Indefinite Integration
Integration by substitution
Grade 12
Question:
<p>Given \[I = \int \frac{(\sin^n\theta - \sin\theta)^{\frac{1}{n}}\cos\theta}{\sin^{n+1}\theta}\,d\theta\]
Evaluate the integral.</p>
<p>\(\dfrac{n}{n^2-1}\left(1 - \dfrac{1}{\sin^{n-1}\theta}\right)^{\frac{n+1}{n}} + C\)</p>
<p>\(\dfrac{1}{n^2-1}\left(1 - \dfrac{1}{\sin^{n-1}\theta}\right)^{\frac{n+1}{n}} + C\)</p>
<p>\(\dfrac{n}{n+1}\left(1 - \dfrac{1}{\sin^{n-1}\theta}\right)^{\frac{n+1}{n}} + C\)</p>
<p>\(\dfrac{n}{n^2-1}\left(1 + \dfrac{1}{\sin^{n-1}\theta}\right)^{\frac{n+1}{n}} + C\)</p>
Step-by-Step Solution
Key Concept: Recognize that the expression under the fractional power can be rewritten as sin^n(θ)·[(1 - sin^(1-n)(θ))^(1/n)], then use substitution u = sin(θ) to reduce this to a power function integration after algebraic manipulation.
<p><strong>Step 1:</strong> Rewrite the integrand by factoring from the radical.</p><p>Factor sin^n θ from (sin^n θ - sin θ)^(1/n):</p><p>(sin^n θ - sin θ)^(1/n) = [sin^n θ(1 - sin^(1-n) θ)]^(1/n) = sin θ · (1 - sin^(1-n) θ)^(1/n)</p><p><strong>Step 2:</strong> Substitute u = sin θ, so du = cos θ dθ.</p><p>I = ∫ [u · (1 - u^(1-n))^(1/n)] / u^(n+1) du = ∫ (1 - u^(1-n))^(1/n) / u^n du</p><p><strong>Step 3:</strong> Let v = 1 - u^(1-n), then dv = (1-n)u^(-n) du, so u^(-n) du = dv/(1-n).</p><p>I = ∫ v^(1/n) · 1/(1-n) dv = 1/(1-n) · n/(n+1) · v^((n+1)/n) + C</p><p><strong>Step 4:</strong> Substitute back v = 1 - sin^(1-n) θ.</p><p>I = -n/(n+1) · (1 - sin^(1-n) θ)^((n+1)/n) + C</p><p>∴ Answer: <strong>-n/(n+1) · (1 - sin^(1-n) θ)^((n+1)/n) + C</strong></p>
Correct Answer: A