Answer
$\frac{1}{2}\sin \left( {{t^2}} \right) - \frac{1}{3}{\sin ^3}\left( {{t^2}} \right) + \frac{1}{{10}}{\sin ^5}\left( {{t^2}} \right) + C$
Work Step by Step
$$\eqalign{
& \int {t{{\cos }^5}\left( {{t^2}} \right)} dt \cr
& {\text{Split }}{\cos ^5}\left( {{t^2}} \right){\text{ as }}{\cos ^4}\left( {{t^2}} \right)\cos \left( {{t^2}} \right) \cr
& \int {t{{\cos }^4}\left( {{t^2}} \right)\cos \left( {{t^2}} \right)} dt \cr
& {\text{Use the property }}{\left( {{a^m}} \right)^n} = {a^{mn}} \cr
& = \int {t{{\left[ {{{\cos }^2}\left( {{t^2}} \right)} \right]}^2}\cos \left( {{t^2}} \right)} dt \cr
& {\text{Rewrite }}{\sin ^2}x{\text{ using }}{\sin ^2}x + {\cos ^2}x = 1 \cr
& = \int {t{{\left[ {1 - {{\sin }^2}\left( {{t^2}} \right)} \right]}^2}\cos \left( {{t^2}} \right)} dt \cr
& {\text{Expand the binomial}} \cr
& = \int {t\left( {1 - 2{{\sin }^2}\left( {{t^2}} \right) + {{\sin }^4}\left( {{t^2}} \right)} \right)\cos \left( {{t^2}} \right)} dt \cr
& {\text{Integrate by the substitution method}} \cr
& {\text{Let }}u = \sin \left( {{t^2}} \right),{\text{ }}du = 2t\cos \left( {{t^2}} \right)dt \cr
& = \int {\left( {1 - 2{u^2} + {u^4}} \right)\left( {\frac{1}{2}} \right)} du \cr
& = \frac{1}{2}\int {\left( {1 - 2{u^2} + {u^4}} \right)} du \cr
& {\text{Use the power rule }}\int {{u^n}} du = \frac{1}{{n + 1}}{u^{n + 1}} + C \cr
& = \frac{1}{2}\left( {u - \frac{2}{3}{u^3} + \frac{1}{5}{u^5}} \right) + C \cr
& = \frac{1}{2}u - \frac{1}{3}{u^3} + \frac{1}{{10}}{u^5} + C \cr
& {\text{Write in terms of }}t,{\text{ substitute }}\sin \left( {{t^2}} \right){\text{ for }}u \cr
& = \frac{1}{2}\sin \left( {{t^2}} \right) - \frac{1}{3}{\sin ^3}\left( {{t^2}} \right) + \frac{1}{{10}}{\sin ^5}\left( {{t^2}} \right) + C \cr} $$