Question

If $b>a,$ then ${\int }_a^b\frac{dx}{\sqrt{(x-a)(b-x)}}]=?$

Answer 1

Finding the value of integral $\mathbf{}{\mathbf{\int }}_{\mathbf{a}}^{\mathbf{b}}\frac{\mathbf{d}\mathbf{x}}{\sqrt{\mathbf{(}\mathbf{x}\mathbf{}\mathbf{-}\mathbf{}\mathbf{a}\mathbf{)}\mathbf{}\mathbf{(}\mathbf{b}\mathbf{}\mathbf{-}\mathbf{}\mathbf{x}\mathbf{)}}}\mathbf{]}\mathbf{}\mathbf{:}$

Given that $b>a,$ and ${\int }_a^b\frac{dx}{\sqrt{(x-a)(b-x)}}]$

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

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

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

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

$={\sin }^{-1}\frac{(b-{\displaystyle \frac{a+b}{2}})}{{\displaystyle \frac{b-a}{2}}}-{\sin }^{-1}\frac{(a-{\displaystyle \frac{a+b}{2}})}{{\displaystyle \frac{b-a}{2}}}$

$={\sin }^{-1}\left(1\right)-{\sin }^{-1}(-1)$

$=\frac{\pi }{2}-(-\frac{\pi }{2})$

$=\pi$

Hence, the value of ${\mathbf{\int }}_{\mathbf{a}}^{\mathbf{b}}\frac{\mathbf{d}\mathbf{x}}{\sqrt{\mathbf{(}\mathbf{x}\mathbf{}\mathbf{-}\mathbf{}\mathbf{a}\mathbf{)}\mathbf{}\mathbf{(}\mathbf{b}\mathbf{}\mathbf{-}\mathbf{}\mathbf{x}\mathbf{)}}}\mathbf{]}\mathbf{}$is $\mathit{\pi }\mathbf{.}$