How can I integrate \sin{x^2} using Taylor series?

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Discussion Overview

The discussion centers on the integration of the function \(\sin{x^2}\) using Taylor series. Participants explore various approaches and identities related to the integration of trigonometric functions, particularly focusing on the limitations of finding an elementary integral for \(\sin{x^2}\).

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant seeks a hint for integrating \(\sin{x^2}\) and mentions using Taylor series for term-by-term integration.
  • Another participant suggests using the identity \(\sin^2{x} = \frac{1}{2}(1 - \cos(2x))\) as a potential approach, although it is not directly applicable to \(\sin{x^2}\).
  • A different participant states that \(\sin{x^2}\) does not have an elementary integral, indicating that a Taylor series might be the only viable method for integration.
  • One participant acknowledges a misunderstanding by initially addressing \(\sin^2{x}\) instead of \(\sin{x^2}\).
  • A later reply agrees with the notion that using Taylor series is the best approach for integrating \(\sin{x^2}\).

Areas of Agreement / Disagreement

Participants generally agree that \(\sin{x^2}\) does not have an elementary integral and that using Taylor series is a reasonable method. However, there is some confusion regarding the applicability of trigonometric identities to this specific integral.

Contextual Notes

There is an acknowledgment of the limitations in finding an elementary integral for \(\sin{x^2}\), and participants note the dependence on the Taylor series approach without resolving the specifics of the integration process.

euler_fan
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I would appreciate a hint as how to integrate the following, after some thought, I used Taylor series and integrated in term.

Thanks

[tex]\int\sin{x^2}[/tex]
 
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Use the identity [itex]\sin^2x=\frac{1}{2}(1-\cos(2x))[/itex]. This is derived from the double angle formula for cosine; [itex]\cos(2x)=\cos^2x-\sin^2x=1-2\sin^2x[/itex]. Rearranging gives the result.
 
sin(x^2) apparently has no elementary integral. hence a taylor series is about all you can do. a discussion of such questions is in a paper on brian conrad's page at umichigan.

sin^2(x) of course does have, as can be seen by using integration by parts, or the trig identity suggested above.
 
Last edited:
Sorry... I misread it as sin^2(x)!
 
euler_fan said:
... after some thought, I used Taylor series and integrated in term.

Good Work, Thats the best you can do.
 

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