$3$\displaystyle{ \displaystyle{ \int \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ {a}^{2}- {x}^{2} } \hspace{2}}dx }={ \sin }^{ - 1 }\frac{\hspace{2}x\hspace{2}}{\hspace{2}a\hspace{2}}+C }$

　 $3$a$ は定数， $3$C$ は積分定数

$3$\displaystyle{{\sin }^{ - 1 }x}$ の微分を参照のこと。

$3$\displaystyle{ { \sin }^{ - 1 }\frac{\hspace{2}x\hspace{2}}{\hspace{2}a\hspace{2}} }$ を微分すると，

$3$\displaystyle{\begin{array}{lll}{ \{ { \sin }^{ - 1 }\frac{\hspace{2}x\hspace{2}}{\hspace{2}a\hspace{2}} \} }^{\prime }& \vspace{6}\\ =\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ 1- { $ \frac{\hspace{2}x\hspace{2}}{\hspace{2}a\hspace{2}} $ }^{2} } \hspace{2}}\cdot { $ \frac{\hspace{2}x\hspace{2}}{\hspace{2}a\hspace{2}} $ }^{\prime }& \vspace{6}\\ =\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ 1- \frac{\hspace{2} {x}^{2} \hspace{2}}{\hspace{2} {a}^{2} \hspace{2}} } \hspace{2}}\cdot $ \frac{\hspace{2}1\hspace{2}}{\hspace{2}a\hspace{2}} $ & \vspace{6}\\ =\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ {a}^{2}- {x}^{2} } \hspace{2}}& \vspace{6}\\ \end{array}}$

となる。よって，上の公式が成り立つ。