Integrate
(cosx)/((2+sinx)(3-sinx)) dx
∫ (cosx)/((2+sinx)(3-sinx)) dx
let sinx = u
then cosx = du/dx
and we get
∫ 1/((2+u)(3-u)) du
Using partial fractions:
let 1/( (2+u)(3-u)) = A/(2+u) + B(3-u)
then (2+u)(3-u) = A(3-u) + B(2+u)
Remember -1 ≤ u ≤ 1, so pick any values of u in that domain
let u = 0 ---> (2)(3) = 3A + 2B
let u = 1 ---> (3)(2) = 2A + 3B
solving these, we get A = 6/5 and B = 6/5
so ∫ 1/((2+u)(3-u)) du = ∫ (6/5)(1/(2+u)) + (6/5)(1/(3-u)) du
= (6/5) (ln(2+u) - ln(3-u) ) + c
recall that u = sinx
so .....
∫ (cosx)/((2+sinx)(3-sinx)) dx = (6/5)(ln(2+sinx) - ln(3-sinx) ) + c