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pi-2-pi-2-cosx-




Question Number 123509 by Jamshidbek2311 last updated on 26/Nov/20
∫_(−(π/)2) ^(π/2) ∣cosx∣=?
π2π2cosx∣=?
Answered by bramlexs22 last updated on 26/Nov/20
= ∫_(−π/2) ^(π/2) ∣cos x∣ dx = 2∫_0 ^(π/2) cos x dx  = 2(sin x)]_0 ^(π/2)  = 2
=π/2π/2cosxdx=2π/20cosxdx=2(sinx)]0π/2=2

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