Question

Someone help me integrate this please >_

Answer 1

∫( (sinx) / (2 - 3cosx)) dx.

From laws of integration:   ∫ f¹(u) / f(u) du = In(f(u)) + constant.

d/dx (2 - 3cosx) =  0 -3(-sinx) = 3sinx.

1/3d/dx(2 - 3cosx) = (1/3)*3sinx = sinx.

∫ ((sinx) / (2 - 3cosx)) dx.      =  ∫ ((1/3) d/dx (2 - 3cosx) / (2 - 3cosx))dx

= 1/3 ∫ (d/dx (2 - 3cosx) / (2 - 3cosx))dx 

= (1/3)ln(2 - 3cosx) + Constant.