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Question #41549

Difficulty: 8/10 | 🟢 Free

The Question

Evaluate the definite integral: I=∫02πx1−cos⁡(2x)2(1+cos⁡2x)dxI = \int_{0}^{2\pi} \frac{x \sqrt{1 - \cos(2x)}}{\sqrt{2} (1 + \cos^2 x)} dx

Options

  • A.

    π2\pi^2

  • B.

    0

  • C.

    π22\frac{\pi^2}{2}

  • D.

    2π22\pi^2

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Symmetry Utilization|Definite Integral Properties (King's Property)|The Quadrant Adjustment Logic

Why This Question Is Tricky

Common Mistakes

This question tests the student's vigilance with 'Quadrant Adjustment Logic' and domain-restricted simplifications.

A common conceptual reflex is to blindly simplify 1−cos⁡(2x)\sqrt{1 - \cos(2x)} as 2sin⁡x\sqrt{2}\sin x instead of the mathematically rigorous 2∣sin⁡x∣\sqrt{2}|\sin x|.

If the absolute value is ignored, applying King's Property (x→2π−xx \to 2\pi - x) introduces a false sign flip because sin⁡(2π−x)=−sin⁡x\sin(2\pi - x) = -\sin x.

This traps the student into the erroneous equation I=−II = -I, resulting in a zero-integral conclusion (00).

Recognizing the absolute value preserves the even symmetry of the trigonometric part across the shifted quadrant, allowing King's Property to correctly eliminate the algebraic xx multiplier and yield the true non-zero result.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterApplication of Integrals - Class 12
ConceptSymmetry Utilization
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree