Answer
$\frac{2}{{\sqrt 3 }} - 1$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /6} {\frac{{\sin t}}{{{{\cos }^2}t}}} dt \cr
& {\text{Let }}u = \cos t,{\text{ }}du = - \sin tdt,{\text{ }}\sin tdt = - du \cr
& {\text{The new limits of integration are:}} \cr
& x = 0 \to u = \cos \left( 0 \right),{\text{ }}u = 1 \cr
& x = \frac{\pi }{6} \to u = \cos \left( {\frac{\pi }{6}} \right),{\text{ }}u = \frac{{\sqrt 3 }}{2} \cr
& {\text{Applying the substitution}} \cr
& \int_0^{\pi /6} {\frac{{\sin t}}{{{{\cos }^2}t}}} dt = \int_1^{\sqrt 3 /2} {\frac{{ - du}}{{{u^2}}}} \cr
& = - \int_1^{\sqrt 3 /2} {\frac{{du}}{{{u^2}}}} \cr
& = - \int_1^{\sqrt 3 /2} {{u^{ - 2}}du} \cr
& {\text{Integrate apply the power rule }}\int {{u^n}du} = \frac{{{u^{n + 1}}}}{{n + 1}}{\text{ + C }} \cr
& = - \left[ {\frac{{{u^{ - 2 + 1}}}}{{ - 2 + 1}}} \right]_1^{\sqrt 3 /2} \cr
& = - \left[ {\frac{{{u^{ - 1}}}}{{ - 1}}} \right]_1^{\sqrt 3 /2} = \left[ {\frac{1}{u}} \right]_1^{\sqrt 3 /2} \cr
& {\text{Evaluating the limits of integration}} \cr
& = \frac{2}{{\sqrt 3 }} - 1 \cr} $$
