## Calculus: Early Transcendentals (2nd Edition)

${ = \frac{{{e^x}\sin x}}{2} + \frac{{{e^x}\cos x}}{2} + C}$
$\begin{gathered} \int_{}^{} {{e^x}\cos xdx} \hfill \\ \hfill \\ set\,\,\,the\,\,substitution \hfill \\ \hfill \\ u = {e^x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,du = {e^x}dx \hfill \\ dv = \cos x\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,v = \sin x \hfill \\ \hfill \\ use\,\,uv - \int_{}^{} {vdu} \hfill \\ \hfill \\ {\text{replacing the values }}{\text{in the equation}} \hfill \\ \hfill \\ = {e^x}\sin x - \int_{}^{} {{e^x}\sin xdx} \hfill \\ {\text{integrate}}\,\,{\text{by}}\,\,{\text{parts}}\,\,{\text{again}} \hfill \\ \hfill \\ u = {e^x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,du = {e^x}dx \hfill \\ du = \sin xdx\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,v = - \cos x \hfill \\ \hfill \\ {\text{replacing the values }}{\text{in the equation}} \hfill \\ \hfill \\ = {e^x}\sin x - \,\,\left[ { - {e^x}\cos x - \int_{}^{} {\,\left( { - \cos x} \right){e^x}dx} } \right] \hfill \\ \hfill \\ = {e^x}\sin x + {e^x}\cos x - \int_{}^{} {{e^x}\cos xdx} \hfill \\ \hfill \\ solving\,\,for\,\,\,\int_{}^{} {{e^x}\cos xdx} \hfill \\ \hfill \\ = 2\int_{}^{} {{e^x}\cos xdx} = {e^x}\sin x + {e^x}\cos x + C \hfill \\ \hfill \\ \int_{}^{} {{e^x}\cos xdx = \frac{{{e^x}\sin x}}{2} + \frac{{{e^x}\cos x}}{2} + C} \hfill \\ \end{gathered}$