(3 Given $x \geqslant 0, y \geqslant 0$, and $x^{2}+y^{2}=1$, then the maximum value of $x(x+y)$ is . $\qquad$
(3) $\frac{\sqrt{2}+1}{2}$ Hint: Let $x=\cos \theta, y=\sin \theta, \theta \in\left[0, \frac{\pi}{2}\right]$, then
$$
\begin{aligned}
x(x+y) & =\cos \theta(\cos \theta+\sin \theta)=\frac{1+\cos 2 \theta}{2}+\frac{1}{2} \sin 2 \theta \\
& =\frac{\sqrt{2}}{2} \sin \left(2 \theta+\frac{\pi}{4}\right)+\frac{1}{2},
\end{aligned}
$$

Therefore, when $\theta=\frac{\pi}{8}$, $x(x+y)$ reaches its maximum value $\frac{\sqrt{2}+1}{2}$.