Question

The value of \( \int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} dx \) is equal to:

• A. \( \sqrt{(1-x^2)} \sin^{-1} x + C \)
• B. \( x \sin^{-1} x + C \)
• C. \( x - \sqrt{(1-x^2)} \sin^{-1} x + C \)
• D. \( \sqrt{(\sin^{-1} x)} + C \)

Hint:

When performing trigonometric substitution in integrals, choose an appropriate substitution to simplify the integrand and make the integration process easier.

Answer 1

The Correct Option is C

Let \( I = \int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \, dx \).

Step 1: Employ trigonometric substitution. Let \( \sin^{-1} x = t \), which implies \( x = \sin t \).

Differentiating yields \( \frac{1}{\sqrt{1 - x^2}} \, dx = dt \). Thus, \( \cos t = \sqrt{1 - x^2} \).

Step 2: Substitute into the integral. Replacing the expressions in the integral gives: \[ I = \int t \sin t \, dt. \] This simplifies to: \[ I = t (-\cos t) - \int (-\cos t) \, dt. \]

Step 3: Simplify the expression. Further simplification results in: \[ I = -t \cos t + \sin t + C. \]

Step 4: Back-substitute \( t = \sin^{-1} x \). Now, substitute \( t = \sin^{-1} x \) into the equation: \[ I = -(\sin^{-1} x) \cos (\sin^{-1} x) + \sin (\sin^{-1} x) + C. \] Using the identities \( \cos (\sin^{-1} x) = \sqrt{1 - x^2} \) and \( \sin (\sin^{-1} x) = x \), we obtain: \[ I = x - \sqrt{1 - x^2} \sin^{-1} x + C. \]

Final Answer: \[ \boxed{x - \sqrt{1 - x^2} \sin^{-1} x + C} \]