Question

Solve $\underset{2}{\overset{6}{\int }}\sqrt{\left(6-x\right)\left(x-2\right)}dx$

Answer 1

We have,

$I={\int }_2^6\sqrt{(6-x)(x-2)}dx$

$I={\int }_2^6\sqrt{6x-12-x^2+2x}dx$

$I={\int }_2^6\sqrt{8x-12-x^2}dx$

$I={\int }_2^6\sqrt{8x-12-16+16-x^2}dx$

$I={\int }_2^6\sqrt{{\left(2\sqrt{7}\right)}^2-{(4-x)}^2}dx$

Let

$t=4-x$

$-dt=dx$

Therefore,

$I=-{\int }_2^{-2}\sqrt{{\left(2\sqrt{7}\right)}^2-t^2}dt$

We know that

$\int \sqrt{a^2-x^2}dx=\frac{1}{2}x\sqrt{a^2-x^2}+\frac{1}{2}{a}^2{\sin }^{-1}\left(\frac{x}{a}\right)+C$

Therefore,

$I=-{[\frac{1}{2}t\sqrt{{\left(2\sqrt{7}\right)}^2-t^2}+\frac{1}{2}{\left(2\sqrt{7}\right)}^2{\sin }^{-1}\left(\frac{t}{2\sqrt{7}}\right)]}_2^{-2}$

$=-(\frac{1}{2}(-2)\sqrt{{\left(2\sqrt{7}\right)}^2-{(-2)}^2}+\frac{1}{2}{\left(2\sqrt{7}\right)}^2{\sin }^{-1}\left(\frac{-2}{2\sqrt{7}}\right))+$

$(\frac{1}{2}\left(2\right)\sqrt{{\left(2\sqrt{7}\right)}^2-{\left(2\right)}^2}+\frac{1}{2}{\left(2\sqrt{7}\right)}^2{\sin }^{-1}\left(\frac{2}{2\sqrt{7}}\right))$

$I=-(-\sqrt{28-4}+\frac{1}{2}\times 28{\sin }^{-1}\left(\frac{-1}{\sqrt{7}}\right))+(\sqrt{28-4}+\frac{1}{2}\times 28{\sin }^{-1}\left(\frac{1}{\sqrt{7}}\right))$

$I=-(-\sqrt{24}+14{\sin }^{-1}\left(\frac{-1}{\sqrt{7}}\right))+(\sqrt{24}+14{\sin }^{-1}\left(\frac{1}{\sqrt{7}}\right))$

$I=2\sqrt{24}-14{\sin }^{-1}\left(\frac{-1}{\sqrt{7}}\right)+14{\sin }^{-1}\left(\frac{1}{\sqrt{7}}\right)$

Hence, the value is $2\sqrt{24}-14{\sin }^{-1}\left(\frac{-1}{\sqrt{7}}\right)+14{\sin }^{-1}\left(\frac{1}{\sqrt{7}}\right)$.