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Question #41491

Difficulty: 8/10 | 🟢 Free

The Question

Evaluate the integral: $I = \int_{0}^{\pi} \frac{(\pi - 2x) \sin(2x)}{\sum_{k=0}^{3} \binom{3}{k} \tan^{2k} x} \, dx$

Options

A.
$\frac{93\pi}{256}$
B.
0
C.
$\frac{35\pi}{256}$
D.
$\frac{\pi}{2}$

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

Substitution (Change of Variable)|Integration by Parts (LIATE Rule)|Properties of Binomial Coefficients|Domain and Range Nuance

Why This Question Is Tricky

Common Mistakes

This question tests the 'Domain and Range Nuance' during integration by substitution.

The denominator simplifies to $\sec^6 x$ , which strongly tempts students to use the substitution $u = \tan x$ .

However, $\tan x$ is discontinuous at $x = \pi/2$ , which lies strictly inside the integration interval $[0, \pi]$ .

A blind substitution maps the limits from $0$ to $0$ , incorrectly yielding an answer of $0$ (Option B).

To correctly evaluate the integral, students must recognize the singularity, simplify the integrand to a continuous function ( $2(\pi - 2x) \sin x \cos^7 x$ ), and use the shift property ( $x = \pi/2 - t$ ) followed by Integration by Parts (LIATE rule) and Wallis' formula.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Integrals - Class 12
Concept	Substitution (Change of Variable)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free