Question

evaluate integral cos(x^2) using the known mclaurin series. i just need to know which approach to take

Answer 1

take the expansion for \(\cos(x)\) and replace each \(x\) by \(x^2\) then integrate term by term

Answer 2

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