Let $I=\int\limits_0^1 \frac{\sin x}{\sqrt{x}} d x$ and $J=\int\limits_0^1 \frac{\cos x}{\sqrt{x}} d x$. Then, which one of the following is true?

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

We have,

$\sin x<x$ for $0<x<1$

$\Rightarrow \frac{\sin x}{\sqrt{x}}<\frac{x}{\sqrt{x}}$ for $0<x<1$

$\Rightarrow \int\limits_0^1 \frac{\sin x}{\sqrt{x}} d x<\int\limits_0^1 \sqrt{x} d x \Rightarrow I<\frac{2}{3}\left[x^{3 / 2}\right]_0^1 \Rightarrow I<\frac{2}{3}$

Also,

$\cos x<1$ for all $x \in(0,1)$

$\Rightarrow \frac{\cos x}{\sqrt{x}}<\frac{1}{\sqrt{x}}$ for all $x \in(0,1)$

$\Rightarrow \int\limits_0^1 \frac{\cos x}{\sqrt{x}} d x<\int\limits_0^1 \frac{1}{\sqrt{x}} d x \Rightarrow J<2\left[x^{1 / 2}\right]_0^1 \Rightarrow J<2$