Answer
$$1$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /3} {\frac{{\sin x}}{{{{\cos }^2}x}}} dx \cr
& {\text{Let }}u = \cos x,{\text{ }}du = - \sin xdx \cr
& {\text{The new limits of integration are:}} \cr
& x = \frac{\pi }{3} \to u = \cos \left( {\frac{\pi }{3}} \right) = \frac{1}{2} \cr
& x = 0 \to u = \cos \left( 0 \right) = 1 \cr
& {\text{Substituting}} \cr
& \int_0^{\pi /3} {\frac{{\sin x}}{{{{\cos }^2}x}}} dx = \int_1^{1/2} {\frac{1}{{{u^2}}}} \left( { - du} \right) \cr
& = \int_1^{1/2} {\left( { - \frac{1}{{{u^2}}}} \right)} du \cr
& {\text{Integrating}} \cr
& = \left[ {\frac{1}{u}} \right]_1^{1/2} \cr
& = \frac{1}{{1/2}} - \frac{1}{1} \cr
& = 1 \cr} $$
