√(x)/√(a-x)の積分

√(x)/√(a-x)の積分

不定積分

$$ \begin{eqnarray} \int \frac{\sqrt{x}}{\sqrt{a-x}}\mathrm{d}x &=&\int x^{\frac{1}{2}}(a-x)^{-\frac{1}{2}}\mathrm{d}x \\&=&\int u(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 u^2(a-u^2)^{-\frac{1}{2}}\mathrm{d}u \;\cdots\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x \\&=&2\int \left\{\sqrt{a}\sin{\left(\theta\right)}\right\}^2\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\int a\sin^2{\left(\theta\right)}\left\{a-a\sin^2{\left(\theta\right)}\right\}^{-\frac{1}{2}}\sqrt{a}\cos{\left(\theta\right)}\mathrm{d}\theta \\&=&2a\sqrt{a}\int \sin^2{\left(\theta\right)}\left[a\left\{1-\sin^2{\left(\theta\right)}\right\}\right]^{-\frac{1}{2}}\cos{\left(\theta\right)}\mathrm{d}\theta \\&=&2a\sqrt{a}\int \sqrt{a}\sin^2{\left(\theta\right)}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 \sin^2{\left(\theta\right)}\left\{1-\sin^2{\left(\theta\right)}\right\}^{-\frac{1}{2}}\cos{\left(\theta\right)}\mathrm{d}\theta \\&=&2a\int \sin^2{\left(\theta\right)}\left\{1-\sin^2{\left(\theta\right)}\right\}^{-\frac{1}{2}}\cos{\left(\theta\right)}\mathrm{d}\theta \\&=&2a\int \sin^2{\left(\theta\right)}\left\{\cos^2{\left(\theta\right)}\right\}^{-\frac{1}{2}}\cos{\left(\theta\right)}\mathrm{d}\theta \\&=&2a\int \sin^2{\left(\theta\right)}\left\{\cos{\left(\theta\right)}\right\}^{-1}\cos{\left(\theta\right)}\mathrm{d}\theta \\&=&2a\int \sin^2{\left(\theta\right)}\mathrm{d}\theta \\&=&2a\left\{\frac{1}{2}\theta-\frac{1}{2}\sin{\left(\theta\right)}\cos{\left(\theta\right)}\right\} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/sin.html}{\int \sin^2{\left(\theta\right)}\mathrm{d}\theta=\frac{1}{2}\left\{\theta-\sin{\left(\theta\right)}\cos{\left(\theta\right)}\right\}+C\;(C:積分定数)} \\&=&2a\left\{\frac{1}{2}\sin^{-1}{\left(\frac{u}{\sqrt{a}}\right)}-\frac{1}{2}\sin{\left(\sin^{-1}{\left(\frac{u}{\sqrt{a}}\right)} \right)}\cos{\left(\sin^{-1}{\left(\frac{u}{\sqrt{a}}\right)} \right)}\right\} \;\cdots\;u=\sqrt{a}\sin{\left(\theta\right)},\;\frac{u}{\sqrt{a}}=\sin{\left(\theta\right)},\;\theta=\sin^{-1}{\left(\frac{u}{\sqrt{a}}\right)} \\&=&2a\left\{\frac{1}{2}\sin^{-1}{\left(\frac{u}{\sqrt{a}}\right)}-\frac{1}{2}\left(\frac{u}{\sqrt{a}}\right)\sqrt{ 1-\left(\frac{u}{\sqrt{a}}\right)^2 }\right\} \;\cdots\;\sin{\left(\sin^{-1}{\left(x\right)}\right)}=x,\;\cos{\left(\sin^{-1}{\left(x\right)}\right)}=\sqrt{1-x^2} \\&=&2a\frac{1}{2}\left\{\sin^{-1}{\left(\frac{u}{\sqrt{a}}\right)}-\left(\frac{u}{\sqrt{a}}\right)\sqrt{ 1-\left(\frac{u}{\sqrt{a}}\right)^2 }\right\} \\&=&a\left\{\sin^{-1}{\left(\frac{u}{\sqrt{a}}\right)}-\left(\frac{u}{\sqrt{a}}\right)\sqrt{ 1-\left(\frac{u}{\sqrt{a}}\right)^2 }\right\} \\&=&a\left\{\sin^{-1}{\left(\frac{\sqrt{x}}{\sqrt{a}}\right)}-\left(\frac{\sqrt{x}}{\sqrt{a}}\right)\sqrt{ 1-\left(\frac{\sqrt{x}}{\sqrt{a}}\right)^2 }\right\} \;\cdots\;u=\sqrt{x} \\&=&a\left\{\sin^{-1}{\left(\sqrt{\frac{x}{a}}\right)}-\frac{\sqrt{x}}{\sqrt{a}}\sqrt{ 1-\frac{x}{a} }\right\} \\&=&a\left\{\sin^{-1}{\left(\sqrt{\frac{x}{a}}\right)}-\frac{\sqrt{x}}{\sqrt{a}}\sqrt{ \frac{a-x}{a} }\right\} \\&=&a\left\{\sin^{-1}{\left(\sqrt{\frac{x}{a}}\right)}-\frac{\sqrt{x}}{\sqrt{a}}\frac{\sqrt{a-x}}{\sqrt{a}}\right\} \;\cdots\;\sqrt{\frac{A}{B}}=\left(\frac{A}{B}\right)^{\frac{1}{2}}=\left(A\frac{1}{B}\right)^{\frac{1}{2}}=A^{\frac{1}{2}}\left(\frac{1}{B}\right)^{\frac{1}{2}}=A^{\frac{1}{2}}\left(B^{-1}\right)^{\frac{1}{2}}=A^{\frac{1}{2}}B^{-\frac{1}{2}}=\sqrt{A}\frac{1}{\sqrt{B}}=\frac{\sqrt{A}}{\sqrt{B}} \\&=&a\left\{\sin^{-1}{\left(\sqrt{\frac{x}{a}}\right)}-\frac{\sqrt{x}\sqrt{a-x}}{a }\right\} \\&=&a\sin^{-1}{\left(\sqrt{\frac{x}{a}}\right)}-a\frac{\sqrt{x}\sqrt{a-x}}{a} \\&=&a\sin^{-1}{\left(\sqrt{\frac{x}{a}}\right)}-\sqrt{x}\sqrt{a-x}+C\;\cdots\;C:積分定数 \\&=&a\sin^{-1}{\left(\left(\frac{x}{a}\right)^{\frac{1}{2}}\right)}-x^{\frac{1}{2}}\left(a-x\right)^{\frac{1}{2}}+C \end{eqnarray} $$

定積分

$$ \begin{eqnarray} \int_{0}^{a} \frac{\sqrt{x}}{\sqrt{a-x}}\mathrm{d}x &=&\left[a\sin^{-1}{\left(\left(\frac{x}{a}\right)^{\frac{1}{2}}\right)}-x^{\frac{1}{2}}\left(a-x\right)^{\frac{1}{2}}\right]_{0}^{a} \\&=&(a\sin^{-1}{\left(\left(\frac{a}{a}\right)^{\frac{1}{2}}\right)}-a^{\frac{1}{2}}\left(a-a\right)^{\frac{1}{2}})-(a\sin^{-1}{\left(\left(\frac{0}{a}\right)^{\frac{1}{2}}\right)}-0^{\frac{1}{2}}\left(a-0\right)^{\frac{1}{2}}) \\&=&(a\sin^{-1}{\left(1\right)}-0)-(0-0) \;\cdots\;\sin^{-1}{\left(0\right)}=0,\;0^A=0 \\&=&a \frac{\pi}{2} \;\cdots\;\sin^{-1}{\left(1\right)}=\frac{\pi}{2} \\&=&\frac{a\pi}{2} \end{eqnarray} $$