Question

Evaluate the integral: \[ \int \sin^{-1} \left( \sqrt{\frac{x - a}{x}} \right) dx \]

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

Hint:

For inverse trigonometric integrals, use suitable trigonometric identities and substitutions to simplify expressions.

Answer 1

The Correct Option is A

Step 1: Use the substitution Let \[ t = \sin^{-1} \left( \sqrt{\frac{x - a}{x}} \right). \] Then, \[ \sin t = \sqrt{\frac{x - a}{x}}. \] Squaring both sides, \[ \sin^2 t = \frac{x - a}{x}. \] Thus, \[ x \sin^2 t = x - a. \] Step 2: Differentiate both sides Using implicit differentiation, \[ 2x \sin t \cos t \frac{dt}{dx} + \sin^2 t = 1. \] Rearranging, \[ \frac{dt}{dx} = \frac{1 - \sin^2 t}{2x \sin t \cos t}. \] Using trigonometric identities, solve the integral, leading to: \[ x \cos^{-1} \left(\sqrt{\frac{a}{x}}\right) - \sqrt{ax - a^2} + C. \]