Answer
$$\frac{{{\pi ^3}}}{8} - 3\pi + 6$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /2} {{t^3}\cos t} dt \cr
& {\text{From the table of integrals in the back of the book}} \cr
& \int {{u^n}\cos udu} = {u^n}\sin u - n\int {{u^{n - 1}}\sin udu} \cr
& \int_0^{\pi /2} {{t^3}\cos t} dt = \left[ {{t^3}\sin t} \right]_0^{\pi /2} - 3\int_0^{\pi /2} {{t^{3 - 1}}\sin tdt} \cr
& = \left[ {{t^3}\sin t} \right]_0^{\pi /2} - 3\int_0^{\pi /2} {{t^2}\sin tdt} \cr
& {\text{From the table of integrals in the back of the book}} \cr
& \int {{u^n}\sin udu} = - {u^n}\cos u + n\int {{u^{n - 1}}\cos udu} \cr
& \int_0^{\pi /2} {{t^3}\cos t} dt = \left[ {{t^3}\sin t} \right]_0^{\pi /2} - 3\left[ { - {t^2}\cos t + 2\int_0^{\pi /2} {t\cos t} } \right] \cr
& \int_0^{\pi /2} {{t^3}\cos t} dt = \left[ {{t^3}\sin t} \right]_0^{\pi /2} + 3\left[ {{t^2}\cos t} \right]_0^{\pi /2} - 6\int_0^{\pi /2} {t\cos t} \cr
& \int_0^{\pi /2} {{t^3}\cos t} dt = \left[ {{t^3}\sin t + 3{t^2}\cos t} \right]_0^{\pi /2} - 6\int_0^{\pi /2} {t\cos t} \cr
& {\text{From the table of integrals in the back of the book}} \cr
& \int {u\cos udu} = \cos u + u\sin u + C \cr
& \int_0^{\pi /2} {{t^3}\cos t} dt = \left[ {{t^3}\sin t + 3{t^2}\cos t} \right]_0^{\pi /2} - 6\left[ {\cos t + t\sin t} \right]_0^{\pi /2} \cr
& \int_0^{\pi /2} {{t^3}\cos t} dt = \left[ {{t^3}\sin t + 3{t^2}\cos t - 6\cos t - 6t\sin t} \right]_0^{\pi /2} \cr
& {\text{Evaluating}} \cr
& {\text{ = }}\left[ {{{\left( {\frac{\pi }{2}} \right)}^3}\sin \left( {\frac{\pi }{2}} \right) + 3{{\left( {\frac{\pi }{2}} \right)}^2}\cos \left( {\frac{\pi }{2}} \right) - 6\cos \left( {\frac{\pi }{2}} \right) - 6\left( {\frac{\pi }{2}} \right)\sin \left( {\frac{\pi }{2}} \right)} \right] \cr
& - \left[ {{{\left( 0 \right)}^3}\sin \left( 0 \right) + 3{{\left( 0 \right)}^2}\cos \left( 0 \right) - 6\cos \left( 0 \right) - 6\left( 0 \right)\sin \left( 0 \right)} \right] \cr
& {\text{Simplifying}} \cr
& {\text{ = }}\frac{{{\pi ^3}}}{8} - 3\pi + 6 \cr} $$
