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Q)

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

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A)
Toolbox:
• (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)

$2I=\int \limits _0 ^ \pi \Large\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}}+\frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x}}dx$

$=\int \limits _0 ^ \pi \Large\frac{e^{\cos x}+e ^{-\cos x}}{e^{\cos x}+e^{-\cos x}}dx$

$=\int \limits _0 ^ \pi dx$

on integrating we get

$2I=[x]_0^\pi$

on applying limits

$2I=\pi-0=\pi$

Therefore $I=\frac{\pi}{2}$