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Question Number 156240 by SANOGO last updated on 09/Oct/21
∫^(π/2) _((−π)/2) ∣sin(x)∣ dx  ∫^π _0 ∣cos(x)∣dx
π2π2sin(x)dx0πcos(x)dx
Commented by SANOGO last updated on 09/Oct/21
merci bien le dur
mercibienledur
Commented by tabata last updated on 09/Oct/21
you are welcome
youarewelcome
Commented by tabata last updated on 09/Oct/21
∫_(−(𝛑/2)) ^( (𝛑/2)) ∣ sin(x) ∣ dx    ∣  sin(x) ∣ is always even function and symetric   a bout zero     ∴ ∫_(− (𝛑/2)) ^(  (𝛑/2))  ∣ sin (x) ∣ dx = 2 ∫_0 ^(  (𝛑/2)) ∣ sin(x) ∣ dx     I = − 2 ( cos (x) )_0 ^(𝛑/2) = − 2 ( − 1 ) = 2    when: ∫_0 ^( 𝛑)  ∣ cos (x) ∣ dx also = 2    ⟨ M . T  ⟩
π2π2sin(x)dxsin(x)isalwaysevenfunctionandsymetricaboutzeroπ2π2sin(x)dx=20π2sin(x)dxI=2(cos(x))0π2=2(1)=2when:0πcos(x)dxalso=2M.T

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