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Question #41424

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = 16 \sin^4(2x) \left( \sqrt{\cos^4 x - \sin^4 x - 1} \right)^2$ be a real-valued function. Evaluate the definite integral: $I = \int_{0}^{\pi/2} f(x) \, dx$

Options

A.
0
B.
$-3\pi$
C.
$3\pi$
D.
$-\frac{3\pi}{2}$

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

Trigonometric Identical Substitutions|Wallis' Formula|Region Ambiguity|The Empty Set Trap

Why This Question Is Tricky

Common Mistakes

This question tests the 'Empty Set / Measure Zero Trap' combined with trigonometric identities and Wallis' formula.

Students often blindly simplify $(\sqrt{y})^2$ to $y$ (yielding $\cos 2x - 1 = -2\sin^2 x$ ) and apply Wallis' formula to the resulting $-512 \sin^6 x \cos^4 x$ , arriving at the trap answer $-3\pi$ .

However, they fail to check the domain of the real-valued integrand.

The square root restricts the valid domain in the given interval to a single point ( $x=0$ ), making the actual integral $0$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Integrals - Class 12
Concept	Trigonometric Identical Substitutions
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free