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Category: Algebra

0-2pi-sgn-cosx-dx-

Question Number 12401 by sin (x) last updated on 21/Apr/17 02πsgn(cosx)dx=? Commented by FilupS last updated on 21/Apr/17 nZ$$\mathrm{for}\:\mathrm{cos}\left({x}\right)>\mathrm{0},\:\:\:\frac{\pi}{\mathrm{2}}\left(\mathrm{4}{n}−\mathrm{1}\right)<{x}<\frac{\pi}{\mathrm{2}}\left(\mathrm{4}{n}+\mathrm{1}\right)…