Answer
$$\frac{3}{{\root 3 \of {\cos x} }} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\left( {\cos x} \right)}^{ - 4/3}}\sin x} dx \cr
& {\text{set }}u = \cos x{\text{ then }}\frac{{du}}{{dx}} = - \sin x,\,\,\,\,\,\,\,\,\frac{{ - du}}{{\sin x}} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {{{\left( {\cos x} \right)}^{ - 4/3}}\left( {\sin x} \right)} dx = \int {{u^{ - 4/3}}\left( {\sin x} \right)} \left( {\frac{{ - du}}{{\sin x}}} \right) \cr
& = - \int {{u^{ - 4/3}}du} \cr
& {\text{integrate by using the power rule for integration}} \cr
& = - \frac{{{u^{ - 1/3}}}}{{ - 1/3}} + C \cr
& {\text{simplifiyng}} \cr
& = \frac{3}{{{u^{1/3}}}} + C \cr
& = \frac{3}{{\root 3 \of u }} + C \cr
& {\text{write in terms of }}x \cr
& = \frac{3}{{\root 3 \of {\cos x} }} + C \cr} $$