int frac{1}{sqrt{2x-x^2}} dx = . . . . . . | vedclass

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Question

(B) To solve the integral $I = \int \frac{1}{\sqrt{2x-x^2}} dx$,we first complete the square for the expression inside the square root: $2x - x^2 = -(

Vedclass · Accepted Answer

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

Vedclass · Answer

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

$\int \frac{1}{\sqrt{2x-x^2}} dx = $ . . . . . . $+ C$.

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