Question

The value of \(\int_{0}^{1} \frac{x^2}{\sqrt{1 - x^2}} \, dx\) is:

• A. \(\pi/6\)
• B. \(\pi/4\)
• C. \(\pi/2\)
• D. \(\pi/8\)

Hint:

Exam Tip:
For integrals with \(\sqrt{1 - x^2}\):

• Substitute \(x = \sin \theta\).
• Or \(x = \cos \theta\).
• Simplify using trigonometric identities.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to evaluate a definite integral involving \(\sqrt{1 - x^2}\). The standard substitution is \(x = \sin \theta\).

Step 2: Key Formula or Approach:
Let \(x = \sin \theta\). Then \(dx = \cos \theta \, d\theta\).
When \(x = 0\), \(\theta = 0\). When \(x = 1\), \(\theta = \pi/2\).
Also, \(\sqrt{1 - x^2} = \sqrt{1 - \sin^2 \theta} = \cos \theta\).

Step 3: Detailed Explanation:
\[ I = \int_{0}^{1} \frac{x^2}{\sqrt{1 - x^2}} \, dx = \int_{0}^{\pi/2} \frac{\sin^2 \theta}{\cos \theta} \cdot \cos \theta \, d\theta = \int_{0}^{\pi/2} \sin^2 \theta \, d\theta \] Now, \(\sin^2 \theta = \frac{1 - \cos 2\theta}{2}\).
So, \[ I = \int_{0}^{\pi/2} \frac{1 - \cos 2\theta}{2} \, d\theta = \frac{1}{2} \left[ \theta - \frac{\sin 2\theta}{2} \right]_{0}^{\pi/2} = \frac{1}{2} \left[ \frac{\pi}{2} - 0 \right] = \frac{\pi}{4} \]

Step 4: Final Answer:
Therefore, option (B) is correct.