Answer
$$ e $$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /4} {\left( {1 + {e^{\tan \theta }}} \right)} {\sec ^2}\theta d\theta \cr
& {\text{Use substitution}}{\text{. Let }}u = \tan \theta,{\text{ so that }}du = {\sec ^2}\theta d\theta \cr
& {\text{The new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}\theta = \pi /4,{\text{ }}u = \tan \left( {\pi /4} \right) = 1 \cr
& \,\,\,\,\,\,{\text{If }}\theta = 0,{\text{ }}u = \tan \left( 0 \right) = 0 \cr
& {\text{Then}} \cr
& \int_0^{\pi /4} {\left( {1 + {e^{\tan \theta }}} \right)} {\sec ^2}\theta d\theta = \int_0^1 {\left( {1 + {e^u}} \right)} du \cr
& {\text{Integrate}} \cr
& \int_0^1 {\left( {1 + {e^u}} \right)} du = \left( {u + {e^u}} \right)_0^1 \cr
& {\text{Use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = \left( {1 + {e^1}} \right) - \left( {0 + {e^0}} \right) \cr
& {\text{Simplifying}} \cr
& = 1 + {e^1} - 0 - 1 \cr
& = e \cr} $$
