Answer
$$\frac{1}{{3\left( {\sin 3x + 1} \right)}} + \frac{1}{3}\ln \left| {\frac{{\sin 3x}}{{\sin 3x + 1}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\cos 3x}}{{\left( {\sin 3x} \right){{\left( {\sin 3x + 1} \right)}^2}}}dx} \cr
& {\text{Make an appropiate }}u{\text{ - substitution }} \cr
& u = \sin 3x,\,\,\,\,\,\,\,du = 3\cos 3xdx,\,\,\,\,\,\,dx = \frac{{du}}{{3\cos 3x}} \cr
& {\text{write in terms of }}u \cr
& \int {\frac{{\cos 3x}}{{\left( {\sin 3x} \right){{\left( {\sin 3x + 1} \right)}^2}}}dx} = \int {\frac{{\cos 3x}}{{u{{\left( {u + 1} \right)}^2}}}\left( {\frac{{du}}{{3\cos 3x}}} \right)} \cr
& = \int {\frac{1}{{u{{\left( {u + 1} \right)}^2}}}\left( {\frac{{du}}{3}} \right)} \cr
& = \frac{1}{3}\int {\frac{1}{{u{{\left( {u + 1} \right)}^2}}}du} \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By formula 67}} \cr
& \left( {67} \right):\,\,\,\,\,\,\int {\frac{{du}}{{u{{\left( {a + bu} \right)}^2}}} = \frac{1}{{a\left( {a + bu} \right)}} + \frac{1}{{{a^2}}}\ln \left| {\frac{u}{{a + bu}}} \right| + C} \cr
& {\text{take }}a = 1,\,\,\,b = 1 \cr
& \frac{1}{3}\int {\frac{1}{{u{{\left( {u + 1} \right)}^2}}}du} = \frac{1}{{3\left( 1 \right)\left( {u + 1} \right)}} + \frac{1}{{3{{\left( 1 \right)}^2}}}\ln \left| {\frac{u}{{u + 1}}} \right| + C \cr
& {\text{simplifying}} \cr
& = \frac{1}{{3\left( {u + 1} \right)}} + \frac{1}{3}\ln \left| {\frac{u}{{u + 1}}} \right| + C \cr
& {\text{write in terms of }}x{\text{, and replace }}\sin 3x{\text{ for }}u \cr
& = \frac{1}{{3\left( {\sin 3x + 1} \right)}} + \frac{1}{3}\ln \left| {\frac{{\sin 3x}}{{\sin 3x + 1}}} \right| + C \cr} $$
