The integral ∫π/4π/12 8cos2x(tanx+cotx)³dx equals?

I=∫π/4π/12 8cos2x(tanx+cotx)³dx
I=∫π/4π/12 8cos2x(sinx/cosx + cosx/sinx)³dx
I=∫π/4π/12 8cos2x(sin²x + cos²x)³/ (sinxcosx)³ dx
I=∫π/4π/12 8cos2x/(sinxcosx)³ dx
I=∫π/4π/12 8cos2x sin³x cos³x / (sinxcosx)³ dx
I=∫π/4π/12 8cos2x sin³x/cos³x * (sinxcosx)³/(sinxcosx)³ dx
I = ∫π/4π/12 8cos2x (sin²x + cos²x)³/ (sin³x cos³x) dx
I = ∫π/4π/12 8cos2x / (sin³x cos³x) dx
I = ∫π/4π/12 8cos2x / (sinxcosx)³ dx
I = ∫π/4π/12 8cos2x / (sin³x cos³x) dx
I = ∫π/4π/12 8cos2x sin³x cos³x / (sin³x cos³x)³ dx
I = ∫π/4π/12 8cos2x/(sin³x cos³x) dx
I = ∫π/4π/12 8cos2x (sin2x)³/ (2³ sin³x cos³x) dx
I = ∫π/4π/12 8cos2x (sin2x)³/8(sinxcosx)³ dx
I = ∫π/4π/12 cos2x (sin2x)³/(sinxcosx)³ dx
Let sin2x = t
2cos2x dx = dt
When x = π/12, t = sin(π/6) = 1/2
When x = π/4, t = sin(π/2) = 1
I = ∫1/2¹ t³/2 dt = 1/2 ∫1/2¹ t³ dt = 1/2 [t⁴/4]1/2¹ = 1/8 [1 - (1/2)⁴] = 1/8 [1 - 1/16] = 1/8 * 15/16 = 15/128