Question

I got the wrong answer...
So posting again.....

Integrate
(cosx)/((2+sinx)(3-sinx)) dx

Answer 1

If I recall, my answer was (6/5)(ln(2+sinx) - ln(3-sinx) ) + c

When I differentiate that using Wolfram, I got
6 cosx/((2+sinx)(3-sinx)) <---- Wolfram expanded the denominator

so its 6 vs 6/5 , so we are apart by a factor of 1/5
I can't find my earlier reply due to Jiskha's non-functioning SEARCH
feature, perhaps you can after finding my post.

so try (1/5)(ln(2+sinx) - ln(3-sinx) ) + c

Answer 2

(cosx)/((2+sinx)(3-sinx)) = cosx/(6 + sinx - sin^2x)
Let u = 6 + sinx - sin^2x
du = cosx - 2sinx cosx
∫(cosx)/((2+sinx)(3-sinx)) dx = ∫du/u + ∫2sinx cosx/(6 + sinx - sin^2x)
now let v = sinx and
∫2sinx cosx/(6 + sinx - sin^2x) = ∫2v/((2+v)(3-v)) dv
Integrate that using partial fractions. You'll end up with some ln(z) functions.
Wolframalpha says that the integral is -2/5 tanh^{^-1} (1-2sinx)/5
Recall that tanh^{^-1}(z) = 1/2 ln (1+z)/(1-z)

Answer 3

hmmm. mine looks overly complicated. Using partial fractions,
1/((2+u)(3-u)) = 1/5 (1/(u+2) - 1/(u-3))
so you now start with
1/5 ∫ cosx/(sinx+2) - cosx/(sinx-3) dx
and mathhelper's quick and easy solution falls right out.