Question

How do you find the integral of `int dx/(e^x +e^-x)` from negative infinity to infinity?

Answer 1

`pi/2`

Explanation:

`f(x)=1/(e^x+e^(-x))`
`f(-x)=1/(e^(-x)+e^(x))`
=>f(x)=f(-x)
=> the given function is even function
son`int_-oo^oo1/(e^x+e^(-x))dx=2int_0^oo1/(e^x+e^(-x))dx=2int_0^ooe^x/(e^(2x)+1)dx=2int_0^ooe^x/((e^x)^2+1)dx`
if `y =tan^-1x`
`dy=1/(x^2+1)dx`
=>`y=int1/(x^2+1)dx=tan^-1x`
if you substitute `x=e^x` then `dtan^-1(e^x)=1/((e^x)^2+1)e^xdx`
=>`2int_0^ooe^x/((e^x)^2+1)dx=2int_0^ood(tan^-1e^(x))dx=2[tan^-1e^x]_0^oo=2[tan^-1oo-tan^-1(1)]=2[pi/2-pi/4]=pi/2`