Definite Integration
Evaluation of definite integrals
Grade 12
Question:
<p>The integral \(\displaystyle\int_{\pi/6}^{\pi/4} \frac{dx}{\sin 2x(\tan^5 x + \cot^5 x)}\) equals:</p>
<p>\(\dfrac{1}{20}\tan^{-1}\left(\dfrac{1}{9\sqrt{3}}\right)\)</p>
<p>\(\dfrac{1}{10}\left(\dfrac{\pi}{4}-\tan^{-1}\left(\dfrac{1}{9\sqrt{3}}\right)\right)\)</p>
<p>\(\dfrac{\pi}{40}\)</p>
<p>\(\dfrac{1}{5}\left(\dfrac{\pi}{4}-\tan^{-1}\left(\dfrac{1}{3\sqrt{3}}\right)\right)\)</p>
Step-by-Step Solution
Key Concept: Convert the denominator using sin 2x = 2sin x cos x and express tan⁵x + cot⁵x in terms of sin x and cos x to create a substitution-friendly form. Recognize that the integral simplifies dramatically when you use the substitution t = tan x.
<p><strong>Step 1:</strong> Rewrite the denominator. Note that sin 2x = 2sin x cos x, so:</p><p>∫ dx/(2sin x cos x(tan⁵x + cot⁵x))</p><p><strong>Step 2:</strong> Express tan⁵x + cot⁵x in terms of sin and cos:</p><p>tan⁵x + cot⁵x = sin⁵x/cos⁵x + cos⁵x/sin⁵x = (sin¹⁰x + cos¹⁰x)/(sin⁵x cos⁵x)</p><p><strong>Step 3:</strong> The integral becomes:</p><p>∫ (sin⁵x cos⁵x)/(2sin x cos x(sin¹⁰x + cos¹⁰x)) dx = ∫ (sin⁴x cos⁴x)/(2(sin¹⁰x + cos¹⁰x)) dx</p><p><strong>Step 4:</strong> Divide numerator and denominator by cos¹⁰x:</p><p>∫ (tan⁴x)/(2(tan¹⁰x + 1)) · sec²x dx</p><p><strong>Step 5:</strong> Substitute t = tan x, dt = sec²x dx:</p><p>∫₁/√₃^₁ (t⁴)/(2(t¹⁰ + 1)) dt</p><p><strong>Step 6:</strong> Substitute u = t⁵, du = 5t⁴dt:</p><p>(1/10) ∫₁/(3√₃)^₁ du/(u² + 1) = (1/10)[arctan u]₁/(3√₃)^₁</p><p><strong>Step 7:</strong> Evaluate: (1/10)[π/4 - π/6] = (1/10)(π/12)</p><p>∴ Answer: <strong>π/120</strong></p>
Correct Answer: B