Which of the following inequalities is/are TRUE?

We know that $\cos x = 1 - \dfrac{{x}^{2}}{2 !} + \dfrac{{x}^{4}}{4 !} - \dfrac{{x}^{6}}{6 !} + \ldots \ldots \ldots \ldots$

$⇒ \cos x \ge 1 - \dfrac{{x}^{2}}{2}$

$∴ {∫}_{0}^{1}xcosxdx \ge {∫}_{0}^{1}x$\left(1 - \dfrac{{x}^{2}}{2}\right)dx = {\left(\dfrac{{x}^{2}}{2} - \dfrac{{x}^{4}}{8}\right)}_{0}^{1} = \dfrac{1}{2} - \dfrac{1}{8} = \dfrac{3}{8}$</p><p>So, option \left(A\right)$ is correct.

Now, we also know that $\sin x = x - \dfrac{{x}^{3}}{3 !} + \dfrac{{x}^{5}}{5 !} - \dfrac{{x}^{7}}{7 !} + \ldots \ldots \ldots \ldots \ldots$

$⇒ \sin x \ge x - \dfrac{{x}^{3}}{3 !}, x \in$\left[0, 1\right]

$∴ {∫}_{0}^{1}xsinxdx \ge {∫}_{0}^{1}x \left(x - \dfrac{{x}^{3}}{6}\right)$dx = {\left(\dfrac{{x}^{3}}{3} - \dfrac{{x}^{5}}{6 \times 5}\right)}_{0}^{1} =\dfrac{1}{3} - \dfrac{1}{30} = \dfrac{9}{30} = \dfrac{3}{10}

So, option \left(B\right) is correct.

Now, ${∫}_{0}^{1}{x}^{2}cosxdx \ge {∫}_{0}^{1}{x}^{2}\left(1 - \dfrac{{x}^{2}}{2}\right)$dx = {\left(\dfrac{{x}^{3}}{3} - \dfrac{{x}^{5}}{2 \times 5}\right)}_{0}^{1} =\dfrac{1}{3} - \dfrac{1}{10} = \dfrac{7}{30}

Also, ${x}^{2}\cos x \le {x}^{2}$

$⇒ {∫}_{0}^{1}{x}^{2}cosxdx \le {∫}_{0}^{1}{x}^{2}dx$

$⇒ {∫}_{0}^{1}{x}^{2}cosxdx \le {\left(\dfrac{{x}^{3}}{3}\right)}_{0}^{1} =$\dfrac{1}{3}

So, \dfrac{7}{30}$\le {∫}_{0}^{1}{x}^{2}cosxdx \le$\dfrac{1}{3}$</p><p>So, option \left(C\right)$ is incorrect.

Now, ${∫}_{0}^{1}{x}^{2}sinxdx \ge {∫}_{0}^{1}{x}^{2}\left(x - \dfrac{{x}^{3}}{3 !}\right)dx = {\left(\dfrac{{x}^{4}}{4} - \dfrac{{x}^{6}}{6 \times 6}\right)}_{0}^{1} = \dfrac{1}{4} - \dfrac{1}{36} = \dfrac{8}{36} = \dfrac{2}{9}$

So, option \left(D\right) is correct.