Question

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

Answer 1

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

Explanation:

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`