Let $I=$$\underset{0}{\overset{1}{\int }}\frac{\sin x}{\sqrt{x}}dx$ and $J=$$\underset{0}{\overset{1}{\int }}\frac{\cos x}{\sqrt{x}}dx$ which one of the following is true 

$$\begin{gathered} \mathrm{I}={\int }_0^1 \frac{\mathrm{Sm} \mathrm{x}}{\sqrt{\mathrm{x}}}\mathrm{dx}, \mathrm{J}={\int }_0^1 \frac{\cos \mathrm{x}}{\sqrt{\mathrm{x}}}\mathrm{dx} \\ \mathrm{Since} Sin \mathrm{x}<x \mathrm{for} \mathrm{x}\in \left(0,\frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow \frac{\mathrm{Sm} \mathrm{x}}{\sqrt{\mathrm{x}}}<\frac{\mathrm{x}}{\sqrt{\mathrm{x}}} \\ \Rightarrow \frac{\mathrm{Sm} \mathrm{x}}{\sqrt{\mathrm{x}}}<\sqrt{\mathrm{x}} \\ {\int }_0^1 \frac{\mathrm{Sm} \mathrm{x}}{\sqrt{\mathrm{x}}}\mathrm{dx} < {\int }_0^1\sqrt{\mathrm{x}}\mathrm{dx}={\left(\frac{{\mathrm{x}}^{3/2}}{3/2}\right)}_0^1 \\ \mathrm{I}<\frac{2}{3}(1-0)=\frac{2}{3} \\ \boxed{\mathrm{I} < \frac{2}{3}} \end{gathered}$$

$$\begin{gathered} \mathrm{J}={\int }_0^1\frac{\cos \mathrm{x}}{\sqrt{\mathrm{x}}}\mathrm{dx} \\ \mathrm{Since} \cos \mathrm{x}<1 \\ \Rightarrow \frac{\cos \mathrm{x}}{\sqrt{\mathrm{x}}}<\frac{1}{\sqrt{\mathrm{x}}} \\ \Rightarrow {\int }_0^1\frac{\cos \mathrm{x}}{\sqrt{\mathrm{x}}}<{\int }_0^1{\mathrm{x}}^{-1/2}\mathrm{dx} \\ \Rightarrow \mathrm{J}<{\left(\frac{{\mathrm{x}}^{1/2}}{1/2}\right)}_0^1={\left(2\sqrt{\mathrm{x}}\right)}_0^1=2 \\ \boxed{\therefore \mathrm{J} < 2} \end{gathered}$$