Answer
$$ - \frac{{\cos 7x}}{{14}} + \frac{{\cos 3x}}{6} + C$$
Work Step by Step
$$\eqalign{
& \int {\sin 2x\cos 5x} dx \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{The integrand has a product of trigonometric functions}} \cr
& {\text{Use formula 40}} \cr
& \left( {40} \right):\,\,\,\,\,\int {\sin mu\cos nu} du = - \frac{{\cos \left( {m + n} \right)u}}{{2\left( {m + n} \right)}} - \frac{{\cos \left( {m - n} \right)u}}{{2\left( {m - n} \right)}} + C \cr
& \int {\sin 2x\cos 5x} dx\,\, \Rightarrow \,\,\,\,m = 2,\,\,\,n = 5 \cr
& {\text{Then by formula 40}} \cr
& \int {\sin 2x\cos 5x} dx = - \frac{{\cos \left( {2 + 5} \right)x}}{{2\left( {2 + 5} \right)}} - \frac{{\cos \left( {2 - 5} \right)x}}{{2\left( {2 - 5} \right)}} + C \cr
& {\text{simplifying}} \cr
& \int {\sin 2x\cos 5x} dx = - \frac{{\cos 7x}}{{14}} - \frac{{\cos 3x}}{{ - 6}} + C \cr
& \int {\sin 2x\cos 5x} dx = - \frac{{\cos 7x}}{{14}} + \frac{{\cos 3x}}{6} + C \cr} $$
