int frac{dx}{sqrt{2x - x^2}} = | vedclass

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Question

(B) To evaluate the integral $I = \int \frac{dx}{\sqrt{2x - x^2}}$,we first complete the square for the expression inside the square root: $2x - x^2 =

Vedclass · Accepted Answer

$\sin^{-1}(x - 1) + c$

Vedclass · Answer

(B) To evaluate the integral $I = \int \frac{dx}{\sqrt{2x - x^2}}$,we first complete the square for the expression inside the square root: $2x - x^2 = -(x^2 - 2x) = -(x^2 - 2x + 1 - 1) = 1 - (x - 1)^2$. Substituting this back into the integral,we get: $I = \int \frac{dx}{\sqrt{1 - (x - 1)^2}}$. Using the standard integration formula $\int \frac{du}{\sqrt{1 - u^2}} = \sin^{-1}(u) + c$,where $u = x - 1$ and $du = dx$,we obtain: $I = \sin^{-1}(x - 1) + c$.

$\int \frac{dx}{\sqrt{2x - x^2}} = $

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