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Question #41446

Difficulty: 8/10 | 🟢 Free

The Question

Let I=∑k=06(6k)(−1)k∫0π/8sin⁡2kxcos⁡12−2kx dxI = \sum_{k=0}^{6} \binom{6}{k} (-1)^k \int_{0}^{\pi/8} \sin^{2k} x \cos^{12-2k} x \, dx The value of II is:

Options

  • A.

    1196+5π128\frac{11}{96} + \frac{5\pi}{128}

  • B.

    5π64\frac{5\pi}{64}

  • C.

    1148+5π64\frac{11}{48} + \frac{5\pi}{64}

  • D.

    5π128\frac{5\pi}{128}

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

Wallis' Formula|General Term Formulation|Arbitrary Constants vs Parameters

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly apply reduction formulas while strictly observing boundary conditions, specifically targeting the 'Arbitrary Constants vs Parameters' trap.

Students often memorize Wallis' formula as a parameter-driven sequence and blindly apply it to integrals of the form ∫cos⁡nx dx\int \cos^n x \, dx.

Here, the binomial expansion condenses the integrand to cos⁡6(2x)\cos^6(2x), and the substitution 2x=t2x=t shifts the upper limit to π/4\pi/4, not π/2\pi/2.

The trap lies in assuming the boundary term 1n[cos⁡n−1tsin⁡t]\frac{1}{n}[\cos^{n-1}t \sin t] vanishes (as it does for π/2\pi/2), leading students to treat the recursive parameter as an independent constant and erroneously select Option D (or B, if they miss the dt/2dt/2 differential factor).

The correct execution requires retaining the non-zero boundary constants at each recursive step.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterIntegrals - Class 12
ConceptWallis' Formula
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree