Taylor polynomial and integration Let $f: \mathbb{R} \to \mathbb{R}$ be defined as $f(x) := \cos^2 x,\  x\in \mathbb{R}$ 
1) Find the Taylor polynomial for $f$ of degree 3 at the point 0 and show that: 
$$\lvert (\cos^2x-(1-x^2) \vert \le \frac{1}{3}x^4, \ \forall x\in\mathbb{R} $$
2) Use 1) to show $$\lvert \int_0^\frac{1}{2} (\cos\sqrt t)^2\,dt - \frac{3}{8}\rvert\le \frac{1}{72} $$

1) I am not going to show the computation, but just the result: 
The Taylor polynomial at degree 3 at 0 is given by:
$$ T_n(x) = 1-x^2$$  
The second part of the question is the remainder term: 
$$\lvert R_n \rvert = \lvert f(x)-T_n(x) \rvert =\lvert \cos^2 x-(1-x^2) \vert \le \frac{1}{3}x^4 $$
and I know that 
$$\lvert R_n \rvert \le \frac{M_n}{(n+1)!}\lvert x-x_0 \rvert^{n+1}$$
Where $M_n$ is given by: 
$$ M_n \ge \max {\lvert f^{n+1}(t) \rvert \lvert t\in [x_0,x]}$$ 
The Taylor polynomial of degree 4 is: $8(\cos^2 x-\sin^2 x)$ at  0 $\;8(\cos^2 x - \sin^2 x) = 8$
It then follows that $\frac{M_n}{(n+1)!}\lvert x-x_0 \rvert^{n+1} = \frac{8}{4!}x^4 = \frac{1}{3}x^4$ 
2) How am I supposed to use this to show that the integral is less than $\frac{1}{72}$?

Let $x=\sqrt{t}$ in (1) and integrate.
$$\int_0^{1/2}(1-t)\,dt=\frac{3}{8}$$
$$\int_0^{1/2}\frac{1}{3}t^2\,dt=\frac{1}{72}$$
