The Taylor series for \(\sin(x^2)\) is

\[ \sin(x^2) = x^2 - \frac{(x^2)^3}{3!} + \frac{(x^2)^5}{5!} - \frac{(x^2)^7}{7!} + \cdots \]

which can be expressed as

\[ \sin(x^2) = x^2 - \frac{x^6}{6} + \frac{x^{10}}{120} - \frac{x^{14}}{5040} + \cdots \]

I will integrate this from 0 to \(\sqrt{\pi}\):

\[ \int_0^{\sqrt{\pi}} \left( x^2 - \frac{x^6}{6} + \frac{x^{10}}{120} - \frac{x^{14}}{5040} \right) \, dx \]

each term will be integrated separatly

  1. \(\int_0^{\sqrt{\pi}} x^2 \, dx = \frac{x^3}{3} \Bigg|_0^{\sqrt{\pi}} = \frac{(\sqrt{\pi})^3}{3} = \frac{\pi^{3/2}}{3}\)
  2. \(\int_0^{\sqrt{\pi}} \frac{x^6}{6} \, dx = \frac{1}{6} \frac{x^7}{7} \Bigg|_0^{\sqrt{\pi}} = \frac{1}{42} (\sqrt{\pi})^7 = \frac{\pi^{7/2}}{42}\)
  3. \(\int_0^{\sqrt{\pi}} \frac{x^{10}}{120} \, dx = \frac{1}{120} \frac{x^{11}}{11} \Bigg|_0^{\sqrt{\pi}} = \frac{1}{1320} (\sqrt{\pi})^{11} = \frac{\pi^{11/2}}{1320}\)
  4. \(\int_0^{\sqrt{\pi}} \frac{x^{14}}{5040} \, dx = \frac{1}{5040} \frac{x^{15}}{15} \Bigg|_0^{\sqrt{\pi}} = \frac{1}{75600} (\sqrt{\pi})^{15} = \frac{\pi^{15/2}}{75600}\)

The terms will be summed up using alternate signs as in the series

\[ \frac{\pi^{3/2}}{3} - \frac{\pi^{7/2}}{42} + \frac{\pi^{11/2}}{1320} - \frac{\pi^{15/2}}{75600} \]

A value can be calculated using the terms.