Six = t ; sinπ/6 = 1/2 ; sinπ/2 = 1 ⇒
{π/6∫π/2} [cos²x/∛sinx] · d(sinx) = {1/2∫1} (1-t²)/∛t ·dt =
= ∫ dt/∛t - ∫ t²dt/∛t = ∫ t^(-1/3) ·dt - ∫ t^(5/3)·dt=
= {1/2 | 1} {[ (t^(-1/3+1))/(-1/3+1)] - [(t^(5/3+1)/(5/3+1)]} =
= 2/3· t^(2/3) - 8/3 · t^(8/3) |1/2⇵1 =
|1
= 2/3 · ∛t² - 8/3·t·∛t² = 2/3·∛t²·(1 - 4t) | =
|1/2
= 2/3 ·1·(1 -4·1) - 2/3·∛(1/4) ·(1 - 4·1/2) =
= -2 +4/3 ·∛(1/4)
Если допустил случайные ошибки исправьте сами