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Question #41492

Difficulty: 8/10 | 🟢 Free

The Question

Evaluate the definite integral: $I = \int_{0}^{2\pi} \frac{6\cos x - 6\sin x + 5}{3 + 2\sin x + 2\cos x} \, dx$

Options

A.
$10\pi$
B.
$0$
C.
$5\pi$
D.
$12\pi$

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

Substitution (Change of Variable)|Logarithmic Integration via Trig Manipulation|Extraneous Solution Rejection

Why This Question Is Tricky

Common Mistakes

The question tests the student's ability to recognize derivative patterns in rational trigonometric integrals to avoid brutal substitutions.

A common trap is applying the Weierstrass substitution $t = \tan(x/2)$ to the entire integral, which converts it into a complex rational algebraic function requiring logarithmic integration via partial fractions.

Because $t = \tan(x/2)$ is discontinuous at $x = \pi$ , integrating blindly from the lower limit $t=0$ to the upper limit $t=0$ yields an extraneous result of $0$ .

This 'Lethal Collision' with the fundamental theorem of calculus punishes algorithmic substitution without domain analysis, a hallmark of JEE Advanced Extraneous Solution Rejection traps.

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