Answer
$$\ln \sqrt 2 $$
Work Step by Step
$$\eqalign{
& \int_{\pi /4}^{\pi /2} {\cot t} dt \cr
& {\text{Use the identity }}\cot t = \frac{{\cos t}}{{\sin t}} \cr
& \int_{\pi /4}^{\pi /2} {\frac{{\cos t}}{{\sin t}}} dt \cr
& {\text{Use substitution:}}\cr
& {\text{Let }}u = \sin t,{\text{ so that }}du = \cos tdt \cr
& {\text{The new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = \pi /2,{\text{ }}u = \sin \left( {\pi /2} \right) = 1 \cr
& \,\,\,\,\,\,{\text{If }}x = \pi /4,{\text{ }}u = \sin \left( {\pi /4} \right) = \sqrt 2 /2 \cr
& {\text{Write the integral in terms of }}u \cr
& \int_{\pi /4}^{\pi /2} {\frac{{\cos t}}{{\sin t}}} dt\int_{\sqrt 2 /2}^1 {\frac{1}{u}} du \cr
& {\text{Integrate }} \cr
& = \left( {\ln \left| u \right|} \right)_{\sqrt 2 /2}^1 \cr
& {\text{Use the fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = \ln 1 - \ln \left( {\frac{{\sqrt 2 }}{2}} \right) \cr
& {\text{Simplifying, we get:}} \cr
& = - \ln \left( {\frac{{\sqrt 2 }}{2}} \right) \cr
& = \ln \sqrt 2 \cr} $$
