Definite Integration
Evaluation by Substitution
Grade 12
Question:
<p>If the value of the integral <div>\[\int_0^{1/2} \frac{x}{(1-x^2)^{3/2}} dx\]</div> is <div>\[\frac{k}{6}\]</div>, then <p>\(k\) is equal to</p> <span style="color: #888;">(JEE Main 2020)</span></p>
<p>(a) \(3\sqrt{2} + \pi\)</p>
<p>(b) \(2\sqrt{3} - \pi\)</p>
<p>(c) \(2\sqrt{3} + \pi\)</p>
<p>(d) \(3\sqrt{2} - \pi\)</p>
Step-by-Step Solution
Key Concept: Use substitution $x = \sin\theta$ to convert the integral and evaluate using trigonometric identities.
<p><strong>Solution:</strong> Let $x = \sin\theta$.</p><p>Then $dx = \cos\theta\, d\theta$</p><p>When $x = 0$: $\theta = 0$</p><p>When $x = 1/2$: $\theta = \pi/6$</p><p>$I = \int_0^{\pi/6} \frac{\sin\theta}{(1-\sin^2\theta)^{3/2}} \cos\theta\, d\theta = \int_0^{\pi/6} \frac{\sin\theta\cos\theta}{\cos^3\theta} d\theta = \int_0^{\pi/6} \frac{\sin\theta}{\cos^2\theta} d\theta$</p><p>$= \int_0^{\pi/6} \tan\theta\sec\theta\, d\theta = [\sec\theta]_0^{\pi/6} = \sec(\pi/6) - \sec(0) = \frac{2}{\sqrt{3}} - 1 = \frac{2\sqrt{3}}{3} - 1$</p><p>Therefore, $k = 2\sqrt{3} - \pi$ is the answer (b).</p>
Correct Answer: b