Which of the following inequalities is/are TRUE?

(1) We know that  $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\mathrm{...}$

So,option (1) is correct.
(2)  $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\mathrm{...}$
 $\therefore \sin x\ge x-\frac{x^3}{6}$
 $\Rightarrow \underset{0}{\overset{1}{\int }}x\sin x\ge \underset{0}{\overset{1}{\int }}x(x-\frac{x^3}{6})dx$
 $\Rightarrow \underset{0}{\overset{1}{\int }}x\sin x\ge \frac{1}{3}-\frac{1}{30}\Rightarrow \underset{0}{\overset{1}{\int }}x\sin xdx\ge \frac{9}{30}=\frac{3}{10}$
So, option  (2) is correct.
(3) Now,  $\cos x<1$
$\Rightarrow x^2\cos x<x^2$ 
 $\Rightarrow \underset{0}{\overset{1}{\int }}{x}^2\cos xdx<\frac{1}{3}$
So option (3) is incorrect.
(4)  $\underset{0}{\overset{1}{\int }}{x}^2\sin xdx\ge \underset{0}{\overset{1}{\int }}{x}^2(x-\frac{x^3}{6})dx$
$\underset{0}{\overset{1}{\int }}{x}^2\sin xdx\ge \frac{1}{4}-\frac{1}{36}$
 $\underset{0}{\overset{1}{\int }}{x}^2\sin xdx\ge \frac{2}{9}.$
So, option (4) is correct.