Greetings! I'm not sure how to approach this question, it's asking that I rewrite this:

∫(sin(x)-x)/x^3 dx

as an infinite series. Any help is appreciated!

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- Mar 15th 2013, 09:46 AMRobertXIVEvaluating an indefinite integral as an infinite series
Greetings! I'm not sure how to approach this question, it's asking that I rewrite this:

∫(sin(x)-x)/x^3 dx

as an infinite series. Any help is appreciated!

- Mar 15th 2013, 11:15 AMHallsofIvyRe: Evaluating an indefinite integral as an infinite series
Write sin(x) as a Taylor's series. Then you can subtract x, divide by $\displaystyle x^3$ and integrate term by term.

- Mar 15th 2013, 02:32 PMRobertXIVRe: Evaluating an indefinite integral as an infinite series
Ok let me see...

sinx as a Taylor series would be the sum of ((-1)^n/(2n+1)!)*x^(2n+1)

so then would we have (((-1)^n/(2n+1)!)*x^(2n+1)/x^3) - (x/x^3)? - Mar 15th 2013, 03:03 PMProve ItRe: Evaluating an indefinite integral as an infinite series
Yes, and surely you can now simplify...

- Mar 15th 2013, 04:30 PMHallsofIvyRe: Evaluating an indefinite integral as an infinite series
It might make more sense if you write it out as $\displaystyle sin(x)= x- x^3/3!+ x^5/5!- x^7/7!+ x^9/9!- \cdot\cdot\cdot$.

Then $\displaystyle sin(x)- x= - x^3/3!+ x^5/5!- x^7/7!+ x^9/9!- \cdot\cdot\cdot$

and $\displaystyle \frac{sin(x)- x}{x^3}= -1+ x^2/5!- x^4/7!+ x^6/9!- \cdot\cdot\cdot$

Integrate that. - Mar 16th 2013, 10:37 AMRobertXIVRe: Evaluating an indefinite integral as an infinite series
Ah! Alright that makes much more sense now. Thank you very much!