Evaluate $int cos(x^2) dx$ using a power series?

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Answer 1 · 2017-11-27T19:32:32

$int \ cos(x^2) \ dx = c+x - (x^5)/(5 * 2!) + (x^9)/(9*4!) - x^13/(13*6!) + ...$

Starting with the Maclaurin Series for $cosx$

$cosx = 1 - x^2/(2!) + x^4/(4!) - x^6/(6!) + ... \ \ \ \ \ \ AA x in RR$

So then:

$cos(x^2) = 1 - (x^2)^2/(2!) + (x^2)^4/(4!) - (x^2)^6/(6!) + ...$
$" " = 1 - (x^4)/(2!) + (x^8)/(4!) - (x^12)/(6!) + ...$

Then we have:

$int \ cos(x^2) \ dx = int \ 1 - (x^4)/(2!) + (x^8)/(4!) - (x^12)/(6!) + ... \ dx$
$" " = x - (x^5)/(5*2!) + (x^9)/(9*4!) - x^13/(13*6!) + ... +c$

Evaluate int cos(x^2) dx int cos(x^2) dx using a power series?