3 studs

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

$\Rightarrow a+b+c+d = 1+1+2+5 = \boxed{9}$

Note that we have 3 distinct terms in the integrand. But it would be convenient to use IBP if we have two terms instead of 3. Hence , we rewrite the given integral as $\displaystyle \dfrac{1}{2} \int_{0}^{\frac{\pi}{2}} e^x\sin 2x \ dx$ using the identity $\sin 2x=2\sin x \cos x$ . Now we conveniently evaluate this integral using Integration by parts:

$u=\sin(2x)\Rightarrow du=2\cos 2x \ dx$ , $dv=e^x \ dx \Rightarrow v=e^x$ , hence we have:

$\begin{aligned} \dfrac{1}{2} \int_{0}^{\frac{\pi}{2}} e^x\sin 2x \ dx &= \dfrac{1}{2}\left( e^x \sin(2x) \bigg |_{0}^{\frac{\pi}{2}} - 2\int_{0}^{\frac{\pi}{2}} e^x\cos 2x \ dx \right) \\ &=\dfrac{1}{2}\left(0 - 2\left( e^x\cos(2x) \bigg |^{\frac{\pi}{2}}_{0} + 2\int_{0}^{\frac{\pi}{2}} e^x\sin 2x \ dx\right)\right) \\ &\text{The above step is by IBP} \ (u=\cos 2x \ , \ dv=e^x \ dx ) \\ &\Rightarrow \dfrac{I}{2}= \dfrac{2e^{\pi/2} + 2 - 4I}{2} \ where \ I=\int_{0}^{\frac{\pi}{2}} e^x\sin 2x \ dx \\ &\Rightarrow 5I=2(e^{\pi/2} + 1) \\ &\Rightarrow I=\dfrac{2(e^{\pi/2} + 1)}{5} \\ &\Rightarrow \dfrac{I}{2} = \dfrac{(e^{\pi/2} + 1)}{5} \\ & \Rightarrow \int_{0}^{\frac{\pi}{2}} e^x\sin x\cos x \ dx = \dfrac{(e^{\pi/2} + 1)}{5} \end{aligned}$

Hence , $a=1 \ , \ b=1 \ , \ c=2 \ , \ d=5 \Rightarrow a+b+c+d=\boxed{9}$ /