Definite Integration
Estimation/Inequality of Definite Integrals
Grade 12
Question:
<p>For \(0 < x < \dfrac{1}{2}\), \(0 < x^{2n} \leq x^2\), which of the following is correct?</p><p>\[\frac{1}{2} < \int_0^{1/2} \frac{dx}{\sqrt{1-x^{2n}}} \leq \frac{\pi}{6} < 1\]</p>
<p>\(\displaystyle\int_0^{1/2} \frac{dx}{\sqrt{1-x^{2n}}} < \frac{\pi}{6}\)</p>
<p>\(\displaystyle\int_0^{1/2} \frac{dx}{\sqrt{1-x^{2n}}} > \frac{1}{2}\)</p>
<p>\(\dfrac{1}{2} < \displaystyle\int_0^{1/2} \frac{dx}{\sqrt{1-x^{2n}}} \leq \dfrac{\pi}{6}\)</p>
<p>\(\displaystyle\int_0^{1/2} \frac{dx}{\sqrt{1-x^{2n}}} = \dfrac{\pi}{6}\)</p>
Step-by-Step Solution
Key Concept: Use the property that for 0 < a < 1, we have ∫₀^π x·f(sin x)dx = (π/2)∫₀^π f(sin x)dx, combined with the substitution technique to evaluate definite integrals involving trigonometric functions and their reciprocals.
<p><strong>Step 1:</strong> Use King's Property: For the integral ∫₀^π x/(a + b sin x)dx where 0 < a < 1, apply the transformation x → π - x.</p><p><strong>Step 2:</strong> Let I = ∫₀^π x/(a + b sin x)dx. Then I = ∫₀^π (π-x)/(a + b sin(π-x))dx = ∫₀^π (π-x)/(a + b sin x)dx (since sin(π-x) = sin x).</p><p><strong>Step 3:</strong> Adding both expressions: 2I = π∫₀^π 1/(a + b sin x)dx.</p><p><strong>Step 4:</strong> For ∫₀^π 1/(a + b sin x)dx with a > |b|, use Weierstrass substitution t = tan(x/2): This gives π/√(a² - b²).</p><p><strong>Step 5:</strong> Therefore I = (π²)/(2√(a² - b²)).</p><p><strong>Step 6:</strong> With a = 1 and b = 1/2: I = (π²)/(2√(1 - 1/4)) = (π²)/(2 · √(3/4)) = (π²)/(√3).</p><p>∴ Answer: A, B</p>
Correct Answer: A, B