Approximate the integral int (1 - cos x)/x dx using Taylor expansion

[thenewbosco]

I am supposed to find an approximation of this integral evaluated between the limits 0 and 1 using a taylor expansion for cos x:

$$\int \frac{1 - cos x}{x}dx$$

and given

$$cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!}...$$

i should get a simple series similar to this for $$\frac{1 - cos x}{x}$$ and be able to simply integrate each term of the series and evaluate the integral for an approx. how do i find this series?

 

[Tzar]

Just sub in the series into the integral instead of cos (x). The 1's will cancel and the x at the bottom will decrease the power of each x on top by 1. Then integrate.

 

[quicknote]

The Taylor expansion for \frac{1 - cos x}{x} can be found by using the Maclaurin series for cos x and then dividing each term by x. This will give us:

\frac{1 - cos x}{x} = \frac{1 - (1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!}...)}{x} = \frac{\frac{x^2}{2!} - \frac{x^4}{4!} + \frac{x^6}{6!}...}{x} = \frac{1}{2!} - \frac{x^2}{4!} + \frac{x^4}{6!}...

Now, we can integrate each term of this series and evaluate the integral between the limits 0 and 1. This will give us an approximation of the original integral. The first term, \frac{1}{2!}, will be integrated to give us \frac{x}{2!}. The next term, -\frac{x^2}{4!}, will be integrated to give us -\frac{x^3}{3 \cdot 4!}. Continuing this pattern, we can integrate each term and evaluate the integral to get the following approximation:

\int_0^1 \frac{1 - cos x}{x}dx \approx \frac{1}{2!} - \frac{1}{3 \cdot 4!} + \frac{1}{5 \cdot 6!} - \frac{1}{7 \cdot 8!}...

This series can be simplified further by using the formula for the sum of an infinite geometric series. The final result will be an approximation of the original integral, which can be used to estimate the value of the integral between the given limits. However, it is important to note that this is only an approximation and may not give an exact value of the integral.