# 009C Sample Final 2, Problem 6

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

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

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

Solution:

(a)

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

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

Step 2:
Now, to express the indefinite integral as a power series, we have

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

(b)

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

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

(a)     ${\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{4n+3}}{(4n+3)(2n+1)!}}}$
(b)     ${\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(4n+3)(2n+1)!}}}$