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Question #40431

Difficulty: 8/10 | 🟢 Free

The Question

Evaluate the integral: $I = \int_{0}^{2\pi} \frac{x \sqrt{1 - \cos 2x}}{1 + \cos^2 x} \, dx$

Options

A.
$\sqrt{2} \pi^2$
B.
$-\frac{\pi^2}{\sqrt{2}}$
C.
$\pi^2$
D.
$\frac{\pi^2}{\sqrt{2}}$

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

ASTC Quadrant Sign Convention|Substitution (Change of Variable)|Integration Constant Trap

Why This Question Is Tricky

Common Mistakes

This question tests the critical ASTC quadrant sign convention and the absolute value trap in definite integration.

Students frequently make the error of writing $\sqrt{1 - \cos 2x} = \sqrt{2}\sin x$ , ignoring that $\sin x$ is negative in the interval $[\pi, 2\pi]$ .

This omission leads to evaluating the integral without the absolute value, which incorrectly yields $-\frac{\pi^2}{\sqrt{2}}$ .

Recognizing the absolute value $|\sin x|$ is mandatory, forcing the use of definite integral properties (King's Rule) and periodicity to correctly evaluate the area across multiple quadrants.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Trigonometric Functions - Class 11
Concept	ASTC Quadrant Sign Convention
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free