Question

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

• A. \(sin^{-1}(x-1)+c\)
• B. \(cos^{-1}(x-1)+c\)
• C. \(tan^{-1}(x-1)+c\)
• D. \(sin^{-1}x+c\)

Hint:

Complete the square under the root to reach the sin inverse standard form.

Answer 1

The Correct Option is A

Step 1: Approach
Check by differentiating option (A).

Step 2: Derivative
$\dfrac{d}{dx}\sin^{-1}(x-1)=\dfrac{1}{\sqrt{1-(x-1)^2}}$.

Step 3: Simplify
$1-(x-1)^2=1-x^2+2x-1=2x-x^2$. So the derivative is $\dfrac{1}{\sqrt{2x-x^2}}$, the integrand.

Step 4: Conclusion
Option (A) is an antiderivative. The cosine inverse option differs in sign and the others have different derivatives.

Final Answer:
Completing the square gives sin inverse of (x - 1), option (A). \[ \boxed{\sin^{-1}(x-1)+c} \]