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The area enclosed between the curves y = sec⁻¹(x), y = cot⁻¹(x) and the line x = 1 can be expressed as which of the following integral forms?

1. integral[1 to (1+sqrt(5))/2] (cot⁻¹ x - sec⁻¹ x) dx
2. integral[0 to a] sec(x) dx + integral[a to pi/4] cot(x) dx - pi/4, where sin(a) = cos²(a)
3. integral[0 to a] sec(x) dx + integral[a to pi/4] cot(x) dx - pi/4 + 1, where sin(a) = cos²(a)
4. integral[1 to (1+sqrt(5))/2] (sec⁻¹ x - cot⁻¹ x) dx

Correct answer: integral[1 to (1+sqrt(5))/2] (cot⁻¹ x - sec⁻¹ x) dx

Solution

For x >= 1, cot⁻¹(x) is above sec⁻¹(x) until they meet at x = (1+sqrt5)/2; the area is the integral of (cot⁻¹ x - sec⁻¹ x) over [1, (1+sqrt5)/2].

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