Answer
$$ - \frac{1}{{32}}\cos 16x + \frac{1}{4}\cos 2x + C$$
Work Step by Step
$$\eqalign{
& \int {\sin 7x\cos 9x} dx \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By the formula 60}} \cr
& \left( {60} \right):\,\,\,\,\,\int {\sin mu\cos nu} du = - \frac{{\cos \left( {m + n} \right)u}}{{2\left( {m + n} \right)}} - \frac{{\cos \left( {m - n} \right)u}}{{2\left( {m - n} \right)}} + C \cr
& {\text{with }}u = x,\,\,\,m = 7{\text{ and }}n = 9 \cr
& \cr
& \int {\sin 7x\cos 9x} dx = - \frac{{\cos \left( {7 + 9} \right)x}}{{2\left( {7 + 9} \right)}} - \frac{{\cos \left( {7 - 9} \right)x}}{{2\left( {7 - 9} \right)}} + C \cr
& {\text{simplifying}} \cr
& \int {\sin 7x\cos 9x} dx = - \frac{{\cos 16x}}{{32}} - \frac{{\cos 2x}}{{ - 4}} + C \cr
& \int {\sin 7x\cos 9x} dx = - \frac{1}{{32}}\cos 16x + \frac{1}{4}\cos 2x + C \cr} $$
