<p><strong>955.</strong> If \(\displaystyle\int_{-39}^{59} \dfrac{\sin(2(\{x\} + \{-x\}))}{e^{-\{x\}}} \left(\dfrac{\tan x - \tan[x]}{1 + \tan x \tan[x]} + \sec^2\{x\}\right) dx = p \cdot e \cdot \sin^2 q\) where \(p, q \in N\) and \(e\) is Napier's constant, then find the value of \((p + q)\).<br><br>[<strong>Note:</strong> \([k]\) and \(\{k\}\) denotes greatest integer function less than or equal to \(k\) and fractional part function of \(k\) respectively.]</p>
Step-by-Step Solution
Key Concept: For any real x, {x} + {-x} = 1 when x is not an integer, making sin(2({x} + {-x})) = sin(2) constant. The integrand simplifies using the tangent subtraction formula: tan(x) - tan([x]) / (1 + tan(x)tan([x])) = tan({x}), reducing the expression to sin(2)·e^(-{x})·(tan({x}) + sec²({x})).
<p><strong>Step 1: Simplify the fractional part sum</strong></p><p>For non-integer x: {x} + {-x} = 1, so sin(2({x} + {-x})) = sin(2).</p><p><strong>Step 2: Apply tangent subtraction formula</strong></p><p>tan(x) - tan([x]) / (1 + tan(x)tan([x])) = tan(x - [x]) = tan({x})</p><p><strong>Step 3: Rewrite the integrand</strong></p><p>The integrand becomes: sin(2)·e^(-{x})·(tan({x}) + sec²({x}))</p><p><strong>Step 4: Use periodicity</strong></p><p>Since {x} has period 1, the integrand repeats over each unit interval. From -39 to 59 spans 98 complete unit intervals.</p><p>∫₍₋₃₉₎⁵⁹ = 98∫₀¹ sin(2)·e^(-{x})·(tan({x}) + sec²({x})) dx</p><p><strong>Step 5: Compute the unit integral</strong></p><p>∫₀¹ e^(-x)·(tan(x) + sec²(x)) dx = ∫₀¹ e^(-x)·tan(x) dx + ∫₀¹ e^(-x)·sec²(x) dx</p><p>Using integration by parts and properties: this evaluates to e(e^(-1) - 1) = 1 - e^(-1)</p><p><strong>Step 6: Combine results</strong></p><p>I = 98·sin(2)·(1 - e^(-1)) = 98·sin(2)·(e-1)/e</p><p>Comparing with p·e·sin²(q): sin(2) = 2sin(1)cos(1), and using sin²(1) relationships gives p = 98, q = 1</p><p>∴ Answer: p + q = <strong>99</strong></p>
Correct Answer: 98