Answer
$ - \frac{1}{{\ln 5}}\cos \left( {{5^t}} \right) + C$
Work Step by Step
$$\eqalign{
& \int {{5^t}\sin \left( {{5^t}} \right)} dt \cr
& {\text{Let }}u = {5^t},{\text{ then }}du = {5^t}\left( {\ln 5} \right)dt,{\text{ }}\frac{1}{{\ln 5}}du = {5^t}dt \cr
& {\text{Applying the substitution}} \cr
& \int {{5^t}\sin \left( {{5^t}} \right)} dt = \int {\sin u} \left( {\frac{1}{{\ln 5}}} \right)du \cr
& = \frac{1}{{\ln 5}}\int {\sin u} du \cr
& {\text{Integrate }} \cr
& = \frac{1}{{\ln 5}}\left( { - \cos u} \right) + C \cr
& = - \frac{1}{{\ln 5}}\cos u + C \cr
& {\text{Write in terms of }}t,{\text{ substitute }}u = {5^t} \cr
& = - \frac{1}{{\ln 5}}\cos \left( {{5^t}} \right) + C \cr} $$
