1/√(x(a-x))の積分

1/√(x(a-x))の積分

不定積分

$$ \begin{eqnarray} \int \frac{1}{\sqrt{x(a-x)}}\mathrm{d}x &=&\int \left\{x(a-x)\right\}^{-\frac{1}{2}}\mathrm{d}x \\&=&\int x^{-\frac{1}{2}}\left(a-x\right)^{-\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 \left[a-\left\{\sqrt{a}\sin{\left(\theta\right)}\right\}^2\right]^{-\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 \left[a-\left\{\sqrt{a}^2\sin^2{\left(\theta\right)}\right\}\right]^{-\frac{1}{2}}\;\cos{\left(\theta\right)}\mathrm{d}\theta \;\cdots\;(AB)^C=A^CB^C \\&=&2\sqrt{a}\int \left\{a-a\sin^2{\left(\theta\right)}\right\}^{-\frac{1}{2}}\;\cos{\left(\theta\right)}\mathrm{d}\theta \\&=&2\sqrt{a}\int \left[a\left\{1-\sin^2{\left(\theta\right)}\right\}\right]^{-\frac{1}{2}}\;\cos{\left(\theta\right)}\mathrm{d}\theta \\&=&2\sqrt{a}\int a^{-\frac{1}{2}}\left\{1-\sin^2{\left(\theta\right)}\right\}^{-\frac{1}{2}}\;\cos{\left(\theta\right)}\mathrm{d}\theta \\&=&2a\sqrt{a}a^{-\frac{1}{2}}\int \left\{1-\sin^2{\left(\theta\right)}\right\}^{-\frac{1}{2}}\;\cos{\left(\theta\right)}\mathrm{d}\theta \\&=&2\int \left\{1-\sin^2{\left(\theta\right)}\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 \left\{\cos^2{\left(\theta\right)}\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 \left\{\cos{\left(\theta\right)}\right\}^{2(-\frac{1}{2})}\;\cos{\left(\theta\right)}\mathrm{d}\theta \;\cdots\;\left(A^B\right)^C=A^{BC} \\&=&2\int \left\{\cos{\left(\theta\right)}\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:積分定数) \\&=&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:積分定数 \end{eqnarray} $$

定積分

$$ \begin{eqnarray} \int_0^a \frac{1}{\sqrt{x(a-x)}}\mathrm{d}x &=&\left[2\sin^{-1}{\left(\sqrt{\frac{x}{a}}\right)}\right]_0^a \\&=&\left\{2\sin^{-1}{\left(\sqrt{\frac{a}{a}}\right)}\right\}-\left\{2\sin^{-1}{\left(\sqrt{\frac{0}{a}}\right)}\right\} \\&=&\left\{2\sin^{-1}{\left(1\right)}\right\}-\left\{2\sin^{-1}{\left(0\right)}\right\} \\&=&\left(2\frac{\pi}{2}\right)-\left(2\cdot0\right) \\&=&\pi-0 \\&=&\pi \end{eqnarray} $$