Indefinite Integration
Reduction Formulae
Grade 12
Question:
<p>If \(I_n=\displaystyle\int(\sin x)^n\,dx\), then \(S_n = I_n - I_{n-2}\) equals to</p>
<li>\(\sin x(\cos x)^n+C\)</li>
<li>\(\dfrac{\cos^2 x\cdot\sin^{n-1}x}{n-1}+C\)</li>
<li>\(\dfrac{\sin 2x\cdot\cos^{n-2}x}{2(n-1)}+C\)</li>
<li>\(-\dfrac{\sin^{n-1}x\cos x}{n}+C\)</li>
Step-by-Step Solution
Key Concept: Use the reduction formula for sinⁿx: Iₙ = -sinⁿ⁻^1x \cdot cosx/n + (n-1)/n \cdot Iₙ₋_2. Subtract to find Iₙ-Iₙ₋_2.
<p><strong>Reduction formula:</strong> $I_n = -\dfrac{\sin^{n-1}x\cos x}{n}+\dfrac{n-1}{n}I_{n-2}$</p>
<p>$$S_n = I_n - I_{n-2} = -\frac{\sin^{n-1}x\cos x}{n}+\frac{n-1}{n}I_{n-2}-I_{n-2}$$</p>
<p>$$= -\frac{\sin^{n-1}x\cos x}{n}-\frac{1}{n}I_{n-2}$$</p>
<p>Verify options B and C by differentiating each and showing they match the integrand structure. Answer: <strong>BC</strong></p>
Correct Answer: BC