# Evaluate the following: $\int\limits_0^1\frac{dx}{e^x+e^{-x}}$

Toolbox:
• $\int \large\frac{dx}{a^2+x^2}=\frac{1}{a} \tan ^{-1} (\frac{x}{a})+c$
• $\tan ^{-1}(1)=\large\frac{\pi}{4}$
• $\tan ^{-1}(\infty)=\large\frac{\pi}{2}$
$\int\limits_0^1\large\frac{dx}{e^x+e^{-x}}$
This can be written as
$\large\int \limits_0^1 \frac{dx}{e^x+\frac{1}{e^x}}$
$e^x+\large\frac{1}{e^x}=\large\frac{e^{2x}+1}{e^x}$
Therefore $I=\int _0^1 \large\frac{e^x}{e^{2x}+1}dx$
Put $e^x=t$
On differentiating with respect to t we get,
$e^xdx=dt$
When we substitute for t the limits also changes
When $x=0, \;then\; t=e^x=e^0=1$
When $x=1, \;then\; t=e^1=e$
Hence on substituting this we get
$I=\int \limits_1^e \large\frac{1}{1+t^2}dx$
On integrating we get
$I=\bigg[\tan ^{-1} t\bigg]_1^e+c$
On applying limits we get
$=\tan ^{-1}e -\tan ^{-1}1$
But $\tan ^{-1} 1$ is $\large\frac{\pi}{4}$
Therefore $I=\tan ^{-1}e-\large\frac{\pi}{4}$

answered Apr 17, 2013 by
edited Apr 18, 2013 by meena.p