Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 32

Evaluate $\displaystyle \int^t_0 e^s \sin (t-s) ds$ by using Integration by parts.
If we let $u = \sin (t-s)$ and $dv = e^s ds$, then
$du = - \cos (t-s) ds \quad v = e^s$

So,

$\begin{equation} \begin{aligned} \int^t_0 e^s \sin (t-s) ds &= uv - \int v du = e^s \sin (t -s) - \int e^s \left( -\cos (t-s) \right) ds\\ \\ &= e^s \sin (t-s) + \int e^s \cos (t-s) ds \end{aligned} \end{equation}$


To evaluate $\displaystyle \int e^s \cos (t - s) ds$, we must use Integration by parts once more...

$\begin{equation} \begin{aligned} \text{Hence, if we let } u_1 &= \cos (t -s ) &&\text{ and }& dv_1 &= e^s ds \text{ , then} \\ \\ du_1 &= \sin (t-s) ds &&& v_1 &= e^s \end{aligned} \end{equation}$


So, $\displaystyle \int e^s \cos (t-s) ds = u_1 v_1 - \int v_1 du_1 = e^s \cos (t-s) - \int e^s \sin (t-s) ds$

Going back to the first equation,
$\displaystyle \int^t_0 e^s \sin (t-s) ds = e^s \sin (t-s) + \left[ e^s \cos (t-s) - \int e^s \sin (t-s) ds \right]$

Combining like terms

$\begin{equation} \begin{aligned} 2 \int^t_0 e^s \sin (t-s) ds &= e^s \sin (t-s) + e^s \cos (t-s)\\ \\ \int^t_0 e^s \sin (t-s) ds &= \frac{e^s \sin (t-s) + e^s \cos (t-s)}{2}\\ \\ &= \frac{e^s}{2} [\sin(t-s) + \cos (t-s)] \end{aligned} \end{equation}$


Evaluating from 0 to $t$, we have

$\begin{equation} \begin{aligned} &= \frac{e^t}{2} - \frac{1}{2} \sin t - \frac{1}{2} \cos t\\ \\ &= \frac{1}{2} \left( e^t - \sin t - \cos t \right) \end{aligned} \end{equation}$