos4x)/8=3/8+\cos2x/2+\cos4x/8$, as required.

$\int\cos^4x\,dx=\int(\frac38+\frac12\cos2x+\frac18\cos4x)\,dx=\frac38x+\frac14\sin2x+\frac1{32}\sin4x+C$.