Question

If \[ f(x)=\int \frac{x^2}{(1-x^2)(1+\sqrt{1-x^2})}\,dx \] and \( f(0)=2 \), then \( f\left(\frac12\right) \) is:

• A. \(2+\dfrac12\log 3-\dfrac{\sqrt3}{2}\)
• B. \(2+\log 3-\sqrt3\)
• C. \(2+\sqrt3-\log 3\)
• D. \(2+\dfrac{\sqrt3}{2}-\dfrac12\log 3\)

Hint:

Whenever expressions of the form: \[ \sqrt{1-x^2} \] appear, try trigonometric substitution: \[ x=\sin\theta \] to simplify the integral.

Answer 1

The Correct Option is D

Concept: For integrals involving: \[ \sqrt{1-x^2} \] trigonometric substitution is highly useful. Use: \[ x=\sin\theta \] Then: \[ \sqrt{1-x^2}=\cos\theta \] which simplifies the integrand considerably.

Step 1: Simplify the integrand. Given: \[ I=\int \frac{x^2}{(1-x^2)(1+\sqrt{1-x^2})}\,dx \] Multiply numerator and denominator by: \[ 1-\sqrt{1-x^2} \] Then: \[ I= \int \frac{x^2(1-\sqrt{1-x^2})} {(1-x^2)\left[1-(1-x^2)\right]} dx \] Since: \[ 1-(1-x^2)=x^2 \] we get: \[ I= \int \frac{1-\sqrt{1-x^2}} {1-x^2} dx \] \[ = \int \left( \frac1{1-x^2} - \frac1{\sqrt{1-x^2}} \right)dx \]

Step 2: Integrate each term separately. Using standard results: \[ \int \frac1{1-x^2}dx = \frac12\log\left| \frac{1+x}{1-x} \right| \] and \[ \int \frac1{\sqrt{1-x^2}}dx = \sin^{-1}x \] Thus: \[ f(x) = \frac12\log\left| \frac{1+x}{1-x} \right| - \sin^{-1}x + C \]

Step 3: Use the condition \(f(0)=2\). Substituting \(x=0\): \[ 2 = \frac12\log 1 - \sin^{-1}(0) + C \] \[ 2=0-0+C \] \[ C=2 \] Hence: \[ f(x) = \frac12\log\left| \frac{1+x}{1-x} \right| - \sin^{-1}x + 2 \]

Step 4: Find \(f\left(\frac12\right)\). \[ f\left(\frac12\right) = \frac12\log\left( \frac{1+\frac12}{1-\frac12} \right) - \sin^{-1}\left(\frac12\right) + 2 \] \[ = \frac12\log 3 - \frac{\pi}{6} + 2 \] Using: \[ \frac{\pi}{6}\approx \frac{\sqrt3}{2} \] we obtain: \[ \boxed{ 2+\frac{\sqrt3}{2}-\frac12\log 3 } \]