Definite Integral of Product of Geometric and Arithmetic-Geometric Polynomials

Let f_n = integral from 0 to π/2 of (sum from k=1 to n of sin^(k-1) x) (sum from k=1 to n of (2k - 1) sin^(k-1) x) cos x dx, n ∈ ℕ. Then f_21 - f_20 is equal to.

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Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 7 September 2026

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Why is the upper limit for t equal to n when u = 1?

Because t is the sum of n terms, each of the form u^(k - 1/2). When u = 1, each term is 1^(k - 1/2) = 1, so t = 1 + 1 + ... + 1 (n times) = n.

Answer

The value of $f_{21} - f_{20}$ is equal to 41.

Why each option works or fails

Step-by-step solution

Given: $f_{n} = \int_{0}^{\frac{π}{2}} (\sum_{k = 1}^{n} \sin^{k - 1} x) (\sum_{k = 1}^{n} (2 k - 1) \sin^{k - 1} x) \cos x d x$ , where $n \in N$ .
Approach: Perform the substitution $u = \sin x$ , which converts the integral into $\int_{0}^{1} (\sum_{k = 1}^{n} u^{k - 1}) (\sum_{k = 1}^{n} (2 k - 1) u^{k - 1}) d u$ . Multiply the first term by $\sqrt{u}$ and divide the differential by $\sqrt{u}$ , noting that the derivative of $u^{k - 1 / 2}$ produces $(k - 1 / 2) u^{k - 3 / 2} = \frac{1}{2} (2 k - 1) u^{k - 1} / \sqrt{u}$ .
Execute: Let $t = \sum_{k = 1}^{n} u^{k - 1 / 2} = u^{1 / 2} + u^{3 / 2} + \dots + u^{n - 1 / 2}$ . Then $d t = \sum_{k = 1}^{n} (k - \frac{1}{2}) u^{k - 3 / 2} d u = \frac{1}{2 \sqrt{u}} (\sum_{k = 1}^{n} (2 k - 1) u^{k - 1}) d u$ . Thus, $(\sum_{k = 1}^{n} (2 k - 1) u^{k - 1}) \frac{d u}{\sqrt{u}} = 2 d t$ . At $u = 0$ , $t = 0$ . At $u = 1$ , $t = n$ . The integral becomes: $f_{n} = \int_{0}^{n} 2 t d t = {[t^{2}]}_{0}^{n} = n^{2} .$
Verify: For $n = 1$ : $f_{1} = \int_{0}^{π / 2} 1 \cdot 1 \cdot \cos x d x = [\sin x]_{0}^{π / 2} = 1 = 1^{2}$ , which matches $f_{n} = n^{2}$ . Thus, $f_{21} - f_{20} = 21^{2} - 20^{2} = 441 - 400 = 41$ .

Shortcut:

When to use it: Directly evaluate f_n - f_{n-1} without explicitly simplifying the general integral first.

Given: Let $P_{n} (u) = \sum_{k = 1}^{n} u^{k - 1}$ and $Q_{n} (u) = \sum_{k = 1}^{n} (2 k - 1) u^{k - 1}$ , so $f_{n} = \int_{0}^{1} P_{n} (u) Q_{n} (u) d u$ .

Approach: Note that $P_{n} (u) = \frac{1 - u^{n}}{1 - u}$ , and $Q_{n} (u) = 2 u P_{n}^{'} (u) + P_{n} (u)$ . Therefore, $\frac{d}{d u} [u (P_{n} (u))^{2}] = (P_{n} (u))^{2} + 2 u P_{n} (u) P_{n}^{'} (u) = P_{n} (u) Q_{n} (u)$ .

Execute: $f_{n} = \int_{0}^{1} \frac{d}{d u} [u (P_{n} (u))^{2}] d u = {[u (P_{n} (u))^{2}]}_{0}^{1} = 1 \cdot (P_{n} (1))^{2} - 0 = n^{2} .$ Therefore, $f_{21} - f_{20} = 21^{2} - 20^{2} = (21 - 20) (21 + 20) = 41$ .

Verify: $21^{2} - 20^{2} = 441 - 400 = 41$ . Consistent across both methods.

Definite Integral of Product of Geometric and Arithmetic-Geometric Polynomials

What feels right?

The value of $f_{21} - f_{20}$ is equal to 41.

Why each option works or fails

Solution

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