Proving $\int_0^\sqrt{2\pi}\sin x^2\,\mathrm dx>0$ 
Prove that
$$\int_0^\sqrt{2\pi} \sin x^{2} \, \mathrm{d}x \gt 0 $$

I thought of changing the variable $y =  x^{2}$. And now I need to change this integral so I can integrate on the interval $[0; \pi]$. But I don't know how to do this so I can evaluate the given interval. What should I do?
 A: Using the substitution you mentioned, $y=x^2$, we get:
$$I=\int_{0}^{2\pi} \frac {\sin y}{2\sqrt y} dy$$
Splitting into two intervals gives:
$$I=\int_{0}^{\pi} \frac {\sin y}{2\sqrt y} dy +\int_{\pi}^{2\pi} \frac {\sin y}{2\sqrt y} dy$$
Let us call the first part $I_1$ and the second $I_2$. Since $I_1$ is positive and $I_2$ is negative (this is because the integrand is positive throughout the interval in $I_1$, and negative throughout the interval in $I_2$), we have to prove $I_1>-I_2$. We have, substituting $t=y-\pi$, $$-I_2=\int_{0}^{\pi} \frac {\sin t}{2\sqrt {t+\pi}} dt$$
Since $\frac {\sin t}{\sqrt t} >\frac {\sin t}{\sqrt {t+\pi}}$ for all $t>0$, the result $I_1>-I_2$ follows.
QED.
A: It's actually a rather fun exercise. The idea is to use the fact that the integral of a positive function produces a positive value.
Start with the substitution $x^2=y$ so that $dx=\frac{dy}{2\sqrt{y}}$ so that the integral becomes
$$\int_0^{2\pi}\frac{\sin(y)}{2\sqrt{y}}\,dy$$
Now you split the integral into the subintervals between each of the points $0,\pi/2,\pi,3\pi/2,2\pi$ and do a translation by $k\pi/2$  so that the integral becomes
$$\int_0^{\pi/2}\sin(y)\left(\frac{1}{2\sqrt{y}}-\frac{1}{2\sqrt{y+\pi/2}}+\frac{1}{2\sqrt{y+\pi}}-\frac{1}{2\sqrt{y+3\pi/2}}\right)\,dy$$
and since the function $\frac{1}{\sqrt{y}}$ is decreasing, each pair is a positive function and so the whole integral will be positive.
