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Question #41460

Difficulty: 8/10 | 🟢 Free

The Question

Evaluate the integral I=∫0π/2cos⁡2xln⁡∣1+2cos⁡2x∣ dxI = \int_0^{\pi/2} \cos 2x \ln|1 + 2\cos 2x| \, dx

Options

  • A.

    π4\frac{\pi}{4}

  • B.

    π2\frac{\pi}{2}

  • C.

    0

  • D.

    −π4-\frac{\pi}{4}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Improper Integrals|Logarithmic Integration via Trig Manipulation|Double and Triple Angle Ratios

Why This Question Is Tricky

Common Mistakes

This problem tests the ability to recognize hidden trigonometric ratios within logarithmic integrals.

Students who attempt direct integration by parts on ln⁡∣1+2cos⁡2x∣\ln|1+2\cos 2x| will encounter a divergent integral ∫11+2cos⁡tdt\int \frac{1}{1+2\cos t} dt that requires Cauchy Principal Value to evaluate, often leading to incorrect conclusions or abandoned attempts.

The intended, elegant path requires identifying 1+2cos⁡2x=sin⁡3xsin⁡x1+2\cos 2x = \frac{\sin 3x}{\sin x} to split the logarithm, followed by exploiting the periodicity and symmetry of ln⁡∣sin⁡t∣\ln|\sin t| to annihilate the 3x3x term via a three-part interval split.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterIntegrals - Class 12
ConceptImproper Integrals
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree