Science:Math Exam Resources/Courses/MATH103/April 2016/Question 08 (c)/Solution 1

By part (a), we have

$\cos(t^{2})=\sum _{n=0}^{\infty }(-1)^{n}{\frac {t^{4n}}{(2n)!}}$

for all real $t.$

Then, by using the technique in the hint, we can integrate term-wise to obtain

${\begin{aligned}\int _{0}^{x}\cos(t^{2})\,dx&=\int _{0}^{x}\left(\sum _{n=0}^{\infty }(-1)^{n}{\frac {t^{4n}}{(2n)!}}\right)dx\\&=\sum _{n=0}^{\infty }\int _{0}^{x}(-1)^{n}{\frac {t^{4n}}{(2n)!}}\,dx\\&=\sum _{n=0}^{\infty }(-1)^{n}{\frac {t^{4n+1}}{(2n)!(4n+1)}}{\bigg |}_{t=0}^{t=x}\\&=\color {blue}\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{4n+1}}{(2n)!(4n+1)}},\quad x\in \mathbb {R} .\end{aligned}}$