# Thread: Use the Maclaurin series of cos(x) to state the rst three non-zero terms of the Macl

1. ## Use the Maclaurin series of cos(x) to state the rst three non-zero terms of the Macl

(a) Use the Maclaurin series of cos(x) to state the rst three non-zero
terms of the Maclaurin series for (1-(cos(x))/(x^2)
x2 .
(b) Explain how you would use your answer in (a) to estimate
(integral from 0 to 1) (1-(cos(x))/(x^2) dx

This is a pretty complicated problem, but I got part a)
I got:

first derivative: -2(1-cos(x))(x^-3) + sinx(x^-2)
second derivative: 6(1-cos(x))(x^-4)-2(sinx)(x^-3)-2sinx(x^-3)+cosx(x^-2)
third derivative: -24(1-cosx)(x^-5)+6(sinx)(x^-4)+6sinx(x^-4)-2cosx(x^-3)+6sinx(x^-4)-2cosx(x^-3)-2cosx(x^-3)-sinx(x^-2)

From here, I know I am supposed to plug in (first derivative + second derivative + third derivative) in place of (1-(cos(x))/(x^2) to get the
(integral from 0 to 1) (1-(cos(x))/(x^2) dx, but I don't know how to integrate the whole thing. Please help! Thank you

2. NO, you are NOT supposed to do that!

Read the problem again- it said "use the MacLaurin series of cos(x)" and you did not do that. The MacLaurin series of cos(x) is $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}$. The first four non-zero terms of that are $\displaystyle 1- \frac{1}{2}x^2+ \frac{1}{24}x^4- \frac{1}{720}x^6$.

Replaced cos(x) in $\displaystyle \frac{1- cos(x)}{x^2}$ and you will see why I gave the first four terms of cos(x) rather than just the first three.

That is just a 6th degree polynomial. Integrate the polynomial between 0 and 1.