Integration #4

$\begin{aligned} I & = \int_0^\pi e^{|\cos x|} \sin x \left(2\sin \left(\frac {\cos x}2\right) + 3 \cos \left(\frac {\cos x}2\right)\right) dx & \small \color{#3D99F6} \text{Let }2u = \cos x \implies 2du = - \sin x \ dx \\ & = 2 \int_{-\frac 12}^\frac 12 e^{|2u|} \left(2\sin u + 3 \cos u \right) du & \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \int_{-\frac 12}^\frac 12 \left(e^{|2u|} \left(2\sin u + 3 \cos u \right) + e^{|-2u|} \left(-2\sin u + 3 \cos u \right) \right) du \\ & = 6 \int_{-\frac 12}^\frac 12 e^{|2u|} \cos u \ du & \small \color{#3D99F6} \text{Since the integral is even} \\ & = 12 \int_0^\frac 12 e^{2u} \cos u \ du & \small \color{#3D99F6} \text{By integration by parts} \\ & = 12 \bigg[e^{2u} \sin u + 2e^{2u} \cos u\bigg]_0^\frac 12 - \color{#3D99F6} 48 \int_0^\frac 12 e^{2u} \cos u \ du \\ & = 12 \left(e \sin \frac 12 + 2e \cos \frac 12 - 2 \right) - \color{#3D99F6} 4I \\ & = \frac {12}5 \left(e \sin \frac 12 + 2e \cos \frac 12 - 2 \right) \\ & \approx 9.77819326 \end{aligned}$

Therefore, $\lfloor I \rfloor = \boxed{9}$ .