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