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pi-pi-sin-x-cos-x-dx-




Question Number 95227 by bobhans last updated on 24/May/20
∫_(−π) ^π  ∣sin x + cos x ∣ dx =?
ππsinx+cosxdx=?
Answered by john santu last updated on 24/May/20
Commented by john santu last updated on 24/May/20
∫_(−π) ^(−(π/4)) (sin x+cos x) dx + ∫_(−(π/4)) ^((3π)/4) (sin x+cos x) dx+∫_((3π)/4) ^π (sin x+cos x) dx =  2∫_(−(π/4)) ^((3π)/4) (sin x+cos x) dx =  2 [ −cos x+sin x ]_(−(π/4)) ^((3π)/4) =  2{(√2)−(−(√2))} = 4(√2)
π4π(sinx+cosx)dx+3π4π4(sinx+cosx)dx+π3π4(sinx+cosx)dx=23π4π4(sinx+cosx)dx=2[cosx+sinx]π43π4=2{2(2)}=42

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