Question

Evaluation of the integral \[ \int \frac{dx}{\sqrt{2x - x^2}} \] results in

• A. \(\sin^{-1}(x-1) + c\)
• B. \(\cos^{-1}(x-1) + c\)
• C. \(\sin^{-1}\left(\frac{x}{2}\right) + c\)
• D. \(\cos^{-1}\left(\frac{x}{2}\right) + c\)

Hint:

Always try to rewrite quadratic expressions under square roots in completed-square form to match inverse trigonometric identities.

Answer 1

The Correct Option is A

Step 1: Rewrite the expression inside the square root.
We have: \[ 2x - x^2 = - (x^2 - 2x) = -[(x^2 - 2x + 1) - 1] = -(x - 1)^2 + 1 \] Thus, \[ \sqrt{2x - x^2} = \sqrt{1 - (x-1)^2} \]
Step 2: Substitute to match the standard integral.
We now have: \[ \int \frac{dx}{\sqrt{1 - (x-1)^2}} \] This is of the form: \[ \int \frac{du}{\sqrt{1 - u^2}} = \sin^{-1}(u) + C \] with \( u = x - 1 \).
Step 3: Write the final result.
\[ \boxed{\sin^{-1}(x - 1) + C} \]