$$\int_0^{\sqrt{2 \pi}} \sin \left(x^2\right) \mathrm{d} x=\int_0^{2 \pi} \frac{\sin t}{2 \sqrt{t}} \mathrm{~d} t=\frac{1}{2} \int_0^\pi \frac{\sin t}{\sqrt{t}} \mathrm{~d} t+\frac{1}{2} \int_\pi^{2 \pi} \frac{\sin t}{\sqrt{t}} \mathrm{~d} t$$

$$\frac{1}{2} \int_\pi^{2 \pi} \frac{\sin t}{\sqrt{t}} \mathrm{~d} t=-\frac{1}{2} \int_0^\pi \frac{\sin u}{\sqrt{u+\pi}} \mathrm{d} u=-\frac{1}{2} \int_0^\pi \frac{\sin t}{\sqrt{t+\pi}} \mathrm{d} t$$

\begin{aligned} & \frac{1}{2} \int_0^\pi \frac{\sin t}{\sqrt{t}} \mathrm{~d} t+\frac{1}{2} \int_\pi^{2 \pi} \frac{\sin t}{\sqrt{t}} \mathrm{~d} t=\frac{1}{2} \int_0^\pi \frac{\sin t}{\sqrt{t}} \mathrm{~d} t-\frac{1}{2} \int_0^\pi \frac{\sin t}{\sqrt{t+\pi}} \mathrm{d} t \\ & =\frac{1}{2} \int_0^\pi\left(\frac{\sin t}{\sqrt{t}}-\frac{\sin t}{\sqrt{t+\pi}}\right) \mathrm{d} t=\frac{1}{2} \int_0^\pi\left(\frac{\sin t \cdot \sqrt{t+\pi}}{\sqrt{t} \cdot \sqrt{t+\pi}}-\frac{\sin t \cdot \sqrt{t+\pi}}{\sqrt{t} \cdot \sqrt{t+\pi}}\right) \mathrm{d} t \\ & =\frac{1}{2} \int_0^\pi \frac{\sin t \cdot(\sqrt{t+\pi}-\sqrt{t})}{\sqrt{t \cdot(t+\pi)}} \mathrm{d} t ; \end{aligned}