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Question #41431

Difficulty: 8/10 | 🟢 Free

The Question

Let $I = \int_{0}^{\pi} x \sin^2 x \left[ \frac{2x}{\pi} \right] dx$ , where $[t]$ denotes the greatest integer less than or equal to $t$ . The value of $I$ is:

Options

A.
$\frac{3\pi^2}{16} - \frac{1}{4}$
B.
$\frac{\pi^2}{8}$
C.
$\frac{\pi^2}{4}$
D.
$\frac{3\pi^2}{16} + \frac{1}{4}$

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

Definite Integral Properties (King's Property)|Integration by Parts (LIATE Rule)|Step Function Discontinuities

Why This Question Is Tricky

Common Mistakes

This question tests the student's vigilance against blindly applying the King's Property shortcut $I = \int_0^\pi x f(\sin x) dx = \frac{\pi}{2} \int_0^\pi f(\sin x) dx$ .

The trap is set by introducing the greatest integer function $[\frac{2x}{\pi}]$ , which breaks the symmetry condition $g(\pi - x) = g(x)$ required for the shortcut.

Students who ignore this will incorrectly apply the shortcut to the entire integrand, leading to the trap answer $\frac{\pi^2}{8}$ .

The correct approach requires recognizing the point discontinuities, manually splitting the interval, and executing integration by parts accurately on the resulting sub-interval.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Integrals - Class 12
Concept	Definite Integral Properties (King's Property)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free