Evaluate ${\int }_4^9 \frac{dx}{\sqrt{(9-x)(x-4)}}.$

${\int }_4^9 \frac{dx}{\sqrt{(x-4)(9-x)}} (x-a)=(b-a){\sin }^2\theta$

$a=4,b=9 x-4=(9-4){\sin }^2\theta$

Put $x=a{\cos }^2\theta +b{\sin }^2\theta x-4=5{\sin }^2\theta$

$dx=(b-a)\sin 2\theta d\theta b-x=(b-a){\cos }^2\theta$

$dx=(9-4)\sin 2\theta d\theta 9-x=5{\cos }^2\theta$

$dx=5\sin 2\theta d\theta$

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

$\text{ L.L. }x=4\Rightarrow \theta =0, \text{ U.L. }x=9\Rightarrow \theta =\frac{\pi }{2}$

${\int }_4^9 \frac{dx}{\sqrt{(x-4)(9-x)}}$

$={\int }_0^{\pi /2} \frac{1}{\sqrt{5{\sin }^2\theta 5{\cos }^2\theta }}\cdot 5\sin 2\theta d\theta$

$={\int }_0^{\pi /2} \frac{5\cdot 2\sin \theta \cos \theta }{5\sin \theta \cos \theta }d\theta ={\int }_0^{\pi /2} 2d\theta$

$=2{\int }_0^{\pi /2} 1d\theta =2[\theta {]}_0^{\pi /2}=2\left[\frac{\pi }{2}-0\right]=\pi$