Answer to Question #239405 in Calculus for sam

Question

Consider the following equation $I=\displaystyle{\int_{0}^{1}\sin{(x^2)}dx}$

Find the integral  $\displaystyle{\int_{-1}^{1}\sin{(x^2)}dx}$ in terms of $I$

Answer 1

$\sin x = \sum \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}$

$\text{then} \; \; \sin (x^2) = \sum \dfrac{(-1)^n x^{4n+2}}{(2n+1)!}$

$\int \sin (x^2) = \int \sum \dfrac{(-1)^n x^{4n+2}}{(2n+1)!} = \sum \dfrac{(-1)^n x^{4n+3}}{(2n+1)!(4n+3)} + K$

$\therefore\int_0^1\sin x^2~dx$

$=\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{4n+3}}{(2n+1)!(4n+3)}\right]_0^1$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n}{(2n+1)!(4n+3)}=I$

$\int_{-1}^1\sin x^2~dx$

$=\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{4n+3}}{(2n+1)!(4n+3)}\right]_{-1}^1$

$=\sum\limits_{n=0}^\infty(2\dfrac{(-1)^n}{(2n+1)!(4n+3)})= 2I$