What is the integral from 0 to t of 9e^ssin(t-s) ?

The integral from 0 to t of 9e^ssin(t-s) is 9/2 e^t-9(1/2 cos(t)+1/2 sin(t))

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integral from 0 to t of 9e^ssin(t-s)

\int_{0}^{t} 9 e^{s} sin (t - s) d s

\frac{9}{2} e^{t} - 9 (\frac{1}{2} cos (t) + \frac{1}{2} sin (t))

Solution steps

\int_{0}^{t} 9 e^{s} sin (t - s) d s

Compute the indefinite integral:

\int 9 e^{s} sin (t - s) d s = 9 (\frac{e ^{s} cos ( - s + t )}{2} + \frac{e ^{s} sin ( - s + t )}{2}) + C

Compute the boundaries:

\int_{0}^{t} 9 e^{s} sin (t - s) d s = \frac{9}{2} e^{t} - 9 (\frac{1}{2} cos (t) + \frac{1}{2} sin (t))

= \frac{9}{2} e^{t} - 9 (\frac{1}{2} cos (t) + \frac{1}{2} sin (t))

What is the integral from 0 to t of 9e^ssin(t-s) ?
The integral from 0 to t of 9e^ssin(t-s) is 9/2 e^t-9(1/2 cos(t)+1/2 sin(t))