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.`
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