## Calculus: Early Transcendentals 8th Edition

$\frac{e^x}{2}(\sin x+ \cos x)+C$
Integrating by parts, we have $I= \int e^{x}\cos xdx= $$e^{x}\sin x-\int e^{x}\sin xdx$$=e^{x}\sin x-I_{1}$ $I_{1}=\int e^{x}\sin xdx$ $= e^{x}(-\cos x)+\int e^{x}\cos xdx$ Substituting the value of $I_{1}$, we get $I= e^{x}\sin x+ e^{x}\cos x- I$ or $2I= e^{x}(\sin x+ \cos x)$ or $I= \frac{e^x}{2}(\sin x+ \cos x)$