# 009C Sample Final 2, Problem 6

(a) Express the indefinite integral  $\int \sin(x^{2})~dx$ as a power series.

(b) Express the definite integral  $\int _{0}^{1}\sin(x^{2})~dx$ as a number series.

Foundations:
What is the power series of  $\sin x?$ The power series of  $\sin x$ is   $\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n+1}}{(2n+1)!}}.$ Solution:

(a)

Step 1:
The power series of  $\sin x$ is   $\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n+1}}{(2n+1)!}}.$ So, the power series of  $\sin(x^{2})$ is

${\begin{array}{rcl}\displaystyle {\sin(x^{2})}&=&\displaystyle {\sum _{n=0}^{\infty }{\frac {(-1)^{n}(x^{2})^{2n+1}}{(2n+1)!}}}\\&&\\&=&\displaystyle {\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{4n+2}}{(2n+1)!}}.}\end{array}}$ Step 2:
Now, to express the indefinite integral as a power series, we have

${\begin{array}{rcl}\displaystyle {\int \sin(x^{2})~dx}&=&\displaystyle {\int \sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{4n+2}}{(2n+1)!}}~dx}\\&&\\&=&\displaystyle {\sum _{n=0}^{\infty }\int {\frac {(-1)^{n}x^{4n+2}}{(2n+1)!}}~dx}\\&&\\&=&\displaystyle {\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{4n+3}}{(4n+3)(2n+1)!}}.}\end{array}}$ (b)

Step 1:
From part (a), we have
$\int \sin(x^{2})~dx=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{4n+3}}{(4n+3)(2n+1)!}}.$ Now, we have
$\int _{0}^{1}\sin(x^{2})~dx=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{4n+3}}{(4n+3)(2n+1)!}}{\bigg |}_{0}^{1}.$ Step 2:
Hence, we have

${\begin{array}{rcl}\displaystyle {\int _{0}^{1}\sin(x^{2})~dx}&=&\displaystyle {\sum _{n=0}^{\infty }{\frac {(-1)^{n}(1)^{4n+3}}{(4n+3)(2n+1)!}}-\sum _{n=0}^{\infty }{\frac {(-1)^{n}(0)^{4n+3}}{(4n+3)(2n+1)!}}}\\&&\\&=&\displaystyle {\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(4n+3)(2n+1)!}}-0}\\&&\\&=&\displaystyle {\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(4n+3)(2n+1)!}}}\end{array}}$ (a)     $\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{4n+3}}{(4n+3)(2n+1)!}}$ (b)     $\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(4n+3)(2n+1)!}}$ 