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Question #41578

Difficulty: 8/10 | 🟢 Free

The Question

Evaluate the integral: $I = \int_{0}^{40\pi} \frac{\sqrt{1-\cos(2x)}}{\sqrt{2}\sin x} \left( \sin^{-1}(\sin x) + \cos^{-1}(\cos x) \right) \, dx$

Options

A.
0
B.
$10\pi^2$
C.
$20\pi^2$
D.
$40\pi^2$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Compound Function Boundaries|Composition Continuity|The Zero-Factor Cancellation Trap|Inverse Function Periodicity

Why This Question Is Tricky

Common Mistakes

This question tests the 'Zero-Factor Cancellation Trap' combined with inverse trigonometric periodicity.

Students commonly simplify $\sqrt{1-\cos(2x)}$ to $\sqrt{2}\sin x$ and cancel it with the denominator, missing the absolute value that generates the $\text{sgn}(\sin x)$ function.

If this trap is triggered, the student integrates $\sin^{-1}(\sin x) + \cos^{-1}(\cos x)$ directly, yielding $20\pi^2$ (since the sine term vanishes over full periods and the cosine term accumulates).

Proper domain analysis reveals that $\text{sgn}(\sin x)$ annihilates the cosine term's accumulation while rectifying the sine term, flipping the result to $10\pi^2$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Integrals - Class 12
Concept	Compound Function Boundaries
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free