\( \int _ { 0 } ^ { t } \sin u e ^ { t - u } d u \)

To evaluate the integral \( \int _ { 0 } ^ { t } \sin u e ^ { t - u } d u \), you can use integration by parts. Set \( v = e^{t-u} \) (hence \( dv = -e^{t-u} du \)) and \( dw = \sin u du \). You will differentiate \( w \) to obtain \( w = -\cos u \). After applying integration by parts, the integral simplifies nicely, and you can find that: \[ \int _{0}^{t} \sin u e^{t-u} du = e^t \left( -\cos(u) \bigg|_{0}^{t} + \int_0^t e^{-u} \cos(u) du \right). \] This method gives you a clear path to evaluating the total integral stepwise! To check your work, you can often utilize properties of the Laplace Transform, especially since the function involves an exponential factor. Think of \( \sin(u) \) and \( e^{-u} \) as inputs in a frequency or decay context, allowing for a neat tie into differential equations typically encountered in physics or engineering problems.