Answer
$$\int\frac{\sin\theta d\theta}{\cos^2\theta+\cos\theta-2}=-\frac{1}{3}\Big(\ln|\cos\theta-1|-\ln|\cos\theta+2|\Big)$$
Work Step by Step
$$I=\int\frac{\sin\theta d\theta}{\cos^2\theta+\cos\theta-2}$$
Set $u=\cos\theta$, which means $$du=-\sin\theta d\theta$$ $$\sin\theta d\theta=-du$$
We then have, $$I=-\int\frac{du}{u^2+u-2}$$ $$I=-\int\frac{du}{(u-1)(u+2)}$$ $$I=-\frac{1}{3}\int\frac{3}{(u-1)(u+2)}du$$ $$I=-\frac{1}{3}\int\frac{(u+2)-(u-1)}{(u-1)(u+2)}du$$ $$I=-\frac{1}{3}\Big(\int\frac{du}{u-1}-\frac{du}{u+2}\Big)$$ $$I=-\frac{1}{3}\Big(\ln|u-1|-\ln|u+2|\Big)$$ $$I=-\frac{1}{3}\Big(\ln|\cos\theta-1|-\ln|\cos\theta+2|\Big)$$
