Mathematics

Mathematics

Problem

The value of the definite integral ∫02πnmax⁡(sin⁡x,sin⁡−1(sin⁡x))dx\int_{0}^{2\pi n} \max(\sin x, \sin^{-1}(\sin x)) dx equals to (where, n∈In \in I ) a. n(π2−4)2\frac{n(\pi^2 - 4)}{2} b. n(π2−4)4\frac{n(\pi^2 - 4)}{4} c. n(π2−8)4\frac{n(\pi^2 - 8)}{4} d. n(π2−2)4\frac{n(\pi^2 - 2)}{4}
Intuitive explanation

The key is recognizing that sin⁡−1(sin⁡x)\sin^{-1}(\sin x) doesn't simply equal xx everywhere—it's restricted to [−π/2,π/2][-\pi/2, \pi/2] by the inverse function's range. So you need to figure out which function is larger in different regions of [0,2πn][0, 2\pi n], then exploit the periodicity of sin⁡x\sin x and the repeating pattern of sin⁡−1(sin⁡x)\sin^{-1}(\sin x) to avoid computing the integral directly. Breaking the integral into one period and using symmetry will collapse the calculation significantly.

Intuitive explanation

The key is recognizing that sin⁡−1(sin⁡x)\sin^{-1}(\sin x) doesn't simply equal xx everywhere—it's restricted to [−π/2,π/2][-\pi/2, \pi/2] by the inverse function's range. So you need to figure out which function is larger in different regions of [0,2πn][0, 2\pi n], then exploit the periodicity of sin⁡x\sin x and the repeating pattern of sin⁡−1(sin⁡x)\sin^{-1}(\sin x) to avoid computing the integral directly. Breaking the integral into one period and using symmetry will collapse the calculation significantly.

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