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Question Number 89946 by jagoll last updated on 20/Apr/20
∫ _(−(π/2)) ^(π/2)  (dx/(1+e^(sin x) ))
$$\int\underset{−\frac{\pi}{\mathrm{2}}} {\overset{\frac{\pi}{\mathrm{2}}} {\:}}\:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{e}^{\mathrm{sin}\:\mathrm{x}} } \\ $$
Answered by john santu last updated on 20/Apr/20
I = ∫_((−π)/2) ^(π/2)  ((dx/(1+e^(sin x) )))   replace x by −x   I = ∫_(π/2) ^(−(π/2))  (((−dx)/(1+e^(−sin x) ))) = ∫_(−(π/2)) ^(π/2) ((dx/(1+e^(−sin x) )))  I+I = ∫_(−(π/2)) ^(π/2)  ((1+e^(sin x) )/(1+e^(sin x) )) dx  2I = ∫_(−(π/2)) ^(π/2)  1dx = π  I = (π/2)
$${I}\:=\:\underset{\frac{−\pi}{\mathrm{2}}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\left(\frac{{dx}}{\mathrm{1}+{e}^{\mathrm{sin}\:{x}} }\right)\: \\ $$$${replace}\:{x}\:{by}\:−{x}\: \\ $$$${I}\:=\:\underset{\frac{\pi}{\mathrm{2}}} {\overset{−\frac{\pi}{\mathrm{2}}} {\int}}\:\left(\frac{−{dx}}{\mathrm{1}+{e}^{−\mathrm{sin}\:{x}} }\right)\:=\:\underset{−\frac{\pi}{\mathrm{2}}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\left(\frac{{dx}}{\mathrm{1}+{e}^{−\mathrm{sin}\:{x}} }\right) \\ $$$${I}+{I}\:=\:\underset{−\frac{\pi}{\mathrm{2}}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{\mathrm{1}+{e}^{\mathrm{sin}\:{x}} }{\mathrm{1}+{e}^{\mathrm{sin}\:{x}} }\:{dx} \\ $$$$\mathrm{2}{I}\:=\:\underset{−\frac{\pi}{\mathrm{2}}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\mathrm{1}{dx}\:=\:\pi \\ $$$${I}\:=\:\frac{\pi}{\mathrm{2}} \\ $$

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