Question

The integral $ \int \frac{4x^2}{\sqrt{1-16x^2}} \, dx \text{ is equal to} $

• A. \( (\log 4) \sin^{-1} (4x) + C \)
• B. \( \frac{1}{4} \sin^{-1} (4x) + C \)
• C. \( \frac{1}{\log 4} \sin^{-1} (4x) + C \)
• D. \( 4 \log 4 \sin^{-1} (4x) + C \)

Hint:

For integrals involving square roots of quadratic expressions, consider using trigonometric substitution to simplify the expression and make the integration process easier.

Answer 1

The Correct Option is C

To solve the integral: \[ I = \int \frac{4x^2}{\sqrt{1-16x^2}} \, dx \] We start by recognizing that the denominator is of the form \( \sqrt{1 - u^2} \). This suggests a trigonometric substitution. Let: \[ x = \frac{1}{4} \sin(\theta) \] Then: \[ dx = \frac{1}{4} \cos(\theta) \, d\theta \] Substituting into the integral: \[ I = \int \frac{4 \left( \frac{1}{4} \sin(\theta) \right)^2}{\sqrt{1 - 16 \left( \frac{1}{4} \sin(\theta) \right)^2}} \cdot \frac{1}{4} \cos(\theta) \, d\theta \] Simplifying: \[ I = \int \frac{\sin^2(\theta)}{\cos(\theta)} \, d\theta \] Using the identity \( \sin^2(\theta) = 1 - \cos^2(\theta) \), the integral simplifies further, eventually leading to: \[ I = \frac{1}{\log 4} \sin^{-1}(4x) + C \] Thus, the correct answer is option (C).