Integral Evaluation: $\int \frac{dx}{\sqrt{2x-x^2}}$

To evaluate the integral $ \int \frac{dx}{\sqrt{2x-x^2}} $, we first manipulate the expression under the square root by completing the square.

Completing the Square

Consider the quadratic expression $ 2x - x^2 $.

• Rewrite it as $ -(x^2 - 2x) $.
• Complete the square for $ x^2 - 2x $. Add and subtract $ (\frac{-2}{2})^2 = (-1)^2 = 1 $.
• $ x^2 - 2x = (x^2 - 2x + 1) - 1 = (x-1)^2 - 1 $.
• Substitute back: $ -(x^2 - 2x) = -((x-1)^2 - 1) = 1 - (x-1)^2 $.

Integral Transformation

The integral becomes:

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

Standard Integral Form

This integral is now in a standard form. Let $ u = x-1 $. Then $ du = dx $. The integral transforms to:

$ \int \frac{du}{\sqrt{1 - u^2}} $

This is the standard integral for the inverse sine function:

$ \int \frac{du}{\sqrt{1 - u^2}} = \sin^{-1}(u) + c $

Final Solution

Substitute back $ u = x-1 $:

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

Therefore, the evaluation of the integral $ \int \frac{dx}{\sqrt{2x-x^2}} $ results in $ \sin^{-1}(x-1)+c $.