If b a,then int_a^b frac{dx}{sqrt{(x-a)(b-x | vedclass

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(D) Let $I = \int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}$.Expanding the denominator: $(x-a)(b-x) = -x^2 + (a+b)x - ab$.Completing the square: $-[x^2 - (a+b)

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$\pi$

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(D) Let $I = \int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}$.Expanding the denominator: $(x-a)(b-x) = -x^2 + (a+b)x - ab$.Completing the square: $-[x^2 - (a+b)x + ab] = -[x^2 - (a+b)x + (\frac{a+b}{2})^2 - (\frac{a+b}{2})^2 + ab] = -[(x - \frac{a+b}{2})^2 - (\frac{b-a}{2})^2] = (\frac{b-a}{2})^2 - (x - \frac{a+b}{2})^2$.Thus,$I = \int_a^b \frac{dx}{\sqrt{(\frac{b-a}{2})^2 - (x - \frac{a+b}{2})^2}}$.Using the standard integral $\int \frac{dx}{\sqrt{A^2 - u^2}} = \sin^{-1}(\frac{u}{A}) + C$,we get:$I = [\sin^{-1}(\frac{x - \frac{a+b}{2}}{\frac{b-a}{2}})]_a^b$.Evaluating at the limits:Upper limit $(x=b)$: $\sin^{-1}(\frac{b - \frac{a+b}{2}}{\frac{b-a}{2}}) = \sin^{-1}(\frac{\frac{b-a}{2}}{\frac{b-a}{2}}) = \sin^{-1}(1) = \frac{\pi}{2}$.Lower limit $(x=a)$: $\sin^{-1}(\frac{a - \frac{a+b}{2}}{\frac{b-a}{2}}) = \sin^{-1}(\frac{\frac{a-b}{2}}{\frac{b-a}{2}}) = \sin^{-1}(-1) = -\frac{\pi}{2}$.Therefore,$I = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi$.

If $b > a$,then $\int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}$ is equal to

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