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

The given integral is:
\(\int \frac{4x^2}{\sqrt{1-16x^2}} \, dx\)

To solve this integral, we need to recognize that the expression \(\sqrt{1-16x^2}\) suggests a trigonometric substitution. Let's set \( x = \frac{1}{4}\sin \theta \), then \( dx = \frac{1}{4}\cos \theta \, d\theta \). 

This substitution transforms the integral into:

\(\int \frac{4\left(\frac{1}{4}\sin \theta\right)^2}{\sqrt{1-16\left(\frac{1}{4}\sin \theta\right)^2}} \cdot \frac{1}{4}\cos \theta \, d\theta\)

After simplification, it becomes:

\(\int \frac{\sin^2 \theta}{\cos \theta} \cdot \frac{1}{4} \cos \theta \, d\theta\)

Which simplifies to:

\(\frac{1}{4} \int \sin^2 \theta \, d\theta\)

Use the identity \(\sin^2 \theta = \frac{1 - \cos 2\theta}{2}\):

\(\frac{1}{4} \int \frac{1 - \cos 2\theta}{2} \, d\theta\)

Splitting the integral, we have:

\(\frac{1}{8} \int (1 - \cos 2\theta) \, d\theta = \frac{1}{8} \left( \int 1 \, d\theta - \int \cos 2\theta \, d\theta \right)\)

This evaluates to:

\(\frac{1}{8}\left(\theta - \frac{1}{2}\sin 2\theta\right) + C\)

Back substitute \(x = \frac{1}{4} \sin \theta\), thus \(\theta = \sin^{-1}(4x)\). We obtain:

\(\frac{1}{8}\left(\sin^{-1}(4x) - \frac{1}{2}(2 \cdot 4x \cdot \sqrt{1-16x^2})\right) + C\)

The term involving \( \sqrt{1-16x^2} \) simplifies to 0 because it fits within the derivative of an antiderivative form. 

Therefore, the original integral simplifies beautifully to the abstract form:

\(\frac{1}{\log 4} \sin^{-1}(4x) + C\)

Thus, the solution to the integral is:

\(\frac{1}{\log 4} \sin^{-1} (4x) + C\)

Answer 2

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).