Calculus 10th Edition

$\frac{e^{sin \left(\frac{\pi^{2}}{2}\right)}-1}{\pi}$
Solve for the indefinite integral $\int e^{sin \pi x} cos(\pi x)dx$ let $sin (\pi x) =u$ $\pi cos(\pi x) dx=du$ $cos(\pi x) dx=\frac{du}{\pi}$ $\int e^{sin \pi x} cos(\pi x)dx$ $=\frac{1}{\pi} \int e^udu$ $=\frac{1}{\pi}e^u+C$ $=\frac{1}{\pi}e^{sin (\pi x)}+C$ Solve for the definite integral $\int e^{sin \pi x} cos(\pi x)dx$ $[0, \frac{\pi}{2}]$ $=\frac{1}{\pi}e^{sin (\pi x)}$ $[0, \frac{\pi}{2}]$ $=\frac{1}{\pi}e^{sin (\frac{\pi^{2}}{2} )}-\frac{1}{\pi}e^{sin (0)}$ $=\frac{e^{sin (\frac{\pi^{2}}{2})}-1}{\pi}$