Indefinite Integration
Reduction Formulae
Grade 12
Question:
<p>If \(I_n = \int \tan^n x\, dx\), then \(I_4 + I_6 = a\tan^5 x + bx^5 + C\), where \(C\) is a constant of integration, then the ordered pair \((a, b)\) is equal to:</p>
<p>\(\left(\dfrac{1}{5}, 0\right)\)</p>
<p>\(\left(-\dfrac{1}{5}, 0\right)\)</p>
<p>\(\left(-\dfrac{1}{5}, 1\right)\)</p>
<p>\(\left(\dfrac{1}{5}, -1\right)\)</p>
Step-by-Step Solution
Key Concept: Use the reduction formula for tan^n x: tan^n x = tan^(n-2) x · sec^2 x - tan^(n-2) x, which allows I_n + I_(n-2) to telescope. The key is recognizing that I_4 + I_6 will involve grouping terms strategically using sec^2 x = 1 + tan^2 x.
<p><strong>Step 1:</strong> Use the reduction formula. For I_n = ∫tan^n x dx, write tan^n x = tan^(n-2) x(sec² x - 1) = tan^(n-2) x · sec² x - tan^(n-2) x</p><p><strong>Step 2:</strong> Therefore I_n = ∫tan^(n-2) x · sec² x dx - I_(n-2). Using substitution u = tan x, the first integral equals tan^(n-1) x/(n-1), so: I_n = tan^(n-1) x/(n-1) - I_(n-2)</p><p><strong>Step 3:</strong> Apply to I_4: I_4 = tan³ x/3 - I_2, where I_2 = tan x - x + C</p><p><strong>Step 4:</strong> Apply to I_6: I_6 = tan⁵ x/5 - I_4</p><p><strong>Step 5:</strong> Add: I_4 + I_6 = (tan³ x/3 - I_2) + (tan⁵ x/5 - I_4). Rearranging: I_4 + I_6 + I_4 = tan⁵ x/5 + tan³ x/3 - I_2, so 2I_4 + I_6 = tan⁵ x/5 + tan³ x/3 - (tan x - x). From I_6 = tan⁵ x/5 - I_4, we get: I_4 + I_6 = I_4 + tan⁵ x/5 - I_4 = tan⁵ x/5 + C</p><p><strong>Step 6:</strong> Comparing with I_4 + I_6 = (1/5)tan⁵ x + 0·x + C, we have a = 1/5 and b = 0.</p><p>∴ Answer: (a, b) = (1/5, 0)</p>
Correct Answer: A