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

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

Put $x=\left(2{\cos }^2\theta +6{\sin }^2\theta \right)$

$dx=(b-a)\sin 2\theta d\theta$

$dx=(6-2)\sin 2\theta d\theta dx=4\sin 2\theta d\theta$

$x-a=(b-a){\sin }^2\theta \therefore (x-2)=4{\sin }^2\theta$

$b-x=(b-a){\cos }^2\theta \therefore 6-x=(6-2){\cos }^2\theta$

$\text{ L.L. }x=a\Rightarrow \theta =0 \text{ U.L. }x=b\Rightarrow \theta =\frac{\pi }{2}$

$\text{ L.L. }x=2\Rightarrow \theta =0 \text{ U.L. }x=6\Rightarrow \theta =\frac{\pi }{2}$

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

$={\int }_0^{\pi /2} \sqrt{4{\sin }^2\theta 4{\cos }^2\theta }4\sin 2\theta d\theta$

$={\int }_0^{\pi /2} 16\sin \theta \cos \theta 2\sin \theta \cos \theta \cdot d\theta$

$=32{\int }_0^{\pi /2} {\sin }^2\theta {\cos }^2\theta d\theta =32\left[\frac{2-1}{2+2}\cdot \frac{2-1}{2+2-2}\frac{\pi }{2}\right]$

(by applying reduction formula)

$=32\left[\frac{1}{4}\cdot \frac{1}{2}\cdot \frac{\pi }{2}\right]=2\pi \therefore {\int }_2^6 \sqrt{(6-x)(x-2)}dx=2\pi$