Calculus: Early Transcendentals, Chapter 5, 5.5, Section 5.5, Problem 29

You need to use the following substitution 5^t=u , such that:
5^t=u=>5^t*ln 5 dt = du => 5^t*dt= (du)/(ln 5)
int5^t*sin(5^t) dt = (1/(ln 5))*int sin u du
(1/(ln 5))*int sin u du = -(1/(ln 5))*cos u + c
Replacing back 5^t for u yields:
int5^t*sin(5^t) dt = (-cos(5^t))/(ln 5)+c
Hence, evaluating the indefinite integral, yields int5^t*sin(5^t) dt = (-cos(5^t))/(ln 5)+c.