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Question #41483

Difficulty: 8/10 | 🟢 Free

The Question

Let $D$ be the region in the first quadrant bounded by the $y$ -axis and the curves $y = \tan^{-1} x$ and $y = \cot^{-1} x$ . Evaluate the double integral: $I = \iint_D \frac{x}{\sin^4 y + \cos^4 y} \, dx \, dy$

Options

A.
$1 - \frac{1}{\sqrt{2}} \ln(\sqrt{2}+1)$
B.
$1 + \frac{1}{\sqrt{2}} \ln(\sqrt{2}+1)$
C.
$\frac{\pi}{4} - \frac{1}{\sqrt{2}} \ln(\sqrt{2}+1)$
D.
$1 - \frac{1}{2\sqrt{2}} \ln(\sqrt{2}+1)$

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

Horizontal vs Vertical Strips|Linearity and Partitioning|Domain Constraint Intersections

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to recognize when switching the order of integration (from vertical to horizontal strips) is necessary to bypass a seemingly non-integrable expression.

Using vertical strips requires evaluating an inner integral that yields a complex inverse trigonometric composition, leading to a formidable integration by parts.

The trap lies in persisting with this orientation.

The correct approach requires partitioning the region horizontally due to the domain constraints of the bounding inverse trigonometric functions ( $x = \tan y$ for $y \in [0, \pi/4]$ and $x = \cot y$ for $y \in [\pi/4, \pi/2]$ ).

This transforms the problem into two symmetric, highly tractable single-variable integrals, rewarding geometric visualization and strategic use of Fubini's Theorem.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Integrals - Class 12
Concept	Horizontal vs Vertical Strips
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free