Answer
$$\int\frac{\sin2x}{1+\cos^2x}dx=-\ln(1+\cos^2x)+c$$
Work Step by Step
To evaluate the integral
$$\int\frac{\sin2x}{1+\cos^2x}dx$$
we will use substitution $1+\cos^2x=t$ which gives us $dt=(1+\cos^2x)'dx=-2\cos x\sin xdx=-\sin2xdx\Rightarrow -dt=\sin2xdx$. Putting this into the integral we get:
$$\int\frac{\sin2x}{1+\cos^2x}dx=\int\frac{1}{t}(-dt)=-\ln|t|+c$$
where $c$ is arbitrary constant. Now we have to express solution in terms of $x$:
$$\int\frac{\sin2x}{1+\cos^2x}dx=-\ln|t|+c=-\ln(1+\cos^2x)+c$$
