Question

Solve:
${\int }_a^b\sqrt{(x-a)(b-x)}dx$

Answer 1

${\int }_a^b\sqrt{(x-4)(b-x)}.dx$

$={\int }_a^b\sqrt{bx-x^2-ab+ax}.dx$

$={\int }_a^b\sqrt{(a+b)x-x^2-ab}.dx$

$={\int }_a^b\sqrt{(\frac{a+b}{2}{)}^2-(x-\frac{a+b}{2}{)}^2-ab}.dx$

$={\int }_a^b\sqrt{(\frac{a^2+b^2}{2}{)}^2-(x-\frac{a+b}{2}{)}^2}.dx$

substitute $x-\frac{a+b}{2}=t.$

$={\int }_a^b\sqrt{(\frac{a^2+b^2}{2}{)}^2-t^2}.dt$

$=\left[\frac{1}{2}\right[(x-\frac{a+b}{2})\sqrt{(\frac{\sqrt{a^2+b^2}}{2}{)}^2-(x-\frac{a+b}{2}{)}^2}+\frac{a^2+b^2}{4}si{n}^{-1}[\frac{x-\frac{(a+b)}{2}}{\frac{\sqrt{a^2+b^2}}{2}}{]}_a^b$

$=\frac{1}{2}\left[\right(x-\frac{a+b}{2})\sqrt{(x-a)(b-x)}+\frac{a^2+b^2}{4}si{n}^{-1}(\frac{2x-a-b}{\sqrt{a^2+b^2}}){]}_a^b$

$=\frac{1}{2}[0+\frac{a^2+b^2}{4}si{n}^{-1}(\frac{2b-ab}{\sqrt{a^2}+b^2}\left)\right]-\left[\frac{a^2+b^2}{4}si{n}^{-1}\frac{a-b}{\sqrt{a^2+b^2}}\right]$

$=\frac{1}{2}\left[\frac{a^2+b^2}{4}si{n}^{-1}\right(\frac{(b-a)}{\sqrt{a^2+b^2}})-\frac{a^2+b^2}{4}si{n}^{-1}(\frac{a-b}{\sqrt{a^2+b^2}}\left)\right]$

$=\frac{1}{2}\frac{a^2+b^2}{4}\left[si{n}^{-1}\right(\frac{a-b}{\sqrt{a^2+b^2}})=si{n}^{-1}(\frac{a-b}{\sqrt{a^2+b^2}}\left)\right]$

$=-\frac{a^2+b^2}{4}si{n}^{-1}\frac{a-b}{\sqrt{a^2+b^2}}$