My eleventh integral problem

$\begin{aligned} \int_0^{\frac{\pi}{2}} e^x \cos^2 x \space dx & = \frac{1}{2} \int_0^{\frac{\pi}{2}} e^x \left(1+\cos (2x)\right) \space dx \\ & = \frac{1}{2} \int_0^{\frac{\pi}{2}} e^x \space dx + \frac{1}{4} \int_0^{\frac{\pi}{2}} e^x \left(e^{2xi} + e^{-2xi} \right) \space dx \\ & = \frac{1}{2} \left[e^x\right]_0^\frac{\pi}{2} + \frac{1}{4} \left[ \frac{e^{(1+2i)x}}{1+2i} + \frac{e^{(1-2i)x}}{1-2i} \right]_0^\frac{\pi}{2} \\ & = \frac{1}{2} \left(e^\frac{\pi}{2} - 1 \right) + \frac{1}{4} \left[\frac{e^x\left(e^{2xi} + e^{-2xi} - 2ie^{2xi} + 2ie^{-2xi} \right)}{1+4} \right]_0^\frac{\pi}{2} \\ & = \frac{1}{2} \left(e^\frac{\pi}{2} - 1 \right) + \frac{1}{2\times 5} \left[e^x \left(\cos (2x) +2 \sin (2x) \right) \right]_0^\frac{\pi}{2} \\ & = \frac{1}{2} \left(e^\frac{\pi}{2} - 1 \right) + \frac{1}{10} \left[e^\frac{\pi}{2} \left(\cos \pi + 2\sin \pi \right) - e^0 \left(\cos 0 + 2\sin 0 \right) \right] \\ & = \frac{1}{2} \left(e^\frac{\pi}{2} - 1 \right) + \frac{1}{10} \left(-e^\frac{\pi}{2} -1 \right) \\ & = \frac{2e^\frac{\pi}{2}-3}{5} \end{aligned}$

$\Rightarrow a + b + c + d + f = 2+1+2+3+5 = \boxed{13}$