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調べたこと実装したことなどを取りとめもなく書きます。

積分3

\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。

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