Answer
$$\frac{6}{{\ln 7}}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /2} {{7^{\cos t}}\sin t} dt \cr
& {\text{Use substitution:}}\cr
& {\text{Let }}u = \cos t,{\text{ then }}du = - \sin tdt,\,\,\,\,\sin tdt = - du \cr
& {\text{changing the limits of integration}} \cr
& \,\,\,\,\,\,{\text{For }}x = \pi /2,{\text{ }}u = \cos \left( {\pi /2} \right) = 0 \cr
& \,\,\,\,\,\,{\text{For }}x = 0,{\text{ }}u = \cos \left( 0 \right) = 1 \cr
& {\text{write the integral in terms of }}u \cr
& \int_0^{\pi /2} {{7^{\cos t}}\sin t} dt = \int_1^0 {{7^u}} \left( { - du} \right) \cr
& = - \int_1^0 {{7^u}} du \cr
& {\text{integrate using }}\int {{a^u}} du = \frac{{{a^u}}}{{\ln a}} + C \cr
& = - \left( {\frac{{{7^u}}}{{\ln 7}}} \right)_1^0 \cr
& {\text{using properties of the integrals}} \cr
& = \left( {\frac{{{7^u}}}{{\ln 7}}} \right)_0^1 \cr
& {\text{use fundamental theorem of calculus }}\cr
& \int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = \frac{{{7^1} - {7^0}}}{{\ln 7}} \cr
& {\text{simplifying}} \cr
& = \frac{6}{{\ln 7}} \cr} $$
