Answer
$(\dfrac{1}{5})\ln|\dfrac{\sin \theta-2}{\sin \theta+3}|+c$
Work Step by Step
Integration by parts formula suggests that $\int p'(x) q(x)=p(x) q(x)-\int p(x) q'(x)dx$
Solve $\int \dfrac{\cos \theta d\theta}{\sin^2 \theta+\sin \theta-6}$
Plug $a=\sin \theta \implies da=\cos \theta d\theta$
Thus, we have $\int \dfrac{da}{a^2+a-6}=\int \dfrac{da}{(a+3)(a-2)}$
Write the integral into partial fractions as follows: $\int \dfrac{da}{(a+3)(a-2)}=\int [-\dfrac{1}{5(a+3)}+\dfrac{1}{5(a-2)} ]da$
This implies that $\dfrac{-\ln |a+3|+\ln |a-2|}{5}+c=(\dfrac{1}{5})\ln|\dfrac{a-2}{a+3}|+c$
Hence, $(\dfrac{1}{5})\ln|\dfrac{a-2}{a+3}|+c=(\dfrac{1}{5})\ln|\dfrac{\sin \theta-2}{\sin \theta+3}|+c$
