Evaluate the following: ^2_1 dx (x-1)(2-x) - Vidyadip

Question

Evaluate the following:
$\int\limits^2_1\frac{\text{dx}}{\sqrt{(\text{x}-1)(2-\text{x})}}$

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Answer

Let $\text{I}=\int\limits^2_1\frac{\text{dx}}{\sqrt{(\text{x}-1)(2-\text{x})}}$ $=\int\limits^2_1\frac{\text{dx}}{\sqrt{2\text{x}-\text{x}^2-2+\text{x}}}$
$=\int\limits^2_1\frac{\text{dx}}{\sqrt{-(\text{x}^2-3\text{x}+2)}}$
$\int\limits^2_1\frac{\text{dx}}{\sqrt{-\bigg[\text{x}^2-2\cdot\frac{3}{2}\text{x}+\Big(\frac{3}{2}\Big)^2+2-\frac{9}{4}\bigg]}}$
$=\int\limits^2_1\frac{\text{dx}}{\sqrt{-\bigg\{\Big(\text{x}-\frac{3}{2}\Big)^2-\Big(\frac{1}{2}\Big)^2\bigg\}}}$
$=\int\limits^2_1\frac{\text{dx}}{\sqrt{\Big(\frac{1}{2}\Big)^2-\Big(\text{x}-\frac{3}{2}\Big)^2}}$ $=\Bigg[\sin^{-1}\bigg(\frac{\text{x}-\frac{3}{2}}{\frac{1}{2}}\bigg)\Bigg]^2_1$
$=\big[\sin^{-1}(2\text{x}-3)\big]^2_1=\sin^{-1}1-\sin^{-1}(-1)$
$=\frac{\pi}{2}+\frac{\pi}{2}$ $\Big[\because\sin\frac{\pi}{2}=1\text{ and }\sin(-\theta)=-\sin\theta\Big]$
$=\pi$

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