読者です 読者をやめる 読者になる 読者になる

bate's blog

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5

\int \sqrt{x^{2}+a} \, dx = \frac{1}{2}(x \sqrt{x^2+a}+a \, \log \| x+\sqrt{x^2+a} \|)+C
Cは積分定数
x=\sqrt{a} \sinh\,t, \hspace{5pt} dx=\sqrt{a} \cosh \, t \, dt, \hspace{5pt} t = \sinh^{-1} \, \frac{x}{\sqrt{a}}
\int \sqrt{x^2+a} \, dx = \int \sqrt{a} \, \sqrt{\sinh^2 \, t+1}(\sqrt{a}\cosh \, t \, dt)
\hspace{50pt}(\cosh^2 t - \sinh^2 t = 1)
\hspace{100pt}=a \, \int \cosh^2 t \, dt
\hspace{50pt}(\cosh^2t = \frac{1+\cosh\,(2t)}{2})
\hspace{100pt}=a\, \int \frac{1+\cosh \, (2t)}{2} \, dt
\hspace{100pt}=a\,(\frac{t}{2}+\frac{\sinh\,(2t)}{4})
\hspace{50pt}(\sinh\,(2t)=2\,\cosh\,t\,\sinh\,t)
\hspace{100pt}=\frac{a}{2}(t+\frac{2\,cosh\,t\,\sinh\,t}{2})
\hspace{100pt}=\frac{a}{2}(t+\cosh\,t\,\sinh\,t)
\hspace{50pt}(\cosh\,t = \sqrt{1+\sinh^2t},\hspace{5pt}\sinh\,t=\frac{x}{\sqrt{a}})
\hspace{100pt}=\frac{a}{2}(\sinh^{-1}\,\frac{x}{\sqrt{a}}+\sqrt{1+(\frac{x}{\sqrt{a}})^2}\,\frac{x}{\sqrt{a}})
\hspace{50pt}(\sinh^{-1}x=\log(x+\sqrt{1+x^2}),\hspace{5pt} \sinh^{-1}\frac{x}{\sqrt{a}}=\log(\frac{x}{\sqrt{a}}+\sqrt{1+(\frac{x}{\sqrt{a}}^2})\,)\,)
\hspace{100pt}=\frac{a}{2}(\log \frac{x+\sqrt{x^2+a}}{\sqrt{a}} + \frac{x}{\sqrt{a}}\sqrt{1+\frac{x^2}{a}})
\hspace{100pt}=\frac{a}{2}(\log \| x+\sqrt{x^2+a} \| - \frac{1}{2}\log\,a + \frac{1}{a}(x\sqrt{x^2+a})\,)
\hspace{100pt}=\frac{1}{2}(x\sqrt{x^2+a} + a\,\log \| x+\sqrt{x^2+a} \| + \frac{a}{2}\log \,a)

広告を非表示にする