If int frac{dx}{sqrt{2ax - x^2}} = f(g(x)) + | vedclass

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(A) Step $1$: Complete the square in the denominator: $2ax - x^2 = -(x^2 - 2ax) = -(x^2 - 2ax + a^2 - a^2) = a^2 - (x - a)^2$.Step $2$: Substitute thi

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$\sin^{-1} x, \frac{x - a}{a}$

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(A) Step $1$: Complete the square in the denominator: $2ax - x^2 = -(x^2 - 2ax) = -(x^2 - 2ax + a^2 - a^2) = a^2 - (x - a)^2$.Step $2$: Substitute this into the integral: $\int \frac{dx}{\sqrt{a^2 - (x - a)^2}}$.Step $3$: Use the standard integral formula $\int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}(\frac{u}{a}) + c$.Step $4$: Here $u = x - a$, so the integral becomes $\sin^{-1}(\frac{x - a}{a}) + c$.Step $5$: Comparing this with $f(g(x)) + c$, we get $f(x) = \sin^{-1}(x)$ and $g(x) = \frac{x - a}{a}$.

If $\int \frac{dx}{\sqrt{2ax - x^2}} = f(g(x)) + c$, where $c$ is the constant of integration, then $f(x)$ and $g(x)$ are respectively equal to:

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