009C Sample Final 2, Problem 6 Detailed Solution

(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.

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

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

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