(5) The maximum value of the function $y=x\left(1+\sqrt{1-x^{2}}\right)$ is
(5) $\frac{3}{4} \sqrt{3}$ Hint: The range of $x$ is $|x| \leqslant 1$. To find the maximum value, we can assume $0 < x \leqslant 1$, let $x=\sin \alpha, \alpha \in\left(0, \frac{\pi}{2}\right]$, then
$$
\begin{array}{c}
y=\sin \alpha(1+\cos \alpha), \\
y^{\prime}=\cos \alpha(1+\cos \alpha)+\sin \alpha(-\sin \alpha)=\cos \alpha+\cos 2 \alpha .
\end{array}
$$

Let $y^{\prime}=0$, i.e., $\cos \alpha+\cos 2 \alpha=0$, we get $2 \alpha=\pi-\alpha$, so $\alpha=\frac{\pi}{3}$, at this time $x=\frac{\sqrt{3}}{2}, y=\frac{3}{4} \sqrt{3}$.