1/√(x(a-x))の積分

1/√(x(a-x))の積分

不定積分

$$\begin{array}{rcl}\int \frac{1}{\sqrt{x(a-x)}}\mathrm{d}x& =& \int {\{x(a-x)\}}^{-\frac{1}{2}}\mathrm{d}x\\ & =& \int x^{-\frac{1}{2}}{(a-x)}^{-\frac{1}{2}}\mathrm{d}x\\ & =& \int u^{-1}(a-u^2{)}^{-\frac{1}{2}}2u\mathrm{d}u\cdots u=\sqrt{x},\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{1}{2\sqrt{x}},\mathrm{d}x=2\sqrt{x}\mathrm{d}u=2u\mathrm{d}u\\ & =& 2\int (a-u^2{)}^{-\frac{1}{2}}\mathrm{d}u\cdots u^{-1}u=1,\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x\\ & =& 2\int {[a-{\{\sqrt{a}\sin \left(\theta \right)\}}^2]}^{-\frac{1}{2}}\sqrt{a}\cos \left(\theta \right)\mathrm{d}\theta \cdots u=\sqrt{a}\sin \left(\theta \right),\frac{\mathrm{d}u}{\mathrm{d}\theta }=\sqrt{a}\cos \left(\theta \right),\mathrm{d}u=\sqrt{a}\cos \left(\theta \right)\mathrm{d}\theta \\ & =& 2\sqrt{a}\int {[a-\{{\sqrt{a}}^2{\sin }^2\left(\theta \right)\}]}^{-\frac{1}{2}}\cos \left(\theta \right)\mathrm{d}\theta \cdots (AB{)}^C=A^C{B}^C\\ & =& 2\sqrt{a}\int {\{a-a{\sin }^2\left(\theta \right)\}}^{-\frac{1}{2}}\cos \left(\theta \right)\mathrm{d}\theta \\ & =& 2\sqrt{a}\int {\left[a\{1-{\sin }^2\left(\theta \right)\}\right]}^{-\frac{1}{2}}\cos \left(\theta \right)\mathrm{d}\theta \\ & =& 2\sqrt{a}\int a^{-\frac{1}{2}}{\{1-{\sin }^2\left(\theta \right)\}}^{-\frac{1}{2}}\cos \left(\theta \right)\mathrm{d}\theta \\ & =& 2a\sqrt{a}{a}^{-\frac{1}{2}}\int {\{1-{\sin }^2\left(\theta \right)\}}^{-\frac{1}{2}}\cos \left(\theta \right)\mathrm{d}\theta \\ & =& 2\int {\{1-{\sin }^2\left(\theta \right)\}}^{-\frac{1}{2}}\cos \left(\theta \right)\mathrm{d}\theta \cdots \sqrt{A}{A}^{-\frac{1}{2}}=A^{\frac{1}{2}}{A}^{-\frac{1}{2}}=A^{-\frac{1}{2}+\frac{1}{2}}=A^0=1\\ & =& 2\int {\{{\cos }^2\left(\theta \right)\}}^{-\frac{1}{2}}\cos \left(\theta \right)\mathrm{d}\theta \cdots {\cos }^2\left(\theta \right)+{\sin }^2\left(\theta \right)=1,{\cos }^2\left(\theta \right)=1-{\sin }^2\left(\theta \right)\\ & =& 2\int {\{\cos \left(\theta \right)\}}^{2(-\frac{1}{2})}\cos \left(\theta \right)\mathrm{d}\theta \cdots {\left(A^B\right)}^C=A^{BC}\\ & =& 2\int {\{\cos \left(\theta \right)\}}^{-1}\cos \left(\theta \right)\mathrm{d}\theta \\ & =& 2\int \mathrm{d}\theta \cdots A^{-1}A=1\\ & =& 2\theta \cdots \int \mathrm{d}x=x+C(C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\\ & =& 2{\sin }^{-1}\left(\frac{u}{\sqrt{a}}\right)\cdots \theta ={\sin }^{-1}\left(\frac{u}{\sqrt{a}}\right)\\ & =& 2{\sin }^{-1}\left(\frac{\sqrt{x}}{\sqrt{a}}\right)\cdots u=\sqrt{x}\\ & =& 2{\sin }^{-1}\left(\sqrt{\frac{x}{a}}\right)+C\cdots C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数}\end{array}$$

定積分

$$\begin{array}{rcl}{\int }_0^a\frac{1}{\sqrt{x(a-x)}}\mathrm{d}x& =& {[2{\sin }^{-1}\left(\sqrt{\frac{x}{a}}\right)]}_0^a\\ & =& \{2{\sin }^{-1}\left(\sqrt{\frac{a}{a}}\right)\}-\{2{\sin }^{-1}\left(\sqrt{\frac{0}{a}}\right)\}\\ & =& \{2{\sin }^{-1}\left(1\right)\}-\{2{\sin }^{-1}\left(0\right)\}\\ & =& \left(2\frac{\pi }{2}\right)-(2\cdot 0)\\ & =& \pi -0\\ & =& \pi \end{array}$$