What is ∫_ -π / 2 ^ π / 2 (e^ x x+e^ x x) d x equal to?

$$\begin{aligned} & I=\int_{-\pi / 2}^{\pi / 2}\left(e^{\cos x} \sin x+e^{\sin x} \cos x\right) d x \\ & =\int_{-\pi / 2}^{\pi / 2} e^{\cos x} \cdot \sin x d x \pm \int_{-\pi / 2}^{\pi / 2} e^{\sin x} \cdot \cos x d x \end{aligned}$$\because \quad e^{\cos x} \cdot \sin x$is odd function.$\therefore \int_{-\pi / 2}^{\pi / 2} e^{\cos x} \cdot \sin x d x=0$Let$\sin x=t, \cos x d x=d t$when$x=\frac{-\pi}{2}, t=-1$, when$x=\frac{\pi}{2}, t=1\therefore \quad $\mathrm{I}$=\int_{-1}^1 e^t d t=e-e^{-1}=\frac{e^2-1}{e}$