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# Evaluate the following:$\int\limits_1^2\frac{dx}{\sqrt{(x-1)(2-x)}}$

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A)
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• $\int \limits_a^b \large\frac{dx}{\sqrt {a^2-x^2}}=\bigg[\sin ^{-1} (\frac{x}{a})+c \bigg]_a^b$
Step 1:
$\large\int\limits_1^2\frac{dx}{\sqrt{(x-1)(2-x)}}$
Consider $(x-1)(2-x)$
On expanding we get $-x^2+2x-2+x$
$=-(x^2-3x+2)$
On completing the squares this can be written as
$-\bigg[(x-\large\frac{3}{2})^2-\frac{9}{4} +2 \bigg]$
$=-\bigg[(x-\large\frac{3}{2})^2-(\frac{1}{4}) \bigg]$
$=\bigg [(\large\frac{1}{2})^2-(x-\frac{3}{2})^2\bigg]$
Now substituting this in I we get,
$\int \limits_1^2 \large\frac{dx}{\sqrt {(1/2)^2-(x-3/2)^2}}$
Clearly this is of the form $\int \limits_a^b \large\frac{dx}{\sqrt {a^2-x^2}}$
$=\sin ^{-1}(x/a)+c$
Here $x=(x-3/2)\; and\; a=\large\frac{1}{2}$
Hence on integrating we get,
$\bigg[ \sin ^{-1} \bigg(\large\frac{x-3/2}{1/2} \bigg)\bigg]_1^2$
$=[\sin ^{-1} \large\frac {(2x-3)}{1} \bigg]_1^2$
Step 2:
On applying limits we get,
$I=[\sin ^{-1}(2 \times 2 -3)- \sin ^{-1} (2-3)]$
$=[\sin ^{-1}(1)-\sin ^{-1}(-1)]$
But $\sin ^{-1}(1)=\large\frac{\pi}{2}\;$$and\;\sin ^{-1}=-\pi/2$
Therefore $I=\large\frac{\pi}{2}+\frac{\pi}{2}=\pi$