We know that $\cos x=1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\ldots$
So, $\cos x>1-\frac{x^{2}}{2}$ for all $x \in(0,1)$
$\Rightarrow x \cos x>x-\frac{x^{3}}{2}$
$\Rightarrow \int\limits_{0}^{1} x \cos x \,d x >\int\limits_{0}^{1}\left(x-\frac{x^{3}}{2}\right) d x$
$\Rightarrow \int\limits_{0}^{1} x \cos x\, d x>\frac{3}{8}($ Option $A$ is correct $)$ $\because x^{2} \cos x < x$ for all $x \in(0,1)$
$\Rightarrow \int\limits_{0}^{1} x^{2} \cos x \,d x < \int\limits_{0}^{1} x d x$
$\Rightarrow \int\limits_{0}^{1} x^{2} \cos x \,d x < \frac{1}{2}($ Option $C$ is incorrect $)$
Again we know that, $\sin x=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\ldots$
So, $\sin x > x-\frac{x^{3}}{6}$ for all $x \in(0,1)$
$\Rightarrow x \sin x > x^{2}-\frac{x^{4}}{6}$
$\int\limits_{0}^{1} x \sin x d x>\int\limits_{0}^{1}\left(x^{2}-\frac{x^{4}}{6}\right) d x$
$\Rightarrow \int\limits_{0}^{1} x \sin d x>\frac{3}{10}$ (Option $B$ is correct) $\because x^{2} \sin x>x^{3}-\frac{x^{5}}{6}$
$\Rightarrow \int\limits_{0}^{1} x^{2} \sin x\, d x>\frac{2}{9}$ (Option $D$ is correct)