`int 5^t sin(5^t) dt` Evaluate the indefinite integral.

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.`

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.`