Answer
$$\frac{1}{{12}}{\tan ^{ - 1}}\frac{{\sin 4x}}{3} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\cos 4x}}{{9 + {{\sin }^2}4x}}dx} ,\,\,\,\,u = \sin 4x \cr
& {\text{Using the given substitution}} \cr
& u = \sin 4x,\,\,\,\,\,\,\,du = 4\cos 4xdx,\,\,\,\,\,\,dx = \frac{1}{{4\cos 4x}}du \cr
& {\text{write in terms of }}u \cr
& \int {\frac{{\cos 4x}}{{9 + {{\sin }^2}4x}}dx} = \int {\frac{{\cos 4x}}{{9 + {u^2}}}\left( {\frac{1}{{4\cos 4x}}du} \right)} \cr
& = \frac{1}{4}\int {\frac{1}{{9 + {u^2}}}du} \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By formula 68}} \cr
& \left( {68} \right):\,\,\,\,\,\,\int {\frac{{du}}{{{a^2} + {u^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\frac{u}{a} + C} \cr
& {\text{take }}a = 3 \cr
& \frac{1}{4}\int {\frac{1}{{9 + {u^2}}}du} = \frac{1}{{4\left( 3 \right)}}{\tan ^{ - 1}}\frac{u}{3} + C \cr
& {\text{simplifying}} \cr
& \frac{1}{4}\int {\frac{1}{{9 + {u^2}}}du} = \frac{1}{{12}}{\tan ^{ - 1}}\frac{u}{3} + C \cr
& {\text{write in terms of }}x{\text{; use }}\sin 4x{\text{ for }}u \cr
& = \frac{1}{{12}}{\tan ^{ - 1}}\frac{{\sin 4x}}{3} + C \cr} $$
