Integrate!

By Integration by Parts

Applying integration by parts twice:

$\begin{aligned} I &= \int_0^\frac \pi 2 e^x \cos x \ dx \\ &= e^x \sin x \bigg|_0^\frac \pi 2 - \int_0^\frac \pi 2 e^x \sin x \ dx \\ &= e^\frac \pi 2 + e^x \cos x \bigg|_0^\frac \pi 2 - \int_0^\frac \pi 2 e^x \cos x \ dx \\ &= e^\frac \pi 2 - 1 - I \\ &=\boxed{ \dfrac {e^\frac \pi 2 - 1}2} \end{aligned}$

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By Complex Numbers

$\begin{aligned} I &= \int_0^\frac \pi 2 e^x \cos x \ dx \\ &= \int_0^\frac \pi 2 e^x \Re \left \{e^{ix}\right \} \ dx \\ &= \Re \left \{\int_0^\frac \pi 2 e^{(1+i)x} \ dx \right \} \\ &= \Re \left \{\frac {e^{(1+i)x}}{1+i} \bigg|_0^\frac \pi 2 \right \} \\ &= \Re \left \{\frac {e^\frac \pi 2e^{i\frac \pi 2}-1}{1+i} \right \} \\ &= \Re \left \{\frac {e^\frac \pi 2\left(\cos \frac \pi 2 + i \sin \frac \pi 2 \right) -1}{1+i} \right \} \\ &= \Re \left \{\frac {\left(i e^\frac \pi 2 -1\right)(1-i)}{(1+i)(1-i)} \right \} \\ &= \Re \left \{\frac {e^\frac \pi 2 -1 + ie^\frac \pi 2 +i}2 \right \} \\ & = \boxed{\dfrac {e^\frac \pi 2 -1}2 } \end{aligned}$