Answer
$$\frac{1}{{b\left( {n + 1} \right)}}{\sin ^{n + 1}}\left( {a + bx} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^n}\left( {a + bx} \right)\cos \left( {a + bx} \right)} dx \cr
& {\text{Integrate by substitution}}{\text{,}} \cr
& {\text{ let }}u = \sin \left( {a + bx} \right),{\text{ }} \cr
& {\text{then }}du = \cos \left( {a + bx} \right)bdx,\,\,\,\cos \left( {a + bx} \right)dx = \frac{{du}}{b} \cr
& {\text{Using the substitution}} \cr
& \int {{{\sin }^n}\left( {a + bx} \right)\cos \left( {a + bx} \right)} dx \cr
& = \int {{u^n}\left( {\frac{{du}}{b}} \right)} \cr
& = \frac{1}{b}\int {{u^n}du} \cr
& = \frac{1}{b}\left( {\frac{{{u^{n + 1}}}}{{n + 1}}} \right) + C \cr
& = \frac{1}{{b\left( {n + 1} \right)}}{u^{n + 1}} + C \cr
& {\text{Substituting back }}u = \sin \left( {a + bx} \right) \cr
& = \frac{1}{{b\left( {n + 1} \right)}}{\sin ^{n + 1}}\left( {a + bx} \right) + C \cr} $$
