Answer
$$ - \frac{1}{6}\ln \left| {\frac{{\cos 2x}}{{3 - \cos 2x}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sin 2x}}{{\left( {\cos 2x} \right)\left( {3 - \cos 2x} \right)}}dx} ,\,\,\,\,u = \cos 2x \cr
& {\text{Using the given substitution}} \cr
& u = \cos 2x,\,\,\,\,\,\,\,du = - 2\sin 2xdx,\,\,\,\,\,\,dx = \frac{{du}}{{ - 2\sin 2x}} \cr
& {\text{write in terms of }}u \cr
& \int {\frac{{\sin 2x}}{{\left( {\cos 2x} \right)\left( {3 - \cos 2x} \right)}}dx} = \int {\frac{{\sin 2x}}{{u\left( {3 - u} \right)}}\left( {\frac{{du}}{{ - 2\sin 2x}}} \right)} \cr
& = \int {\frac{1}{{u\left( {3 - u} \right)}}\left( {\frac{{du}}{{ - 2}}} \right)} \cr
& = - \frac{1}{2}\int {\frac{u}{{u\left( {3 - u} \right)}}du} \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By formula 65}} \cr
& \left( {65} \right):\,\,\,\,\,\,\int {\frac{{du}}{{u\left( {a + bu} \right)}} = \frac{1}{a}\ln \left| {\frac{u}{{a + bu}}} \right| + C} \cr
& {\text{take }}a = 3,\,\,\,b = - 1{\text{ }} \cr
& - \frac{1}{2}\int {\frac{u}{{u\left( {3 - u} \right)}}du} = - \frac{1}{{2\left( 3 \right)}}\ln \left| {\frac{u}{{3 - u}}} \right| + C \cr
& {\text{simplifying}} \cr
& - \frac{1}{2}\int {\frac{u}{{u\left( {3 - u} \right)}}du} = - \frac{1}{6}\ln \left| {\frac{u}{{3 - u}}} \right| + C \cr
& {\text{write in terms of }}x{\text{ replace }}\cos 2x{\text{ for }}u \cr
& = - \frac{1}{6}\ln \left| {\frac{{\cos 2x}}{{3 - \cos 2x}}} \right| + C \cr} $$
