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OpenStudy (anonymous):

solve integral [sin(x)/x ]dx

OpenStudy (anonymous):

\[\int\limits_{}^{}( \sin \theta / \theta) d \theta\]

OpenStudy (anonymous):

From Taylor series we have: sin x = Σ (-1)^n * x^(2n+1)/(2n+1)! =>Sin(x)/x= Σ (-1) ^n * x^ (2n+1)/ (2n+1)! * x =>Sin(x)/x= Σ (-1)^n *x^(2n)/(2n+1)! ∫[sin⁡(x)/x] dx = Σ (-1)^n* x^(2n+1)/[(2n+1)*(2n+1)!] + C

OpenStudy (anonymous):

This might be a trick question. [sin x]/x is known to be 1. So find integral of 1.

OpenStudy (anonymous):

No, lim x ->0 [sinx]/x=1. we can not say [sinx]/x=1

OpenStudy (anonymous):

Yes, after a little research, civil.ad89 working is valid. You may include in explanation: integral can not be reached through elementary working and must attempt power series.

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