Example 9. Find the integral $\int \sqrt{a^{2}-x^{2}} d x(a>0)$.
Solution. Apply the trigonometric substitution

$$
x=a \cos t
$$

## We have

$$
d x=-a \sin t d t
$$

Therefore,

$$
\begin{gathered}
\int \sqrt{a^{2}-x^{2}} d x=\int \sqrt{a^{2}-a^{2} \cos ^{2} t}(-a \sin t) d t= \\
=-a^{2} \int \sqrt{1-\cos ^{2} t} \sin t d t=-a^{2} \int \sin ^{2} t d t=-a^{2} \int \frac{1-\cos 2 t}{2} d t= \\
=-\frac{a^{2}}{2} \int d t+\frac{a^{2}}{2} \int \cos 2 t d t=-\frac{a^{2}}{2} t+\frac{a^{2}}{4} \sin 2 t+C .
\end{gathered}
$$

Returning to the original variable $x$. Since $x=a \cos t$, then

$$
\frac{x}{a}=\cos t, t=\arccos \frac{x}{a}
$$

Further,

$$
\begin{gathered}
\sin 2 t=2 \sin t \cos t=2 \sqrt{1-\cos ^{2} t} \cos t= \\
=2 \sqrt{1-\frac{x^{2}}{a^{2}}} \cdot \frac{x}{a}=2 \cdot \frac{x}{a} \sqrt{\frac{a^{2}-x^{2}}{a^{2}}}=\frac{2 x}{a^{2}} \sqrt{a^{2}-x^{2}}
\end{gathered}
$$

Thus, we finally obtain

$$
\begin{gathered}
\int \sqrt{a^{2}-x^{2}} d x=-\frac{a^{2}}{2} t+\frac{a^{2}}{4} \sin 2 t+C= \\
=-\frac{a^{2}}{2} \arccos \frac{x}{a}+\frac{a^{2}}{4} \cdot \frac{2 x}{a^{2}} \sqrt{a^{2}-x^{2}}+C= \\
=-\frac{a^{2}}{2} \arccos \frac{x}{a}+\frac{x}{2} \sqrt{a^{2}-x^{2}}+C
\end{gathered}
$$