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Evaluate:\[\int \limits _0 ^ \pi \Large\frac{e^{\cos x}}{[e^{\cos x}+e^{-\cos x}]}dx\]

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  • (i) $\int \limits_a^b f(x)dx=F(b)-F(a)$
  • (ii) $ \int \limits_a^b f(x)dx=\int \limits_a^b f(a-x) dx$
  • (iii) $ \cos (\pi-x)=\cos x$
Given $\int \limits _0 ^ \pi \Large\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}}dx$------(1)
By applying the property $ \int \limits_a^b f(x)dx=\int \limits_a^b f(a-x) dx$
$I=\int \limits _0 ^ \pi \Large\frac{e^{\cos (\pi-x)}}{e^{\cos (\pi-x)}+e^{-\cos (\pi-x)}}dx$
But $\cos (\pi-x)=-\cos x $
Therefore $I=\int \limits _0 ^ \pi \Large\frac{e^{-\cos x}}{e^{-\cos x}+e^{-(-\cos x)}}dx=\int \limits _0 ^ \pi \Large\frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x}}dx$------(2)
Adding equ(1) and equ (2)
$2I=\int \limits _0 ^ \pi \Large\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}}+\frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x}}dx$
$=\int \limits _0 ^ \pi \Large\frac{e^{\cos x}+e ^{-\cos x}}{e^{\cos x}+e^{-\cos x}}dx$
$=\int \limits _0 ^ \pi dx$
on integrating we get
on applying limits
Therefore $ I=\frac{\pi}{2}$



answered Mar 14, 2013 by meena.p
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