I am looking for alternative solutions to the following problem:
Let $f=\sqrt2 \cos(\frac{\pi}{4}+2x)$
1) Find the Taylor polynomial for $f$ of degree 3 at the point 0 and show that: $$\lvert f(x)-T_3(x) \vert \le \frac{2\sqrt2}{3}x^4, \ \forall x\in\mathbb{R} $$
2) Use 1) to show $$\lvert \int_{-a}^a f(x)dx -2a+\frac{4}{3}a^5 \lvert \le \frac{4\sqrt2}{15}a^5$$
I found the error in 1) to be: $\frac{2\sqrt2}{3}x^4$, and I solved 2) by substituting $f(x)$ with $T_3(x)+\frac{2\sqrt2}{3}x^4$ and then solving the integral and the left hand side of the inequality.
I noticed the the right hand side in: $\lvert \int_{-a}^a f(x)dx -2a+\frac{4}{3}a^5 \lvert \le \frac{4\sqrt2}{15}a^5$ is the fifth term of the taylor series for the indefinite integral of $\int f(x)$, maybe this is somehow of use?
I hope, you can help.
Best regards, Christoffer.