Answer
$$\ln \left( 2 \right)$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /12} {6\tan 3x} dx \cr
& = 6\int_0^{\pi /12} {\tan 3x} dx \cr
& {\text{Integrate with the formula }}\cr
& \int {\tan ax} dx = - \frac{1}{a}\ln \left| {\cos ax} \right| + C \cr
& = - \frac{6}{3}\left( {\ln \left| {\cos 3x} \right|} \right)_0^{\pi /12} \cr
& {\text{Use the fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = - 2\left( {\ln \left| {\cos 3\left( {\frac{\pi }{{12}}} \right)} \right| - \ln \left| {\cos 3\left( 0 \right)} \right|} \right) \cr
& {\text{Simplifying, we get:}} \cr
& = - 2\left( {\ln \left| {\frac{{\sqrt 2 }}{2}} \right| - \ln \left| 1 \right|} \right) \cr
& = - \ln \left( {\frac{{{{\sqrt 2 }^2}}}{{{2^2}}}} \right) \cr
& = \ln \left( 2 \right) \cr} $$
