Chứng minh rằng : $\int_{0}^{\pi}e^{\sin^{2}x}\text{dx} > \frac{3\pi}{2}$.

Chứng minh rằng :
$\int_{0}^{\pi}e^{\sin^{2}x}\text{dx} > \frac{3\pi}{2}$.
*Chú ý : $e = \lim_{n \to \infty}\left ( 1 + \frac{1}{n} \right )^{n}$.
Giải
Xét phép đổi biến $t=\pi -x  \implies dt=-dx$.

Do đó :

$$\begin{array}{rcl} \int\limits_{\frac{\pi }{2}}^\pi  {{e^{{{\sin }^2}x}}dx}  &=& \int\limits_0^{\frac{\pi }{2}} {{e^{{{\sin }^2}t}}\left( { - dt} \right)} \\ &=& \int\limits_0^{\frac{\pi }{2}} {{e^{{{\sin }^2}x}}dx} \\ \Rightarrow \int\limits_0^\pi  {{e^{{{\sin }^2}x}}dx}  &=& \int\limits_0^{\frac{\pi }{2}} {{e^{{{\sin }^2}x}}dx}  + \int\limits_{\frac{\pi }{2}}^\pi  {{e^{{{\sin }^2}x}}dx} \\ &=& 2\int\limits_0^{\frac{\pi }{2}} {{e^{{{\sin }^2}x}}dx} \end{array}$$

Bây giờ áp dụng tính chất $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$ ,ta sẽ có ngay :
$$\int\limits_0^\pi  {{e^{{{\sin }^2}x}}dx}  = 2\int\limits_0^{\frac{\pi }{2}} {{e^{{{\sin }^2}x}}dx}  = 2\int\limits_0^{\frac{\pi }{2}} {{e^{{{\cos }^2}x}}dx}$$

Theo BĐT Cauchy-Schwarz cho tích phân thì :
$$\int\limits_0^\pi  {{e^{{{\sin }^2}x}}dx}  = 2\sqrt {\int\limits_0^{\frac{\pi }{2}} {{e^{{{\sin }^2}x}}dx} .\int\limits_0^{\frac{\pi }{2}} {{e^{{{\cos }^2}x}}dx} }  \ge 2\int\limits_0^{\frac{\pi }{2}} {{e^{\frac{{{{\sin }^2}x}}{2}}}.{e^{\frac{{{{\cos }^2}x}}{2}}}dx}  = \pi \sqrt e  > \frac{{3\pi }}{2}$$