int frac{dx}{{(2ax - x^2)}^{1/2}} = a^n sin^{ | vedclass

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(C) The given integral is $I = \int \frac{dx}{\sqrt{2ax - x^2}}$.Completing the square in the denominator: $2ax - x^2 = -(x^2 - 2ax) = -(x^2 - 2ax + a

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$0$

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(C) The given integral is $I = \int \frac{dx}{\sqrt{2ax - x^2}}$.Completing the square in the denominator: $2ax - x^2 = -(x^2 - 2ax) = -(x^2 - 2ax + a^2 - a^2) = a^2 - (x - a)^2$.So,$I = \int \frac{dx}{\sqrt{a^2 - (x - a)^2}} = \int \frac{dx}{a \sqrt{1 - (\frac{x-a}{a})^2}}$.Let $u = \frac{x-a}{a} = \frac{x}{a} - 1$,then $du = \frac{dx}{a}$,so $dx = a \, du$.Substituting these,$I = \int \frac{a \, du}{a \sqrt{1 - u^2}} = \int \frac{du}{\sqrt{1 - u^2}} = \sin^{-1}(u) + C = \sin^{-1} \left( \frac{x}{a} - 1 \right) + C$.Comparing this with the given formula $a^n \sin^{-1} \left( \frac{x}{a} - 1 \right)$,we see that $a^n = 1$,which implies $a^n = a^0$.Therefore,$n = 0$.

$\int \frac{dx}{{(2ax - x^2)}^{1/2}} = a^n \sin^{-1} \left( \frac{x}{a} - 1 \right)$. In this formula,$n =$ . . . . . . .

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