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Question #40454

Difficulty: 8/10 | 🟢 Free

The Question

Let $D$ be the domain of the real-valued function $y(x)$ defined implicitly by the equation $y^3 + y = \sqrt{\cos(2x)}$ for $x \in [0, \pi]$ . Evaluate the integral $I = \int_D \frac{x \sin x}{1 + \cos^2 x} \, dx$

Options

A.
$\pi \left( \frac{\pi}{4} - \tan^{-1}\left(\frac{1}{\sqrt{2}}\right) \right)$
B.
$\frac{\pi^2}{4}$
C.
$\pi \tan^{-1}\left(\frac{1}{\sqrt{2}}\right)$
D.
$\frac{\pi^2}{8}$

Check Your Answer

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

Symmetry and Periodicity|Definite Integral Properties (King's Property)|Implicit Differentiation Implicit Domain

Why This Question Is Tricky

Common Mistakes

This question tests the interaction between implicit domain constraints and definite integral properties (King's Property).

Students often mechanically apply King's Property $\int_0^\pi x f(\sin x, \cos x) dx = \frac{\pi}{2} \int_0^\pi f(\sin x, \cos x) dx$ without verifying the domain of integration, which leads directly to the trap option B ( $\pi^2/4$ ).

The implicit relation $y^3 + y = \sqrt{\cos(2x)}$ truncates the valid interval $[0, \pi]$ into a disjoint domain $D = [0, \pi/4] \cup [3\pi/4, \pi]$ .

Applying the symmetry property over this restricted, disjoint domain requires careful substitution ( $x = \pi - t$ ) on the second interval to cancel the $x$ term, rewarding students who analyze domain validity before applying integral shortcuts.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Trigonometric Functions - Class 11
Concept	Symmetry and Periodicity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free