Answer
$$\ln 4$$
Work Step by Step
$$\eqalign{
& \int_{ - \pi /2}^{\pi /6} {\frac{{\cos t}}{{1 - \sin t}}} dt \cr
& {\text{Use substitution:}}\cr
&{\text{Let }}u = 1 - \sin t,{\text{ so that }}du = - \cos tdt \cr
& {\text{The new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}t = \pi /6,{\text{ then }}u = 1 - \sin \left( {\pi /6} \right) = 1/2 \cr
& \,\,\,\,\,\,{\text{If }}t = - \pi /2,{\text{ then }}u = 1 - \sin \left( { - \pi /6} \right) = 2 \cr
& {\text{Then}} \cr
& \int_{ - \pi /2}^{\pi /6} {\frac{{\cos t}}{{1 - \sin t}}} dt = \int_2^{1/2} {\frac{{ - 1}}{u}} du \cr
& {\text{integrate}} \cr
& = - \left( {\ln \left| u \right|} \right)_2^{1/2} \cr
& {\text{use the fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = - \left( {\ln \left| {1/2} \right| - \ln \left| 2 \right|} \right) \cr
& {\text{simplifying, we get:}} \cr
& = - \left( { - \ln 2 - \ln 2} \right) \cr
& = 2\ln 2 \cr
& = \ln 4 \cr} $$
