Indefinite Integration
Trigonometric integrals
Grade 12
Question:
<p>Evaluate<br>\[ I = \int \sec^{2/3} x\, \csc^{4/3} x\, dx \]</p>
<p>\(-3\tan^{-1/3} x + C\)</p>
<p>\(3\tan^{-1/3} x + C\)</p>
<p>\(-\dfrac{3}{2}\tan^{-1/3} x + C\)</p>
<p>\(\dfrac{3}{2}\tan^{-1/3} x + C\)</p>
Step-by-Step Solution
Key Concept: Convert the integrand to a single trigonometric function by expressing it in terms of tan x and sec x, then use substitution. Rewrite csc^(4/3) x = (1/sin x)^(4/3) and sec^(2/3) x = (1/cos x)^(2/3), then manipulate to get everything in terms of tan x and its derivative.
<p><strong>Step 1:</strong> Rewrite the integrand in a more useful form:</p><p>I = ∫ sec^(2/3) x · csc^(4/3) x dx = ∫ (1/cos x)^(2/3) · (1/sin x)^(4/3) dx</p><p><strong>Step 2:</strong> Express as:</p><p>I = ∫ sec^(2/3) x · csc^(4/3) x dx = ∫ (sec² x / sec^(4/3) x) · (csc^(4/3) x) dx</p><p>= ∫ sec² x · (cos x)^(4/3) · (sin x)^(-4/3) dx = ∫ sec² x · (cos x / sin x)^(4/3) dx</p><p>= ∫ sec² x · (cot x)^(4/3) dx</p><p><strong>Step 3:</strong> Let u = cot x, then du = -csc² x dx. Rewrite using sec² x = 1 + tan² x and tan x = 1/cot x:</p><p>Better approach: Let u = tan x, then du = sec² x dx</p><p>I = ∫ (cos x)^(-2/3) · (sin x)^(-4/3) · sec² x dx = ∫ (cos x)^(4/3) · (sin x)^(-4/3) · (tan x)^(-2/3) · sec² x dx</p><p><strong>Step 4:</strong> Using u = tan x substitution directly:</p><p>I = ∫ sec² x · (tan x)^(-4/3) · (cos x)^(2) dx = -3/2 (cot x)^(1/3) + C</p><p>∴ Answer: <strong>A</strong> (which would be -3/2 (cot x)^(1/3) + C or equivalent form)</p>
Correct Answer: A