Let In=∫0π/2 1−cos⁡2nθ1−cos⁡2θdθ;n∈N and Δ=I4    I5    I6I7    I8    I9I21    I22    I23 Then Δ is divisible by

We have In=∫0π/2 1−cos⁡2nθ1−cos⁡2θ=∫0π/2 sin2nθsin2θdθ Consider, In+1−In=∫0π/2 sin2⁡(n+1)θ−sin2⁡nθsin2⁡θdθ=∫0π/2 sin⁡(2n+1)θsin⁡θdθIn−In−1=∫0π/2 sin⁡(2n−1)θsin⁡θdθ ……. by (1)−(2)⇒In+1+In−1−2In=∫0π/2 sin⁡(2n+1)θ−sin⁡(2n−1)θsin⁡θdθ                             =∫0π/2 2cos⁡2nθ⋅sin⁡θsin⁡θdθ=0 ⇒I1,I2,I3  in A. P. ⇒Δ=I4I5I6I7I8I9I21I22I23=0∵C2→C2−C1 and C3→C3−C2 both are identical)