Answer
$$ - \frac{1}{5}\cos {\left( {x + 2} \right)^5} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\left( {x + 2} \right)}^4}\sin {{\left( {x + 2} \right)}^5}} dx \cr
& {\text{set }}u = {\left( {x + 2} \right)^5}{\text{ then }}\frac{{du}}{{dx}} = 5{\left( {x + 2} \right)^4},\,\,\,\,\,\,\,\,\frac{{du}}{{5{{\left( {x + 2} \right)}^4}}} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {{{\left( {x + 2} \right)}^4}\sin {{\left( {x + 2} \right)}^5}} dx = \int {{{\left( {x + 2} \right)}^4}\sin u} \left( {\frac{{du}}{{5{{\left( {x + 2} \right)}^4}}}} \right) \cr
& {\text{cancel the common factors}} \cr
& = \int {\sin u} \left( {\frac{{du}}{5}} \right) \cr
& = \frac{1}{5}\int {\sin u} du \cr
& {\text{integrate by using the Basic Trigonometric integral }}\int {\sin x} dx = - \cos x + C \cr
& = - \frac{1}{5}\cos u + C \cr
& {\text{write in terms of }}x \cr
& = - \frac{1}{5}\cos {\left( {x + 2} \right)^5} + C \cr} $$
