Evaluate ∫limits_1^2 dx/(√((x - 1)(2 - x)))

The correct answer is Option (2) → $\pi$

Let $I = \int\limits_{1}^{2} \frac{dx}{\sqrt{(x - 1)(2 - x)}} = \int\limits_{1}^{2} \frac{dx}{\sqrt{2x - x^2 - 2 + x}}$

$= \int\limits_{1}^{2} \frac{dx}{\sqrt{-(x^2 - 3x + 2)}}$

$= \int\limits_{1}^{2} \frac{dx}{\sqrt{-\left[ x^2 - 2 \cdot \frac{3}{2}x + \left( \frac{3}{2} \right)^2 + 2 - \left( \frac{3}{2} \right)^2 \right]}}$

$= \int\limits_{1}^{2} \frac{dx}{\sqrt{-\left\{ \left( x - \frac{3}{2} \right)^2 - \left( \frac{1}{2} \right)^2 \right\}}}$

$= \int\limits_{1}^{2} \frac{dx}{\sqrt{\left( \frac{1}{2} \right)^2 - \left( x - \frac{3}{2} \right)^2}} = \left[ \sin^{-1} \left( \frac{x - \frac{3}{2}}{\frac{1}{2}} \right) \right]_1^2$

$= [\sin^{-1} (2x - 3)]_1^2 = \sin^{-1} 1 - \sin^{-1}(-1)\qquad \left[ ∵\sin^{-1} (1) = \frac{\pi}{2} \text{ and } \sin^{-1}(-1) = \sin^{-1} \sin \left( -\frac{\pi}{2} \right) \right]$

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