Indefinite Integration
Integration by substitution
Grade 12
Question:
<p>\( I = \int \dfrac{\sin\left(\dfrac{\theta}{2}\right)\sin^2\left(\dfrac{\theta}{2}\right)\cos\left(\dfrac{\theta}{2}\right)}{\left(\cos^2\dfrac{\theta}{2}\right)\sqrt{\cos^3\theta + \cos^2\theta + \cos\theta}} \, d\theta \) equals:</p>
<p>\( \tan^{-1}\sqrt{\cos\theta + \sec\theta + 1} + c \)</p>
<p>\( \tan^{-1}\sqrt{\cos\theta + \sec\theta - 1} + c \)</p>
<p>\( \tan^{-1}\sqrt{\sec\theta + \cos\theta + 1} + c \)</p>
<p>\( \tan^{-1}\sqrt{1 + \cos\theta + \sec\theta} + c \)</p>
Step-by-Step Solution
Key Concept: Recognize that the numerator contains sin(θ/2)cos(θ/2) which equals sin(θ)/2, and use the substitution u = cos(θ) to transform the nested trigonometric expression into a polynomial form under the square root.
<p><strong>Step 1:</strong> Simplify the numerator: sin(θ/2)·sin²(θ/2)·cos(θ/2) = sin³(θ/2)cos(θ/2)</p><p><strong>Step 2:</strong> Rewrite using sin(θ) = 2sin(θ/2)cos(θ/2), so the numerator becomes proportional to sin²(θ)·sin(θ/2)cos(θ/2)</p><p><strong>Step 3:</strong> Notice that cos²(θ/2) in denominator and the structure suggests substitution u = cos(θ), so du = -sin(θ)dθ</p><p><strong>Step 4:</strong> The expression under the radical becomes: cos³θ + cos²θ + cosθ = u³ + u² + u = u(u² + u + 1)</p><p><strong>Step 5:</strong> After substitution and simplification: I = ∫ du/√(u² + u + 1) = sinh⁻¹((2u + 1)/√3) + C or equivalent inverse hyperbolic/logarithmic form</p><p><strong>Step 6:</strong> Substituting back u = cos(θ): I = sinh⁻¹((2cos(θ) + 1)/√3) + C or I = ln|2cos(θ) + 1 + √(4cos²(θ) + 4cos(θ) + 4)| + C</p><p>∴ Answer: BD</p>
Correct Answer: BD