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0-pi-e-cos-x-e-cos-x-e-cos-x-dx-




Question Number 125050 by bramlexs22 last updated on 08/Dec/20
  ∫_0 ^π  (e^(cos x) /(e^(cos x) +e^(−cos x) )) dx =?
π0ecosxecosx+ecosxdx=?
Answered by liberty last updated on 08/Dec/20
replace x by π−x   I=∫_0 ^π  (e^(cos x) /(e^(cos x) +e^(−cos x) )) dx   I=∫_π ^0  (e^(−cos x) /(e^(−cos x) +e^(cos x) )) (−dx)   I= ∫_0 ^π  (e^(−cos x) /(e^(−cos x) +e^(cos x) )) dx   we get 2I = ∫_0 ^π  ((e^(cos x) +e^(−cos x) )/(e^(cos x) +e^(−cos x) )) dx   2I = ∫_0 ^π  dx ; I = (π/2)
replacexbyπxI=π0ecosxecosx+ecosxdxI=0πecosxecosx+ecosx(dx)I=π0ecosxecosx+ecosxdxweget2I=π0ecosx+ecosxecosx+ecosxdx2I=π0dx;I=π2

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