Revised:

∫ sin 2y dy /(sin^2 y + 2)^2

= ∫ 2 sin y cos y dy /(sin^2 y +2)^2

Let u= sin^2 y

du = 2 sin y cos y dy

∫ sin 2y dy /(sin^2 y + 2)^2 = ∫ du /(u+2)^2

= -1/(u+2)

= -1/(sin^2 y+2)

F(y) = -1/(sin^2 y +2)

F(pi/6) = -4/9

F(0) = -1/2

F(pi/6)-F(0) = -4/9+1/2 = 1/18

Let u = sin^2(y) + 2

du = 2sin(y)cos(y)dy

= sin(2y) dy

dy = du/sin(2y)

∫(sin2y/(sin^2y+2)^2 )dy = ∫[ sin(2y)/u^2 * du/sin(2y)]du

= ∫ 1/u^2 du

= ∫u^-2 du

= -1/u

= -1/(sin^2(y) + 2) from 0 to π/6

= -1/[ (1/2)^2 + 2] - [ -1/2] ............. sin(∫π/6) = 1/2

= -1/[1/4 + 2] + 1/2

= -4/9 + 1/2

= 1/18