We need compare $I=\int_0^1 \frac{\sin x}{\sqrt x}\,dx,\qquad J=\int_0^1 \frac{\cos x}{\sqrt x}\,dx.$

The options are clearly intended to compare $I$ with $\dfrac23$ and $J$ with $2$.

1. Estimate $I$

For $x\in[0,1]$, we use the standard inequality $\sin x < x \quad (x>0).$ Therefore, $\frac{\sin x}{\sqrt x} < \frac{x}{\sqrt x}=\sqrt x.$ Integrating from $0$ to $1$, $I=\int_0^1 \frac{\sin x}{\sqrt x}\,dx < \int_0^1 \sqrt x\,dx.$ Now, $\int_0^1 \sqrt x\,dx=\int_0^1 x^{1/2}\,dx=\left[\frac{2}{3}x^{3/2}\right]_0^1=\frac23.$ Hence, $I<\frac23.$

So the first comparison is: $I<\frac23.$

2. Estimate $J$

For $x\in[0,1]$, we know $\cos x < 1 \quad (x>0).$ Therefore, $\frac{\cos x}{\sqrt x} < \frac{1}{\sqrt x}.$ Integrating, $J=\int_0^1 \frac{\cos x}{\sqrt x}\,dx < \int_0^1 \frac{1}{\sqrt x}\,dx.$ Now, $\int_0^1 \frac{1}{\sqrt x}\,dx=\int_0^1 x^{-1/2}\,dx=\left[2\sqrt x\right]_0^1=2.$ Hence, $J<2.$

So the second comparison is: $J<2.$

3. Match with the options

We found: $I<\frac23 \quad \text{and} \quad J<2.$ This matches Option B.

4. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.