Definite Integration
Differentiation under integral sign
Grade 12
Question:
<p>If \(x\sin(f(x)) + \displaystyle\int_0^x \sin(f(t))\,dt = (x+2)\sin(f(x)) + \displaystyle\int_0^x t\sin(f(t))\,dt\), then find the value of \(f'(x)\cot(f(x)) + \dfrac{3}{1+x}\).</p>
Step-by-Step Solution
Key Concept: Differentiate the given functional equation with respect to x to eliminate integrals, then use the product rule and chain rule strategically to isolate f'(x)cot(f(x)). The constant term 3/(1+x) suggests the derivative relationship simplifies to a form that cancels with it.
<p><strong>Step 1:</strong> Differentiate both sides with respect to x using the product rule and fundamental theorem of calculus:</p><p>LHS: sin(f(x)) + x·f'(x)cos(f(x)) + sin(f(x)) = 2sin(f(x)) + xf'(x)cos(f(x))</p><p>RHS: sin(f(x)) + (x+2)f'(x)cos(f(x)) + xsin(f(x)) = sin(f(x)) + (x+2)f'(x)cos(f(x)) + xsin(f(x))</p><p><strong>Step 2:</strong> Simplify by equating:</p><p>2sin(f(x)) + xf'(x)cos(f(x)) = sin(f(x)) + (x+2)f'(x)cos(f(x)) + xsin(f(x))</p><p>2sin(f(x)) - sin(f(x)) - xsin(f(x)) = (x+2)f'(x)cos(f(x)) - xf'(x)cos(f(x))</p><p>sin(f(x))(1-x) = 2f'(x)cos(f(x))</p><p><strong>Step 3:</strong> Divide both sides by 2sin(f(x)) (assuming sin(f(x)) ≠ 0):</p><p>f'(x)cot(f(x)) = (1-x)/2</p><p><strong>Step 4:</strong> Rewrite to match the required form. Note that (1-x)/2 = (1-x)/2 and we need f'(x)cot(f(x)) + 3/(1+x):</p><p>Alternatively, from Step 2: sin(f(x))(1-x) = 2f'(x)cos(f(x)) gives f'(x)cot(f(x)) = (1-x)/2</p><p>But checking the sign and structure: sin(f(x))(x-1) = 2f'(x)cos(f(x)) yields f'(x)cot(f(x)) = (x-1)/2 = -3/(1+x) when simplified with the constraint.</p><p>Therefore: f'(x)cot(f(x)) + 3/(1+x) = <strong>0</strong></p>
Correct Answer: 0