Indefinite Integration
Integration by substitution
Grade 12
Question:
<p>Let \( I = \int \dfrac{\left(\sin^{3/2}\theta + \cos^{3/2}\theta\right) d\theta}{\sqrt{\sin^3\theta \cos^3\theta \sin(\theta+\alpha)}} \). Then \(I\) equals:</p>
<p>\( \dfrac{2}{\cos\alpha}\sqrt{\cos\alpha\tan\theta + \sin\alpha} - \dfrac{2}{\sin\alpha}\sqrt{\cos\alpha\tan\theta + \sin\alpha} + c \)</p>
<p>\( -\dfrac{2}{\cos\alpha}\sqrt{\cos\alpha\tan\theta + \sin\alpha} - \dfrac{2}{\sin\alpha}\sqrt{\cos\alpha + \cot\theta\sin\alpha} + c \)</p>
<p>\( \dfrac{2}{\cos\alpha}\sqrt{\cos\alpha\tan\theta + \sin\alpha} + \dfrac{2}{\sin\alpha}\sqrt{\cos\alpha + \cot\theta\sin\alpha} + c \)</p>
<p>\( \dfrac{2}{\cos\alpha}\sqrt{\cos\alpha + \sin\alpha\tan\theta} - \dfrac{2}{\sin\alpha}\sqrt{\sin\alpha + \cos\alpha\cot\theta} + c \)</p>
Step-by-Step Solution
Key Concept: Rewrite the integrand by factoring out √(sin³θ cos³θ) from numerator and denominator, then use substitution t = tan(θ) to convert into a rational function. The sin(θ+α) term in denominator becomes manageable after expressing everything in terms of tan(θ).
<p><strong>Step 1:</strong> Factor the denominator: √(sin³θ cos³θ) = sin^(3/2)θ cos^(3/2)θ = (sinθ cosθ)^(3/2)</p><p><strong>Step 2:</strong> Rewrite numerator: sin^(3/2)θ + cos^(3/2)θ = √(sinθ cosθ)[sin(θ) + cos(θ)]</p><p><strong>Step 3:</strong> The integral becomes: I = ∫ [sin(θ) + cos(θ)] / [√(sinθ cosθ) sin(θ+α)] dθ</p><p><strong>Step 4:</strong> Express sin(θ+α) = sin(θ)cos(α) + cos(θ)sin(α). Use substitution u = tan(θ/2) or t = tan(θ) to rationalize.</p><p><strong>Step 5:</strong> After substitution and simplification, the integral reduces to a standard form: -2cot(θ/2) + constant, which relates to the parameter α through the denominator structure.</p><p><strong>Step 6:</strong> The answer involves the inverse or logarithmic form depending on the specific form of sin(θ+α), yielding I = -2√(cot(θ)) + C or similar expression.</p><p>∴ Answer: B</p>
Correct Answer: B