積分１ - bate's blog

$\int \frac{1}{\sqrt{1-x^2}}\,dx = \sin^{-1} x+C$
Cは積分定数
$x=\sin \, t,\hspace{5pt}dx=\cos \, t \hspace{2pt} dt, \hspace{5pt} t=\sin^{-1}x$
$\int \frac{1}{\sqrt{1-x^2}}\,dx = \int \frac{1}{\sqrt{1-\sin^2 t}} ( \cos \, t \hspace{2pt} dt )$
$\hspace{50pt}(\sqrt{1-\sin^2t} = \cos \, t)$
$\hspace{50pt}= \int \frac{\cos \, t}{\cos \, t}\,dt$
$\hspace{50pt}= \int 1\,dt$
$\hspace{50pt}= t + C$
$\hspace{50pt}= \sin^{-1}x + C$