Question

The value of the integral ${\int }_{\alpha }^{\beta }\frac{dx}{\sqrt{(x-\alpha )(\beta -x)}}$ for $\beta >\alpha$, is

• A. ${\sin }^{-1}\alpha /\beta$
• B. $\pi /2$
• C. ${\sin }^{-1}\beta /2\alpha$
• D. $\pi$

Answer 1

Answer: (D)

$\pi$

Given : ${\int }_{\alpha }^{\beta }\frac{dx}{\sqrt{(x-\alpha )(\beta -x)}}$

$I={\int }_{\alpha }^{\beta }\frac{dx}{\sqrt{(x-\alpha )(\beta -x)}}={\int }_{\alpha }^{\beta }\frac{dx}{\sqrt{-\alpha \beta +x(\beta +\alpha )-x^2}}$

$I={\int }_{\alpha }^{\beta }\frac{dx}{\sqrt{\frac{{(\alpha +\beta )}^2}{4}-\alpha \beta -{[x-\frac{\beta +\alpha }{2}]}^2}}$

$I={\int }_{\alpha }^{\beta }\frac{dx}{\sqrt{\frac{(\beta -\alpha {)}^2}{4}-{[x-\frac{(\beta +\alpha )}{2}]}^2}}$

we know that, $\int \frac{1}{\sqrt{a^2-x^2}}dx={\sin }^{-1}\left(\frac{x}{a}\right)+c$

$I={\sin }^{-1}{\left[\frac{x-\left(\frac{\beta +\alpha }{2}\right)}{\left(\frac{\beta -\alpha }{2}\right)}\right]}_{\alpha }^{\beta }={\sin }^{-1}\left[\frac{\left(\frac{\beta -\alpha }{2}\right)}{\left(\frac{\beta +\alpha }{2}\right)}\right]-{\sin }^{-1}\left[\frac{\left(\frac{\alpha -\beta }{2}\right)}{\left(\frac{\beta -\alpha }{2}\right)}\right]$

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

$I=\frac{\pi }{2}+\frac{\pi }{2}=\pi$

Hence the correct answer is $\pi$