# Integrate: $\int\limits _0^{\pi} e^x \sin x dx$

$\begin {array} {1 1} (a)\;\frac{e^{\pi}+1}{2} \\ (b)\;\frac{e^{\pi}-1}{2} \\ (c)\;\frac{e^{\pi}+1}{\sqrt 2} \\ (d)\;\frac{e^{\pi}-1}{\sqrt 2} \end {array}$

$\bigg [\large\frac{e^x}{2} $$(\sin x - \cos x) \bigg]_0^{\pi} => \large\frac{1}{2}$$(e^{\pi} +1)$
$\large\frac{e^{\pi}+1}{2}$
Hence a is the correct answer.