## Thomas' Calculus 13th Edition

$$\frac{2}{{3\ln 3}}$$
\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}