Question

What is `int_(0)^(1) (e^(2x) - e^(-2x)) / (e^(2x) + e^(-2x))dx`?

Answer 1

Let `(e^(2x) + e^(-2x)) = t`

`dt = [2e^(2x) -2e^(2x)]dx`

`dt/2=[e^(2x)-e^(-2x)]dx`

The given integral is, `int_0^1 (e^(2x)-e^(-2x))/(e^(2x)+e^(-2x))dx`

`=>int 1/t dt/2`

`=>1/2int dt/t`

`=>1/2[logt]`

`=>1/2 [log(e^(2x) + e^(-2x))]_0^1`

`=>1/2 [log (e^2+1/(e^2))] - 1/2 log 2`

`1/2 [log (e^2/2+1/(2e^2))]`

Answer 2

`int_0^1 \ (e^(2x)-e^(-2x))/(e^(2x)+e^(-2x)) \ dx =1/2 \ ln ((e^2+1/e^2)/2) ~~ 0.66250`

Explanation:

We seek

`I=int_0^1 \ (e^(2x)-e^(-2x))/(e^(2x)+e^(-2x)) \ dx`

Noting that:

`sinh(x)=(e^(x)-e^(-x))/2` and `cosh(x)=(e^(x)+e^(-x))/2`

Then we can wrote:

`I=int_0^1 \ sinh(2x)/cosh(2x) \ dx`

`\ \ =int_0^1 \ tanh(2x) \ dx`

`\ \ =[1/2ln(cosh2x)]_0^1`

`\ \ =1/2{ln(cosh2)-ln(cosh0))`

`\ \ =1/2{ln((e^2+1/e^2)/2)-ln((1+1)/2)}`

`\ \ =1/2 \ ln ((e^2+1/e^2)/2)`

`\ \ ~~ 0.66250`