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Question #41215

Difficulty: 8/10 | 🟢 Free

The Question

Evaluate the integral: I=∫02πx(cos⁡−1(cos⁡x)+∣sin⁡−1(sin⁡x)∣)sin⁡4x+cos⁡4x dxI = \int_{0}^{2\pi} \frac{x \left( \cos^{-1}(\cos x) + \left| \sin^{-1}(\sin x) \right| \right)}{\sin^4 x + \cos^4 x} \, dx

Options

  • A.

    3π32\frac{3\pi^3}{\sqrt{2}}

  • B.

    22π32\sqrt{2}\pi^3

  • C.

    3π322\frac{3\pi^3}{2\sqrt{2}}

  • D.

    π32\frac{\pi^3}{\sqrt{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

The Quadrant Adjustment Logic|Definite Integral Properties (King's Property)|Modulus Function Symmetry|Region Ambiguity

Why This Question Is Tricky

Common Mistakes

This question tests the advanced application of King's Property in conjunction with the piecewise definitions of inverse trigonometric functions.

The primary trap is 'Region Ambiguity': students often incorrectly assume sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x over the entire interval [0,π][0, \pi], which leads to an incorrect evaluation of the second split integral.

By forcing the evaluation of sin⁡−1(sin⁡x)\sin^{-1}(\sin x) on [0,π][0, \pi], the student must recognize its symmetry around π/2\pi/2 to correctly fold the integral.

Furthermore, the modulus function acts as a psychological deterrent, though it simplifies elegantly once the domain is restricted to [0,π][0, \pi] where sin⁡−1(sin⁡x)\sin^{-1}(\sin x) is strictly non-negative.

Failing to apply the 2π2\pi multiplier correctly during the folding process is another lethal trap.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterInverse Trigonometric Functions - Class 12
ConceptThe Quadrant Adjustment Logic
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree