Answer
$$\frac{{{{\sin }^8}x}}{8} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sin }^7}x\cos x} dx \cr
& {\text{or we can write as}} \cr
& = \int {{{\left( {\sin x} \right)}^7}\cos x} dx \cr
& {\text{set }}u = \sin x{\text{ then }}\frac{{du}}{{dx}} = \cos x,\,\,\,\,\,\,\,\,\frac{{du}}{{\cos x}} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {{{\left( {\sin x} \right)}^7}\cos x} dx = \int {{u^7}\cos x} \left( {\frac{{du}}{{\cos x}}} \right) \cr
& = \int {{u^7}} du \cr
& {\text{integrate by using the power rule for integration}} \cr
& = \frac{{{u^8}}}{8} + C \cr
& {\text{write in terms of }}x \cr
& = \frac{{{{\left( {\sin x} \right)}^8}}}{8} + C \cr
& or \cr
& = \frac{{{{\sin }^8}x}}{8} + C \cr} $$
