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Question #41169

Difficulty: 8/10 | 🟢 Free

The Question

Let f(x)=sin⁡−1(2x1+x2)+cos⁡−1(1−x21+x2)f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) + \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right). Evaluate the integral I=∫1/33f(x) dxI = \int_{1/\sqrt{3}}^{\sqrt{3}} f(x) \, dx

Options

  • A.

    7π39+2ln⁡(23)\frac{7\pi\sqrt{3}}{9} + 2\ln\left(\frac{2}{3}\right)

  • B.

    10π39−2ln⁡3\frac{10\pi\sqrt{3}}{9} - 2\ln 3

  • C.

    2π33\frac{2\pi\sqrt{3}}{3}

  • D.

    7π39+2ln⁡(32)\frac{7\pi\sqrt{3}}{9} + 2\ln\left(\frac{3}{2}\right)

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

Variable Transformation Subsitution|Argument and Principal Value|Principal Value Branch Boundaries

Why This Question Is Tricky

Common Mistakes

This question tests the student's vigilance regarding the principal value branch boundaries of inverse trigonometric functions.

A common trap is mechanically applying the identity sin⁡−1(sin⁡α)=α\sin^{-1}(\sin \alpha) = \alpha without checking if α∈[−π/2,π/2]\alpha \in [-\pi/2, \pi/2].

Here, the substitution x=tan⁡θx = \tan \theta maps the integration interval such that the argument 2θ2\theta crosses the boundary π/2\pi/2 at x=1x=1.

Failing to split the integral and apply the correct branch sin⁡−1(sin⁡2θ)=π−2θ\sin^{-1}(\sin 2\theta) = \pi - 2\theta for x>1x > 1 leads to the incorrect trap answer.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterInverse Trigonometric Functions - Class 12
ConceptVariable Transformation Subsitution
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree