Find the following integral: int frac{dx}{sqr | vedclass

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Question

(A) To evaluate the integral $\int \frac{dx}{\sqrt{2x-x^2}}$,we first complete the square in the denominator:$2x - x^2 = -(x^2 - 2x) = -(x^2 - 2x + 1

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

Vedclass · Answer

(A) To evaluate the integral $\int \frac{dx}{\sqrt{2x-x^2}}$,we first complete the square in the denominator:$2x - x^2 = -(x^2 - 2x) = -(x^2 - 2x + 1 - 1) = 1 - (x-1)^2$So,the integral becomes $\int \frac{dx}{\sqrt{1-(x-1)^2}}$Let $t = x-1$,then $dt = dx$Substituting these into the integral,we get $\int \frac{dt}{\sqrt{1-t^2}}$Using the standard integral formula $\int \frac{dt}{\sqrt{1-t^2}} = \sin^{-1}(t) + C$Substituting back $t = x-1$,we get $\sin^{-1}(x-1) + C$

Find the following integral: $\int \frac{dx}{\sqrt{2x-x^2}}$

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