By expansion, we have the following inequality when $0 < x < \pi$:

$$e^{\sin^2x}=1+\sin^2 x + \frac{\sin^4 x}{2^2} + \cdots > 1+\sin^2x$$

Therefore,

$$\int_0^{\pi} e^{\sin^2 x}dx > \int_0^{\pi} (1+\sin^2x) dx = \pi +\int_0^{\pi}\sin^2xdx$$

Meanwhile,

$$\int_0^{\pi}\sin^2 x dx=\int_0^{\pi}\frac{1-\cos 2x}{2}dx=\frac{\pi}{2}-\left.\frac{1}{4}\sin 2x\right|_0^{\pi}=\frac{\pi}{2}$$

Therefore, we find that 

$$\int_0^{\pi}e^{\sin^2x}dx > \frac{3\pi}{2}$$$