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Question #41166

Difficulty: 8/10 | 🟢 Free

The Question

Evaluate the integral: I=∫−2π2π(x51+x2+cos⁡2x+∣(sin⁡−1(sin⁡x))2−π216∣)dxI = \int_{-2\pi}^{2\pi} \left( \frac{x^5}{1 + x^2 + \cos^2 x} + \left| \left(\sin^{-1}(\sin x)\right)^2 - \frac{\pi^2}{16} \right| \right) dx

Options

  • A.

    π34\frac{\pi^3}{4}

  • B.

    π32\frac{\pi^3}{2}

  • C.

    3π38\frac{3\pi^3}{8}

  • D.

    π38\frac{\pi^3}{8}

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

Composition Inverse Identity Traps|Even and Odd Function Symmetry|The Modulus Trap

Why This Question Is Tricky

Common Mistakes

This question constructs a 'Lethal Collision' by nesting an inverse trigonometric identity within a modulus function, combined with an odd-function distractor.

The primary trap is the assumption that sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x over the entire domain [−2π,2π][-2\pi, 2\pi], which would lead to an incorrect integration of ∣x2−π2/16∣|x^2 - \pi^2/16|.

To solve it efficiently, students must sequentially apply even/odd symmetry, identify the π\pi-periodicity of the squared inverse function, and use reflection symmetry about π/2\pi/2 to restrict the domain to [0,π/2][0, \pi/2].

Finally, they must correctly split the integral at the modulus root x=π/4x = \pi/4, testing their precision in handling non-differentiable points within simplified intervals.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterInverse Trigonometric Functions - Class 12
ConceptComposition Inverse Identity Traps
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree