29.6. Calculate $\int \sqrt{1-x^{2}} d x$.
29.6. Let $x=\sin y$. Then, according to problem 29.5

$$
\begin{aligned}
\int \sqrt{1-x^{2}} d x & =\int \cos ^{2} y d y=\int \frac{1+\cos 2 y}{2} d y=\frac{y}{2}+\frac{1}{4} \int \cos 2 y d(2 y)= \\
& =\frac{y}{2}+\frac{\sin 2 y}{4}+C=\frac{\arcsin x}{2}+\frac{x \sqrt{1-x^{2}}}{2}+C
\end{aligned}
$$