Why Isn't $2$ Factored Out?

We have $I = \int_{0}^{\pi} e^x(6\cos 3x - 8\sin 3x) \,dx = \int_{0}^{\pi} \left(6e^x\cdot \cos 3x - 8e^x \sin 3x\right)\,dx \\ I = \int_{0}^{\pi} 6e^x\cdot \cos 3x - \int_{0}^{\pi} 8e^x \sin 3x \,dx = 6I_1 -8 I_2$ We can directly use the general formula for evaluating the integrals $I_1$ and $I_2$ . However, lets evaluate without using the general formula. $I_1 = \int e^x\cos 3x= e^x\cos 3x - \int\left(-3\sin 3x \cdot e^x\right)\,dx\\ I_1 = e^x\cos 3x +3\left(e^x\sin 3x -3\int(e^x\cdot \cos 3x)\,dx\right) = e^x\cdot \sin 3x +3e^x\cdot \cos 3x + 9I_1$ $\implies I_1 =\dfrac{ 3 e^x\cdot \sin 3x +e^x\cdot \cos 3x}{10}$ .On the same manner using Integration by parts we can evaluate integral $I_2$ as $I_2 = \dfrac{e^x\cdot \sin3x -3e^x\cdot \cos 3x}{10}$ Hence $I = \dfrac{6}{10}\left(3 e^x\cdot \sin 3x +e^x\cdot \cos 3x\right) - \dfrac{8}{10}\left(e^x\cdot \sin3x -3e^x\cdot \cos 3x\right) \\ I = e^x\left(\sin 3x + 3\cos 3x\right)\large |_{0}^{\pi} = -3e^{\pi}-3 \approx \boxed{-72.4220}$