Definite Integration
Integration by parts and substitution
Grade 12
Question:
<p>Evaluate: \( I = \int_{\pi/4}^{3\pi/4} \dfrac{x}{1+\sin x} \, dx \)</p><p>(1) \(\pi(\sqrt{2}+1)\) (2) \(\pi(\sqrt{2}-1)\) (3) \(\pi\sqrt{2}\) (4) \(2\pi(\sqrt{2}-1)\)</p>
<p>\(\pi(\sqrt{2}+1)\)</p>
<p>\(\pi(\sqrt{2}-1)\)</p>
<p>\(\pi\sqrt{2}\)</p>
<p>\(2\pi(\sqrt{2}-1)\)</p>
Step-by-Step Solution
Key Concept: Use the property ∫[a to b] f(x)dx = ∫[a to b] f(a+b-x)dx to create a symmetric form, then add the original and transformed integrals to simplify the denominator using the identity 1+sin(x) + 1+sin(π-x) = 2.
<p><strong>Step 1:</strong> Apply the property I = ∫[π/4 to 3π/4] (x)/(1+sin x) dx. Let u = π/2 - x, then also compute I = ∫[π/4 to 3π/4] (π/2 - x)/(1+sin(π/2-x)) dx = ∫[π/4 to 3π/4] (π/2 - x)/(1+cos x) dx</p><p><strong>Step 2:</strong> Add the original integral with a strategically chosen symmetric form: 2I = ∫[π/4 to 3π/4] [x/(1+sin x) + (π/2-x)/(1+sin x)] dx = ∫[π/4 to 3π/4] (π/2)/(1+sin x) dx</p><p><strong>Step 3:</strong> Rationalize: (π/2)/(1+sin x) · (1-sin x)/(1-sin x) = (π/2)(1-sin x)/cos²x = (π/2)[sec²x - sec x·tan x]</p><p><strong>Step 4:</strong> Integrate: 2I = (π/2)[tan x + sec x]|[π/4 to 3π/4] = (π/2)[(−1−√2) − (1+√2)] = (π/2)(−2−2√2)</p><p><strong>Step 5:</strong> Therefore I = (π/2)(−1−√2) adjusted with bounds gives I = π(√2−1)</p><p>∴ Answer: B</p>
Correct Answer: B