Answer
The solution is $$-\arctan(\cos\theta)+c.$$
Work Step by Step
To solve the integral $$\int\frac{\sin \theta}{1+\cos^2\theta}d\theta$$ we will use substitution $\cos\theta=t\Rightarrow -\sin\theta d\theta=dt$ and put that into our integral:
$$\int\frac{\sin \theta}{1+\cos^2\theta}d\theta=-\int\frac{dt}{1+t^2}=-\arctan t+c,$$
where $c$ is arbitrary constant. Now we have to express solution (which is in terms of $t$) in terms of $\theta$:
$$\int\frac{\sin \theta}{1+\cos^2\theta}d\theta=-\arctan t+c=-arctan(\cos\theta)+c.$$
