Question

If $b>a$ , then $\underset{a}{\overset{b}{\int }}\frac{dx}{\sqrt{\left(x-a\right)\left(b-x\right)}}$ is equal to

• A. $\frac{\pi }{2}$
• B. $\pi$
• C. $\frac{\pi }{2}(b-a)$
• D. $\frac{\pi }{4}(b-a)$

Answer 1

Answer: (B)

$\underset{a}{\overset{b}{\int }}\frac{dx}{\sqrt{(x-a)(b-x)}}={\int }_a^b\frac{1}{\sqrt{-x^2+(a+b)x-ab}}dx$
$=\underset{a}{\overset{b}{\int }}\frac{1}{\sqrt{{\left(\frac{b-a}{2}\right)}^2-{\left(x-\frac{a+b}{2}\right)}^2}}dx$
$={\left[{\sin }^{-1}\left(\frac{x-\frac{a+b}{2}}{\frac{b-a}{2}}\right)\right]}_a^b$
$={\sin }^{-1}1-{\sin }^{-1}(-1)$
$=\frac{\pi }{2}+\frac{\pi }{2}=\pi$
In above integral if limits are not given, then it will be solved by substituting $x=a{\cos }^2\theta +b{\sin }^2\theta$