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

Question

Charles Anctil · Accepted Answer

# 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!) + ... # Next up is: # 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#

Aurora Alvarado · Answer

undefined To evaluate $ \int \cos(x^2) \, dx $ using a power series, we can express $ \cos(x^2) $ as a power series using the Maclaurin series for $ \cos(x) $. The Maclaurin series for $ \cos(x) $ is $ \cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} $.

Substituting $ x^2 $ for $ x $, we have $ \cos(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n (x^2)^{2n}}{(2n)!} $. Simplifying, we get $ \cos(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n}}{(2n)!} $.

Now, we can integrate $ \cos(x^2) $ term by term to find $ \int \cos(x^2) \, dx $.

$ \int \cos(x^2) \, dx = \int \left( \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n}}{(2n)!} \right) \, dx $

$ = \sum_{n=0}^{\infty} \left( \int \frac{(-1)^n x^{4n}}{(2n)!} \, dx \right) $

$ = \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+1}}{(4n+1)(2n)!} + C $

where $ C $ is the constant of integration.