This reads ∫cos⁻³(2θ)sin(2θ)dθ for [0,30].
Let u=cos(2θ), du=-2sin(2θ)dθ, so sin(2θ)dθ=-½du.
The integral becomes -½∫u⁻³du=-½[-u⁻²/2]=¼u⁻² for u in [cos(0),cos(60)]=[1,½], where 1 is the low limit and ½ is the high limit.
This comes to ¼(4-1)=¾.
The limits apply to θ, not to u, because cosine cannot be outside the range [-1,1].