Example 2.14. $\int_{0}^{(\pi / 2)^{2}} \sin \sqrt{x} d x$.
Solution. First, let's show an incorrect solution leading to a non-integrable function. Transform the integrand as follows:

$$
\begin{gathered}
\int_{0}^{(\pi / 2)^{2}} \frac{\sin \sqrt{x} \cdot \sqrt{x} d x}{\sqrt{x}}=\left|\begin{array}{l}
u=\sqrt{x}, d u=\frac{d x}{2 \sqrt{x}} \\
d v=\frac{\sin \sqrt{x} d x}{\sqrt{x}}, v=-2 \cos \sqrt{x}
\end{array}\right|= \\
\quad=\left.\sqrt{x}(-2 \cos \sqrt{x})\right|_{0}^{(\pi / 2)^{2}}-\int_{0}^{(\pi / 2)^{2}} \frac{\cos \sqrt{x} d x}{\sqrt{x}}
\end{gathered}
$$

Since $\cos x=1$, and $\frac{1}{\sqrt{x}}$ does not exist at $x=0$, the integrand $\frac{\cos \sqrt{x}}{\sqrt{x}}$ on the interval $\left[0,\left(\frac{\pi}{2}\right)^{2}\right]$ is not integrable in the sense of the definitions provided at the beginning of this section.

To correctly compute the given definite integral, we will use the method of substitution and set $x=z^{2}, d x=2 z d z$, the value $x_{0}=0$ will correspond to $z_{0}=0$, and the value $x_{1}=\left(\frac{\pi}{2}\right)^{2}-z_{1}=\frac{\pi}{2}$. After the specified substitution, the original integral will take the form:

$$
\begin{gathered}
I=\int_{0}^{\pi / 2} \sin z(2 z d z)=\left\lvert\, \begin{array}{l}
u=z, d u=d z \\
d v=\sin z d z, v=-\cos z
\end{array}\right. \\
\quad=2\left(-\left.z \cos z\right|_{0}^{\pi / 2}-\int_{0}^{\pi / 2}(-\cos z) d z\right) \\
=-\left.2(-z \cos z+\sin z)\right|_{0}^{\pi / 2}=2
\end{gathered}
$$