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Integration Questions

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

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

Answered by liberty last updated on 08/Dec/20

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