Answer
$$\frac{2}{{3\ln 3}}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /4} {{{\left( {\frac{1}{3}} \right)}^{\tan t}}} {\sec ^2}tdt \cr
& {\text{Use substitution:}}\cr
& {\text{Let }}u = \tan t,{\text{ then }}du = {\sec ^2}tdt \cr
& {\text{changing the limits of integration}} \cr
& \,\,\,\,\,\,{\text{For }}x = \pi /4,{\text{ }}u = \tan \left( {\pi /4} \right) = 1 \cr
& \,\,\,\,\,\,{\text{For }}x = 0,{\text{ }}u = \tan \left( 0 \right) = 0 \cr
& {\text{write the integral in terms of }}u \cr
& \int_0^{\pi /4} {{{\left( {\frac{1}{3}} \right)}^{\tan t}}} {\sec ^2}tdt = \int_0^1 {{{\left( {\frac{1}{3}} \right)}^u}} du \cr
& {\text{integrate using }}\int {{a^u}} du = \frac{{{a^u}}}{{\ln a}} + C\cr
& {\text{ for this exercise set }}a = \frac{1}{3} \cr
& = \left( {\frac{{{{\left( {1/3} \right)}^u}}}{{\ln \left( {1/3} \right)}}} \right)_0^1 \cr
& = \left( {\frac{{{{\left( {1/3} \right)}^u}}}{{ - \ln 3}}} \right)_0^1 \cr
& {\text{use fundamental theorem of calculus: }}\cr
& \int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = - \frac{1}{{\ln 3}}\left( {{{\left( {\frac{1}{3}} \right)}^1} - {{\left( {\frac{1}{3}} \right)}^0}} \right) \cr
& {\text{simplifying}} \cr
& = - \frac{1}{{\ln 3}}\,\left( {\frac{1}{3} - 1} \right) \cr
& = - \frac{1}{{\ln 3}}\,\left( { - \frac{2}{3}} \right) \cr
& = \frac{2}{{3\ln 3}} \cr} $$
