積分３ - bate's blog

$\int \sqrt{1-x^2} \, dx = \frac{1}{2}(x\sqrt{1-x^2}+\sin^{-1}x)+C$
Cは積分定数
$I = \int \sqrt{1-x^2} \, dx$
$x=\sin \, t, \hspace{5pt} dx=\cos \, t \hspace{2pt}dt, \hspace{5pt} t=\sin^{-1}x$
$I = \int \sqrt{1-\sin^{2}t}(\cos \, t \, dt )$
$\hspace{20pt}(\sqrt{1-sin^2t}=\cos\,t)$
$\hspace{8pt} = \int \cos^{2} \,dt$
$\hspace{20pt}(\cos \, t = (\sin \, dt)')$
$\hspace{8pt} = \int (\sin \, t)' \cos \, t \, dt$
$\hspace{20pt}(\int f'(t)g(t) \, dt = f(t)g(t) - \int f(t)g'(t) \, dt)$
$\hspace{8pt} = \sin \, t \, \cos \, t - \int \sin \, t (-\sin \, t) \, dt$
$\hspace{8pt} = \sin \, t \, \cos \, t + \int \sin^{2}t \, dt$
$\hspace{20pt}(\sin^{2}t = 1-\cos^{2}t)$
$\hspace{8pt} = \sin \, t \, \cos \, t + \int (1-\cos^{2}) \, dt$
$\hspace{8pt} = \sin \, t \, \cos \, t + \int 1 \, dt - \int \cos^{2} \, dt$
$\hspace{8pt} = \sin \, t \, \cos \, t + t - I + C$
$I = \sin \, t \, \cos \, t + t - I +C$
$\therefore 2I = \sin \, t \, \cos \, t + t + C$
$I = \frac{1}{2}(x \sqrt{1-x^2}+\sin^{-1}x) + C$
なぜなら,
$\cos \, t = \sqrt{1-\sin^{2}t} = \sqrt{1-x^2}$
積分定数がアレですが気にしないで下さい。