Answer
If $a\gt0$ is continuous on $[a, b]$, show that:
$$
\int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} \frac{x}{a}+C
$$
Let $u=x / a $ and $d u=d x / a, x$. So the Substitution Rule gives
$$
\begin{aligned}
R.H.S &.=\int \frac{d x}{\sqrt{a^{2}-x^{2}}} \\
&=\int \frac{d x}{a \sqrt{1-(x / a)^{2}}} \\
&=\int \frac{d u}{\sqrt{1-u^{2}}} \\
&=\sin ^{-1} u+C \\
&=\sin ^{-1} \frac{x}{a}+C\\
& =L.H.S.
\end{aligned}
$$
Work Step by Step
If $a\gt0$ is continuous on $[a, b]$, show that:
$$
\int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} \frac{x}{a}+C
$$
Let $u=x / a $ and $d u=d x / a, x$. So the Substitution Rule gives
$$
\begin{aligned}
R.H.S &.=\int \frac{d x}{\sqrt{a^{2}-x^{2}}} \\
&=\int \frac{d x}{a \sqrt{1-(x / a)^{2}}} \\
&=\int \frac{d u}{\sqrt{1-u^{2}}} \\
&=\sin ^{-1} u+C \\
&=\sin ^{-1} \frac{x}{a}+C\\
& =L.H.S.
\end{aligned}
$$
