Answer
$$\frac{{{{\sin }^{31}}2x}}{{31\left( {{2^{31}}} \right)}} + C$$
Work Step by Step
$$\eqalign{
& \int {\left( {{{\cos }^{32}}x{{\sin }^{30}}x - {{\cos }^{30}}x{{\sin }^{32}}x} \right)dx} \cr
& {\text{Factoring the integrand, the common factor is }}{\sin ^{30}}x{\cos ^{30}}x \cr
& = \int {{{\sin }^{30}}x{{\cos }^{30}}x\left( {{{\cos }^2}x - {{\sin }^2}x} \right)dx} \cr
& {\text{Use the identity }}{\cos ^2}x - {\sin ^2}x = \cos 2x \cr
& = \int {{{\sin }^{30}}x{{\cos }^{30}}x\left( {\cos 2x} \right)dx} \cr
& {\text{Recall that }}\sin 2\theta = 2\sin \theta \cos \theta ,{\text{ then}} \cr
& = \frac{1}{{{2^{30}}}}\int {{2^{30}}{{\sin }^{30}}x{{\cos }^{30}}x\left( {\cos 2x} \right)dx} \cr
& = \frac{1}{{{2^{30}}}}\int {{{\left( {\sin 2x} \right)}^{30}}\left( {\cos 2x} \right)dx} \cr
& = \frac{1}{{{2^{31}}}}\int {{{\left( {\sin 2x} \right)}^{30}}\left( {2\cos 2x} \right)dx} \cr
& {\text{Integrating}} \cr
& = \frac{1}{{{2^{30}}}}\left( {\frac{{{{\left( {\sin 2x} \right)}^{31}}}}{{31}}} \right) + C \cr
& {\text{Simplifying}} \cr
& = \frac{{{{\sin }^{31}}2x}}{{31\left( {{2^{31}}} \right)}} + C \cr} $$
