Prove that $1<\int_{0}^{\frac{\pi}{2}}\sqrt{\sin x}dx<\sqrt{\frac{\pi}{2}}$using integration.

You can use Holder's inequality: $$ \int_0^{\pi/2} \sqrt{\sin(x)}\, dx \leq \left(\frac{\pi}{2}\right)^{1/2} \left(\int_0^{\pi/2} \sin(x)\, dx\right)^{1/2} = \left(\frac{\pi}{2}\right)^{1/2}\,. $$

The upper bound can be obtained Jensen's inequality $$ \frac{2}{\pi}\int_0^{\pi/2} \sqrt{\sin(x)}\,dx \leq \Big(\frac{2}{\pi}\int_0^{\pi/2} \sin(x)\, dx\Big)^{1/2} = \sqrt{\frac{2}{\pi}} $$ from where

$$\int_0^{\pi/2} \sqrt{\sin(x)}\,dx\leq \sqrt{\frac{\pi}{2}} $$

The lower bound follows from $0\leq \sin x\leq \sqrt{\sin x}$ in $[0,\pi/2]$.

We can do better: using this post $$ \int _0^{\pi/2} \sqrt{\sin(x)} \,dx = \sqrt{\frac{2}{\pi}}\Gamma\left(\frac{3}{4}\right)^2 \approx 1.19 $$

If you want a numerical answer, OTOH we have $$ \int _0^{\pi/2}\sqrt{\sin(x)}\,dx > \int _0^{\pi/2} \sin(x)\,dx = 1 $$OTOH, by Cauchy-Bunyakovsky-Schwarz $$ \left(\int _0^{\pi/2}\sqrt{\sin(x)}\,dx\right)^2\leq \left(\int _0^{\pi/2} 1\,dx \right)\cdot \left(\int _0^{\pi/2} \sin(x)\,dx \right) = \frac{\pi}{2} $$