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Question #41493

Difficulty: 8/10 | 🟢 Free

The Question

Let $I = \int_{0}^{\pi/2} 16x (\sin 4x - \sin 2x) \text{sgn}(\cos 3x \sin x) \, dx$ , where $\text{sgn}(t)$ denotes the signum function. The value of $I$ is:

Options

A.
$8\pi - 3\sqrt{3}$
B.
$4\pi - 3\sqrt{3}$
C.
$8\pi + 3\sqrt{3}$
D.
$-6\pi$

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

Integration by Parts (LIATE Rule)|Compound Angle Expansion|Double and Triple Angle Ratios|Signum Function Discontinuity

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to recognize the implicit absolute value function formed by the product of a function and its signum, $f(x)\text{sgn}(f(x)) = |f(x)|$ .

The primary trap lies in the compound angle factorization $\sin 4x - \sin 2x = 2\cos 3x \sin x$ , which is required to identify the hidden root at $x = \pi/6$ within the integration limits.

Students who fail to identify this discontinuity will integrate directly from $0$ to $\pi/2$ , falling for the lethal distractor $-6\pi$ .

Furthermore, the problem demands high algebraic stamina to correctly re-evaluate the Integration by Parts (LIATE) boundaries across the piecewise split without dropping the alternating signs.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Integrals - Class 12
Concept	Integration by Parts (LIATE Rule)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free