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

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SUMMARY

The integral of the function (1 - cos x)/x can be approximated using a Taylor expansion for cos x. The Taylor series for cos x is given by cos x = 1 - (x^2)/2! + (x^4)/4! - (x^6)/6!. By substituting this series into the integral, the expression simplifies, allowing for the integration of each term in the series. The resulting series provides an effective method for approximating the integral from 0 to 1.

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  • Taylor series expansion
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  • Understanding of trigonometric functions
  • Knowledge of limits in calculus
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thenewbosco
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I am supposed to find an approximation of this integral evaluated between the limits 0 and 1 using a taylor expansion for cos x:

[tex]\int \frac{1 - cos x}{x}dx[/tex]

and given

[tex]cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!}...[/tex]

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

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