Question

Evaluate the integral
${\int }_1^3\frac{d_X}{\sqrt{(x-1)(3-x)}}$

• A. $\pi$
• B. $-\pi$
• C. $\pi /2$
• D. $0$

Answer 1

Answer: (A)

$\pi$

${\int }_1^3\frac{dx}{\sqrt{(x-1)(3-x)}}$
${\int }_1^3\frac{dx}{\sqrt{-3+4x-x^2}}={\int }_1^3\frac{dx}{\sqrt{1-(x-2{)}^2}}$
$={\int }_1^3\frac{dx}{\sqrt{1-(x-2{)}^2}}={[si{n}^{-1}\left(\frac{x-2}{1}\right)+c]}_1^3$
$=si{n}^{-1}(x-2{)}_1^3$
$=si{n}^{-1}\left(1\right)-si{n}^{-1}(-1)$
$=\frac{\pi }{2}-\left(\frac{-\pi }{2}\right)$
$=\pi$