<p>The value of \(\int_{-\pi/2}^{\pi/2} (q\sin^3 x + r\sin^5 x + s\sin^5 x)\,dx\) depends on</p>
Step-by-Step Solution
Key Concept: Use the property that odd functions integrate to zero over symmetric intervals [-a, a]. Decompose the integrand into odd and even parts to identify which terms vanish.
<p><strong>Step 1:</strong> Identify the parity of each term in the integrand.</p><p>• sin³x is an odd function (since sin(-x) = -sin(x), so sin³(-x) = -sin³(x))</p><p>• sin⁵x is an odd function (similarly, sin⁵(-x) = -sin⁵(x))</p><p><strong>Step 2:</strong> Apply the odd function property over symmetric interval.</p><p>For any odd function f(x): ∫₋ₐᵃ f(x)dx = 0</p><p>Therefore:</p><p>∫₋π/₂^(π/2) q·sin³x dx = 0</p><p>∫₋π/₂^(π/2) r·sin⁵x dx = 0</p><p>∫₋π/₂^(π/2) s·sin⁵x dx = 0</p><p><strong>Step 3:</strong> Evaluate the total integral.</p><p>∫₋π/₂^(π/2) (q·sin³x + r·sin⁵x + s·sin⁵x)dx = 0 + 0 + 0 = <strong>0</strong></p><p>The integral is <strong>independent of q, r, and s</strong> and depends on <strong>nothing</strong> (equals zero).</p><p>∴ Answer: D</p>
Correct Answer: D