int frac{1}{sqrt{4x-x^2}} dx = . . . . . . | vedclass

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Question

(B) To solve the integral $I = \int \frac{1}{\sqrt{4x-x^2}} dx$,we first complete the square for the quadratic expression inside the square root.$4x -

Vedclass · Accepted Answer

$\sin^{-1}\left(\frac{x-2}{2}\right)$

Vedclass · Answer

(B) To solve the integral $I = \int \frac{1}{\sqrt{4x-x^2}} dx$,we first complete the square for the quadratic expression inside the square root.$4x - x^2 = -(x^2 - 4x) = -(x^2 - 4x + 4 - 4) = -( (x-2)^2 - 4 ) = 4 - (x-2)^2$.Substituting this back into the integral,we get:$I = \int \frac{1}{\sqrt{4 - (x-2)^2}} dx$.Using the standard integral formula $\int \frac{1}{\sqrt{a^2 - u^2}} du = \sin^{-1}(\frac{u}{a}) + c$,where $a = 2$ and $u = x-2$:$I = \sin^{-1}\left(\frac{x-2}{2}\right) + c$.Thus,the correct option is $B$.

$\int \frac{1}{\sqrt{4x-x^2}} dx = $ . . . . . . $+ c$.

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