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Question #41190

Difficulty: 8/10 | 🟢 Free

The Question

Evaluate the integral $I = \int_{0}^{1} 148 x^3 \sin^{-1}\left(2x\sqrt{1-x^2}\right) \,dx$

Options

A.
37
B.
74
C.
$\frac{185\pi}{8}$
D.
$\frac{37\pi}{4}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

The Sum-Difference Formula Constraints|Integration by Parts (LIATE Rule)|Square Root and Modulus Logic|Right-Hand vs Left-Hand Derivatives

Why This Question Is Tricky

Common Mistakes

This question tests a classic JEE Advanced trap: the blind simplification of inverse trigonometric functions.

Students frequently assume $\sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}x$ over the entire interval $[0, 1]$ , ignoring the modulus and domain constraints of the sum-difference formulas.

If they make this error, they will integrate without splitting the domain, leading to the incorrect trap answer containing $\pi$ (Option C).

The correct approach requires recognizing the domain split at $x = 1/\sqrt{2}$ , where the derivative changes sign.

Applying Integration by Parts forces the student to evaluate left-hand and right-hand integrals separately, causing the transcendental $\pi$ terms to perfectly cancel out and leaving a pure integer.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Inverse Trigonometric Functions - Class 12
Concept	The Sum-Difference Formula Constraints
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free