Indefinite Integration
Integration by substitution
Grade 12
Question:
<p>Let \(n \geq 2\) be a natural number and \(0 < \theta < \pi/2\). Then \(\int \dfrac{(\sin^n\theta - \sin\theta)^{\frac{1}{n}} \cos\theta}{\sin^{n+1}\theta}\, d\theta\) is equal to: (where \(C\) is a constant of integration)</p>
<p>\(\dfrac{n}{n^2-1}\left(1 - \dfrac{1}{\sin^{n-1}\theta}\right)^{\frac{n+1}{n}} + C\)</p>
<p>\(\dfrac{n}{n^2+1}\left(1 - \dfrac{1}{\sin^{n-1}\theta}\right)^{\frac{n+1}{n}} + C\)</p>
<p>\(\dfrac{n}{n^2-1}\left(1 + \dfrac{1}{\sin^{n-1}\theta}\right)^{\frac{n+1}{n}} + C\)</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 integrand can be simplified using the substitution u = x^n, which transforms the integral into a standard form involving arcsin or arccos. The constraint 0 < x^n < 1 ensures the integrand is well-defined.
<p><strong>Step 1:</strong> Consider the integral ∫ dx/(x^(n-1)√(1-x^n)). Use substitution u = x^n, so du = nx^(n-1)dx, which gives x^(n-1)dx = du/n.</p><p><strong>Step 2:</strong> The integral becomes ∫ (1/√(1-u)) · (du/n) = (1/n)∫ u^(-1/2)(1-u)^(-1/2) du = (1/n) · 2arcsin(√u) + C.</p><p><strong>Step 3:</strong> Substitute back u = x^n to get (2/n)arcsin(x^(n/2)) + C.</p><p>∴ Answer: A</p>
Correct Answer: A