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Question #41529

Difficulty: 8/10 | 🟢 Free

The Question

Evaluate the integral: I=∫02π∣2tan⁡(x2)(1−tan⁡2(x2))e−tan⁡2(x/2)∣sec⁡2(x2)dxI = \int_{0}^{2\pi} \left| 2 \tan\left(\frac{x}{2}\right) \left( 1 - \tan^2\left(\frac{x}{2}\right) \right) e^{-\tan^2(x/2)} \right| \sec^2\left(\frac{x}{2}\right) dx

Options

  • A.

    8e\frac{8}{e}

  • B.

    4e\frac{4}{e}

  • C.

    0

  • D.

    16e\frac{16}{e}

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

Integrating f(x) + xf'(x) forms|Tangent Half-Angle Substitution|Modulus Function Breakdown

Why This Question Is Tricky

Common Mistakes

The question tests the application of the Tangent Half-Angle (Weierstrass) substitution combined with the f(t)+tf′(t)f(t) + tf'(t) integration form under a modulus.

Students often blindly apply the substitution t=tan⁡(x/2)t = \tan(x/2) over [0,2π][0, 2\pi] without splitting the integral at the discontinuity x=πx = \pi, leading to a trivial zero bound [t2e−t2]00=0[t^2 e^{-t^2}]_0^0 = 0.

Furthermore, the modulus forces an analysis of the sign of t(1−t2)t(1-t^2), which changes at t=1t=1 and t=−1t=-1 (corresponding to x=π/2x=\pi/2 and 3π/23\pi/2).

Failing to account for these sign inversions and the infinite bounds at x=πx=\pi invalidates the direct use of the reverse product rule, trapping students who ignore piecewise boundary conditions.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterIntegrals - Class 12
ConceptIntegrating f(x) + xf'(x) forms
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree