## Calculus: Early Transcendentals 8th Edition

$\displaystyle \frac{e^{t}-\sin t-\cos t}{2}$
Integration by parts: $\displaystyle \int udv=uv-\int vdu$ The idea is to choose a relatively easy $dv$ to integrate, and a u whose $u'$ does not complicate matters (best case: makes thing simpler). ---- $\left[\begin{array}{ll} u=\sin(t-s) & dv=e^{s}ds\\ & \\ du=-\cos(t-s)ds, & v=e^{s} \end{array}\right]$ $I=\displaystyle \int_{0}^{t}e^{s}\sin(t-s)ds=uv|_{0}^{t}-\int_{0}^{t}vdu$ $= =[e^{s}\displaystyle \sin(t-s)]_{0}^{t}+\int_{0}^{t}e^{s}\cos(t-s)ds$ $=e^{t}\displaystyle \sin 0-e^{0}\sin t+\int_{0}^{t}e^{s}\cos(t-s)ds$ $=-\displaystyle \sin t+\int_{0}^{t}e^{s}\cos(t-s)ds$ By parts, again, $\left[\begin{array}{ll} u=\cos(t-s) & dv=e^{s}ds\\ & \\ du=\sin(t-s)ds, & v=e^{s} \end{array}\right]$ $\displaystyle \int_{0}^{t}e^{s}\cos(t-s)ds =[e^{s}\displaystyle \cos(t-s)]_{0}^{t}-\int_{0}^{t}e^{s}\sin(t-s)ds$ $=e^{t}\cos 0-e^{0}\cos t-I$ ($I$ is the initial integral), so $I=-\sin t+e^{t}-\cos t-I$ $2I=e^{t}-\sin t-\cos t$ $I=\displaystyle \frac{e^{t}-\sin t-\cos t}{2}$