We know when $f(x)$>$g(x)$

$\int \limits_a^bf(x)$>$\int \limits_a^bg(x)$

We know for $0<x<1$, $ \cos x <1 $

$ \frac{\cos x}{\sqrt x}$< $\frac{1}{\sqrt x}$ implies $\int \limits_0^1\frac{\cos x}{\sqrt x}\mathrm dx$<$\int \limits_0^1\frac{1}{\sqrt x}\mathrm dx $

$\int \limits_0^1\frac{1}{\sqrt x}\mathrm dx = 2$

$\int \limits_0^1\frac{\cos x}{\sqrt x}\mathrm dx $<$2$

$J$<$2$

Again we claim $x-s\sin x$>$0$ for $0 \leq x\leq 1$

Let $f(x)=x-\sin x$

$f'(x)=1-\cos x\geq 0$

hence $f(x)$ is monotonic increasing.

Therefore $x-\sin x $> $0$, $x\epsilon [0,1]$

So,$x$>$sinx$

$\sqrt x$ > $\frac{\sin x}{\sqrt x}$ $x\epsilon [0,1]$

integrating both sides with limits $0$ to $1$ we get;

$\int \limits_0^1\frac{\sin x}{\sqrt x} \mathrm dx $<$\frac{2}{3}$

$I$<$\frac{2}{3}$

Therefore,$I<\frac{2}{3}$ and $J<2$