Solution. The formal application of the trigonometric substitution $x=\sin t$ is impossible here, as with such a substitution, the variable $x$ cannot cover the interval $[0,4]$ for any $t$. It is preferable to use the substitution

$$
\begin{aligned}
& z=1-x^{2}, x d x=-0.5 d z, x_{1}=0 \rightarrow z_{1}=1, x_{2}=4 \rightarrow z_{2}=-15 \\
& I=\frac{1}{2} \int_{1}^{-15} z \frac{1}{5}(-d z)=\left.\frac{1}{2} \frac{5}{6} z^{\frac{6}{5}}\right|_{1} ^{-15}=\frac{5}{12}\left((-15)^{\frac{6}{5}}-1^{\frac{6}{5}}\right) \approx 10.33
\end{aligned}
$$

In the following examples, we will demonstrate how to correctly apply integration by parts formulas in definite integrals.
Example 2.12. $I=\int_{0}^{4} x\left(1-x^{2}\right)^{1 / 5} d x$.

