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Question #41517

Difficulty: 8/10 | 🟢 Free

The Question

Let $\vec{a}(\theta) = \cos\theta \hat{i} + \sin\theta \hat{j} + \cos\theta \hat{k}$ and $\vec{b}(\theta) = \sin\theta \hat{i} + \cos\theta \hat{j} + \sin\theta \hat{k}$ . If $I_n = \int_0^{\pi/4} |\vec{a}(\theta) \times \vec{b}(\theta)|^n d\theta$ for $n \ge 2$ , which of the following reduction formulae holds true?

Options

A.
$n I_n = 2(n-1) I_{n-2}$
B.
$n I_n = \sqrt{2}(n-1) I_{n-2}$
C.
$n I_n = (2n-1) I_{n-1}$
D.
$n I_n = 4(n-1) I_{n-2}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Reduction Formulae|Geometric Interpretation of Cross Product|Lagrange Identity

Why This Question Is Tricky

Common Mistakes

The question tests the application of Lagrange's identity to simplify a vector-based integrand for a reduction formula.

The primary trap lies in the transition from $|\vec{a} \times \vec{b}|^2$ to $|\vec{a} \times \vec{b}|^n$ .

Students frequently compute $|\vec{a} \times \vec{b}|^2 = 2\cos^2(2\theta)$ and erroneously substitute this directly as the base of the power $n$ , effectively integrating $(2\cos^2(2\theta))^n$ instead of $(\sqrt{2}\cos(2\theta))^n$ .

This missing scalar square root leads to an incorrect index shift in Walli's formula (using $J_{2n}$ instead of $J_n$ ), resulting in the trap option $n I_n = (2n-1) I_{n-1}$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Integrals - Class 12
Concept	Reduction Formulae
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free