Definite Integration
Properties of definite integrals
Grade 12
Question:
<p>The value of \(\displaystyle\int_0^{\pi/2} \frac{\sin^3 x}{\sin x + \cos x}\,dx\) is:</p>
<p>\(\dfrac{\pi - 2}{8}\)</p>
<p>\(\dfrac{\pi - 1}{4}\)</p>
<p>\(\dfrac{\pi - 2}{4}\)</p>
<p>\(\dfrac{\pi - 1}{2}\)</p>
Step-by-Step Solution
Key Concept: Use the property that ∫₀^(π/2) f(x)dx = ∫₀^(π/2) f(π/2 - x)dx, then add the original and transformed integrals to eliminate the denominator through trigonometric identities.
<p><strong>Step 1:</strong> Let I = ∫₀^(π/2) (sin³x)/(sin x + cos x) dx</p><p><strong>Step 2:</strong> Apply the property: replace x with (π/2 - x):<br/>I = ∫₀^(π/2) (cos³x)/(sin x + cos x) dx</p><p><strong>Step 3:</strong> Add the two equations:<br/>2I = ∫₀^(π/2) (sin³x + cos³x)/(sin x + cos x) dx</p><p><strong>Step 4:</strong> Factor the numerator using a³ + b³ = (a + b)(a² - ab + b²):<br/>sin³x + cos³x = (sin x + cos x)(sin²x - sin x cos x + cos²x)<br/>= (sin x + cos x)(1 - sin x cos x)</p><p><strong>Step 5:</strong> Simplify:<br/>2I = ∫₀^(π/2) (1 - sin x cos x) dx<br/>= ∫₀^(π/2) 1 dx - ∫₀^(π/2) sin x cos x dx<br/>= π/2 - ∫₀^(π/2) (sin 2x)/2 dx<br/>= π/2 - [(-cos 2x)/4]₀^(π/2)<br/>= π/2 - [(-1)/4 - (-1)/4]<br/>= π/2 - 0 = π/2</p><p><strong>Step 6:</strong> Therefore I = π/4</p><p>∴ Answer: <strong>C (π/4)</strong></p>
Correct Answer: C