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Question #41465

Difficulty: 8/10 | 🟢 Free

The Question

Evaluate the definite integral: I=∫0πsin⁡2xln⁡(cos⁡2x)cos⁡3x dxI = \int_{0}^{\pi} \frac{\sin 2x \ln(\cos^2 x)}{\cos^3 x} \, dx

Options

  • A.

    -8

  • B.

    -4

  • C.

    0

  • D.

    \text{Does not exist}

Check Your Answer

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

Logarithmic Integration via Trig Manipulation|Integration by Parts (LIATE Rule)|Logarithmic Differentiation Domain

Why This Question Is Tricky

Common Mistakes

This question tests the student's vigilance regarding the domain of integration and hidden singularities.

A common trap in JEE Advanced is applying the Newton-Leibniz formula (or Integration by Parts) blindly across an interval where the integrand or its antiderivative is undefined.

Here, the integrand has a vertical asymptote at x=π/2x = \pi/2 because ln⁡(cos⁡2x)\ln(\cos^2 x) is undefined and cos⁡3x=0\cos^3 x = 0.

If a student ignores this domain restriction and blindly applies IBP, they will find an antiderivative F(x)=2sec⁡xln⁡(cos⁡2x)+4sec⁡xF(x) = 2\sec x \ln(\cos^2 x) + 4\sec x and incorrectly evaluate F(π)−F(0)=−8F(\pi) - F(0) = -8.

Recognizing the singularity and testing for convergence reveals that the improper integral diverges to −∞-\infty.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterIntegrals - Class 12
ConceptLogarithmic Integration via Trig Manipulation
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree