Definite Integration
Integration by Parts — Reduction Formula
Grade 12
Question:
<p>Let \(I_n = \displaystyle\int_0^{\pi/2}\sin^n x\,dx\). Which reduction formula is correct?</p>
<li>\(I_n = \dfrac{n}{n-1}I_{n-2}\)</li>
<li>\(I_n = n\cdot I_{n-1}\)</li>
<li>\(I_n = \dfrac{n-1}{n}I_{n-2}\)</li>
<li>\(I_n = I_{n-2}-\dfrac{1}{n}\)</li>
Step-by-Step Solution
Key Concept: IBP on sinⁿx = sinⁿ⁻^1x \cdot sin x gives Iₙ = ((n-1)/n) \cdot Iₙ₋_2 (Wallis reduction).
<div class='solution'>
<p>IBP: $u=\sin^{n-1}x$, $dv=\sin x\,dx$:</p>
<p>$$I_n = [-\sin^{n-1}x\cos x]_0^{\pi/2} + (n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2 x\,dx$$</p>
<p>$$= 0 + (n-1)\int_0^{\pi/2}\sin^{n-2}x(1-\sin^2 x)\,dx = (n-1)(I_{n-2}-I_n)$$</p>
<p>$$I_n + (n-1)I_n = (n-1)I_{n-2} \Rightarrow nI_n = (n-1)I_{n-2}$$</p>
<p>$$\boxed{I_n = \frac{n-1}{n}I_{n-2}}$$</p>
</div>
Correct Answer: ['C']