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Question Number 125050 by bramlexs22 last updated on 08/Dec/20

  ∫_0 ^π  (e^(cos x) /(e^(cos x) +e^(−cos x) )) dx =?

$$\:\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{{e}^{\mathrm{cos}\:{x}} }{{e}^{\mathrm{cos}\:{x}} +{e}^{−\mathrm{cos}\:{x}} }\:{dx}\:=?\: \\ $$

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)

$${replace}\:{x}\:{by}\:\pi−{x}\: \\ $$$${I}=\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{{e}^{\mathrm{cos}\:{x}} }{{e}^{\mathrm{cos}\:{x}} +{e}^{−\mathrm{cos}\:{x}} }\:{dx}\: \\ $$$${I}=\underset{\pi} {\overset{\mathrm{0}} {\int}}\:\frac{{e}^{−\mathrm{cos}\:{x}} }{{e}^{−\mathrm{cos}\:{x}} +{e}^{\mathrm{cos}\:{x}} }\:\left(−{dx}\right)\: \\ $$$${I}=\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{{e}^{−\mathrm{cos}\:{x}} }{{e}^{−\mathrm{cos}\:{x}} +{e}^{\mathrm{cos}\:{x}} }\:{dx}\: \\ $$$${we}\:{get}\:\mathrm{2}{I}\:=\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{{e}^{\mathrm{cos}\:{x}} +{e}^{−\mathrm{cos}\:{x}} }{{e}^{\mathrm{cos}\:{x}} +{e}^{−\mathrm{cos}\:{x}} }\:{dx}\: \\ $$$$\mathrm{2}{I}\:=\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\:{dx}\:;\:{I}\:=\:\frac{\pi}{\mathrm{2}} \\ $$

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