## Calculus 8th Edition

$$\int\frac{\sin x}{1+\cos^2x}dx=-\arctan(\cos x)+c$$
To evaluate the integral $$\int\frac{\sin x}{1+\cos^2x}dx$$ we will use substitution $\cos x=t$ which gives us $-\sin xdx=dt\Rightarrow \sin xdx=-dt$. Putting this into the integral we get: $$\int\frac{\sin x}{1+\cos^2x}dx=\int\frac{1}{1+t^2}(-dt)=-\arctan t+c$$ where $c$ is arbitrary constant. Now we have to express solution in terms of $x$: $$\int\frac{\sin x}{1+\cos^2x}dx=-\arctan t+c=-\arctan(\cos x)+c$$