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

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Question

(A) Step $1$: Complete the square for the quadratic expression $2x-x^2$.$2x-x^2 = -(x^2-2x) = -(x^2-2x+1-1) = -( (x-1)^2 - 1 ) = 1 - (x-1)^2$.Step $2$

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

Vedclass · Answer

(A) Step $1$: Complete the square for the quadratic expression $2x-x^2$.$2x-x^2 = -(x^2-2x) = -(x^2-2x+1-1) = -( (x-1)^2 - 1 ) = 1 - (x-1)^2$.Step $2$: Rewrite the integral using the completed square.$\int \frac{1}{\sqrt{1-(x-1)^2}} dx$.Step $3$: Use the standard integral formula $\int \frac{1}{\sqrt{1-u^2}} du = \sin^{-1}(u) + c$.Let $u = x-1$, then $du = dx$.Therefore, $\int \frac{1}{\sqrt{1-u^2}} du = \sin^{-1}(u) + c = \sin^{-1}(x-1) + c$.

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

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