Answer
$$ - \frac{{{{\cos }^9}x}}{9} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\cos }^8}x\sin x} dx \cr
& {\text{set }}u = \cos x{\text{ then }}\frac{{du}}{{dx}} = - \sin x,\,\,\,\,\,\,\,\,\frac{{ - du}}{{\sin x}} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {{{\left( {\cos x} \right)}^8}\left( {\sin x} \right)} dx = \int {{u^8}\left( {\sin x} \right)} \left( {\frac{{ - du}}{{\sin x}}} \right) \cr
& = - \int {{u^8}du} \cr
& {\text{integrate by using the power rule for integration}} \cr
& = - \frac{{{u^9}}}{9} + C \cr
& {\text{write in terms of }}x \cr
& = - \frac{{{{\left( {\cos x} \right)}^9}}}{9} + C \cr
& or \cr
& = - \frac{{{{\cos }^9}x}}{9} + C \cr} $$
