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I-0-2npi-max-sinx-sin-1-sinx-dx-




Question Number 152140 by Ar Brandon last updated on 26/Aug/21
I=∫_0 ^(2nπ) max(sinx, sin^(−1) (sinx))dx
$${I}=\int_{\mathrm{0}} ^{\mathrm{2n}\pi} \mathrm{max}\left(\mathrm{sin}{x},\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}{x}\right)\right){dx} \\ $$

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