#### Answer

$$\int\frac{\cos ydy}{\sin^2y+\sin y-6}=\frac{1}{5}\ln|\sin y-2|-\frac{1}{5}\ln|\sin y+3|+C$$

#### Work Step by Step

$$I=\int\frac{\cos ydy}{\sin^2y+\sin y-6}$$
Set $u=\sin y$, which means $$du=\cos ydy$$
We then have, $$I=\int\frac{du}{u^2+u-6}$$ $$I=\int\frac{du}{(u-2)(u+3)}$$ $$I=\frac{1}{5}\int\frac{5}{(u-2)(u+3)}du$$ $$I=\frac{1}{5}\int\frac{(u+3)-(u-2)}{(u-2)(u+3)}du$$ $$I=\frac{1}{5}\int\frac{1}{u-2}du-\frac{1}{5}\int\frac{1}{u+3}du$$ $$I=\frac{1}{5}\ln|u-2|-\frac{1}{5}\ln|u+3|+C$$ $$I=\frac{1}{5}\ln|\sin y-2|-\frac{1}{5}\ln|\sin y+3|+C$$