Answer
$$\frac{1}{5}\ln \left| {\frac{{\sin y - 2}}{{\sin y + 3}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\cos ydy}}{{{{\sin }^2}y + \sin y - 6}}} \cr
& {\text{let }}x = \sin y,\,\,\,dx = \cos ydy \cr
& {\text{Write the integral in terms of }}x \cr
& \int {\frac{{\cos ydy}}{{{{\sin }^2}y + \sin y - 6}}} = \int {\frac{{dx}}{{{x^2} + x - 6}}} \cr
& \cr
& {\text{Solve the integral by the method of partial fractions.}} \cr
& {\text{The form of the partial fraction decomposition is}} \cr
& \frac{1}{{{x^2} + x - 6}} = \frac{1}{{\left( {x + 3} \right)\left( {x - 2} \right)}} = \frac{A}{{x + 3}} + \frac{B}{{x - 2}} \cr
& {\text{Multiplying by }}\left( {x + 3} \right)\left( {x - 2} \right){\text{, we have}} \cr
& {\text{1}} = A\left( {x - 2} \right) + B\left( {x + 3} \right) \cr
& {\text{if we set }}x = - 3, \cr
& {\text{1}} = A\left( { - 5} \right) + B\left( 0 \right) \cr
& A = - 1/5 \cr
& {\text{if we set }}x = 2, \cr
& {\text{1}} = A\left( 0 \right) + B\left( 5 \right) \cr
& B = 1/5 \cr
& {\text{then}} \cr
& \frac{1}{{\left( {x + 3} \right)\left( {x - 2} \right)}} = \frac{A}{{x + 3}} + \frac{B}{{x - 2}} = \frac{{ - 1/5}}{{x + 3}} + \frac{{1/5}}{{x - 2}} \cr
& \int {\frac{{dx}}{{{x^2} + x - 6}}} = \int {\left( {\frac{{ - 1/5}}{{x + 3}} + \frac{{1/5}}{{x - 2}}} \right)dx} \cr
& {\text{integrating}} \cr
& = - \frac{1}{5}\ln \left| {x + 3} \right| + \ln \left| {x - 2} \right| + C \cr
& = \frac{1}{5}\ln \left| {\frac{{x - 2}}{{x + 3}}} \right| + C \cr
& {\text{Write the solution in terms of }}x,{\text{ substitute }}\sin y{\text{ for }}x \cr
& = \frac{1}{5}\ln \left| {\frac{{\sin y - 2}}{{\sin y + 3}}} \right| + C \cr} $$
