Let's determine the area enclosed by the curves $y=\sin x$ and $y=(4 / \pi)^{2} \sin (\pi / 4) x^{2}$ (the latter being a quadratic function)!
For the intersection points of the two curves:

$$
\sin x=\binom{4}{\pi}^{2} \sin \frac{\pi}{4} x^{2}
$$

From here,

$$
x=0 \quad \text { and } \quad x=\frac{\pi}{4} \text {. }
$$

The sought area:

$$
\begin{gathered}
t=\int_{0}^{\frac{\pi}{4}}\left(\sin x-\frac{16}{2} \sin \frac{\pi}{4} x^{2}\right) d x=\left[-\cos x-\frac{1}{3} \cdot \frac{16}{\pi^{2}} \sin \frac{\pi}{4} x^{3}\right]_{0}^{\frac{\pi}{4}}= \\
=-\frac{\sqrt{2}}{2}-\frac{1}{3} \cdot \frac{16}{\pi^{2}} \sin \frac{\pi}{4} \cdot \frac{\pi^{3}}{4^{3}}+1=1-\frac{\sqrt{2}}{2}\left(1+\frac{\pi}{12}\right)
\end{gathered}
$$