Question

Evaluate the integral? : `int_(-pi/2)^(pi/2) cosx/(e^x+1) dx`

Answer 1

`int_(-pi/2)^(pi/2) cosx/(e^x+1)*dx=1`

Explanation:

`I=int_(-pi/2)^(pi/2) cosx/(e^x+1)*dx`

After using `x=-y` and `dx=-dy` transforms,

`I=int_(pi/2)^(-pi/2) cos(-y)/(e^(-y)+1)*(-dy)`

`=int_(-pi/2)^(pi/2) cos(-y)/(e^(-y)+1)*dy`

`=int_(-pi/2)^(pi/2) cosy/(e^(-y)+1)*dy`

`=int_(-pi/2)^(pi/2) (e^y*cosy)/(e^y+1)*dy`

`=int_(-pi/2)^(pi/2) (e^x*cosx)/(e^x+1)*dx`

After collecting 2 integrals,

`2I=int_(-pi/2)^(pi/2) ((e^x+1)*cosx)/(e^x+1)*dx`

`=int_(-pi/2)^(pi/2) cosx*dx`

`=int_0^(pi/2) 2cosx*dx`

=`[2sinx]_0^(pi/2)`

=`2`

Hence `I=int_(-pi/2)^(pi/2) cosx/(e^x+1)*dx=1`

Answer 2

`int_(-pi/2)^(pi/2) \ cosx/(e^x+1) \ dx = 1`

Explanation:

Let:

`I = int_(-pi/2)^(pi/2) \ cosx/(e^x+1) \ dx` ..... [A]

We cannot find an elementary solution to the indefinite integral but despite this we can readily evaluate this definite integral If we perform a trivial substitution:

`u= -x => (du)/dx = -1`

And using this substitution we can transform the limits of integration:

When `x = { (-pi/2), (pi/2) :} => u = { (pi/2), (-pi/2) :}`

Then substitution into the integral [A] gives:

`I = int_(pi/2)^(-pi/2) \ cos(-u)/(e^(-u)+1) \ (-1) \ du`
`\ \ = - int_(pi/2)^(-pi/2) \ cos(-u)/(e^(-u)+1) \ du`

Using the properties:

`\ \ \ cos(-A) = cos (A)`
`int_a^b \ f(x) \ dx = -int_b^a \ f(x) \ dx`

We have:

`I = int_(-pi/2)^(pi/2) \ cos(u)/(e^(-u)+1) \ du`
`\ \ = int_(-pi/2)^(pi/2) \ e^u/e^u \ cos(u)/(e^(-u)+1) \ du`
`\ \ = int_(-pi/2)^(pi/2) \ (e^u cosu)/(1+e^u) \ du` ..... [B]

If we add Eq [A] to Eq[B], we get:

`I + I = int_(-pi/2)^(pi/2) \ cosx/(e^x+1) \ dx + int_(-pi/2)^(pi/2) \ (e^u cosu)/(1+e^u) \ du`

And as indefinite integrals are independent of the variable of integration, we have:

`2I = int_(-pi/2)^(pi/2) \ cost/(e^t+1) \ dt + int_(-pi/2)^(pi/2) \ (e^t cost)/(1+e^t) \ dt`

`\ \ \ = int_(-pi/2)^(pi/2) \ cost/(1+e^t) + (e^t cost)/(1+e^t) \ dt`

`\ \ \ = int_(-pi/2)^(pi/2) \ (cost + e^t cost)/(1+e^t) \ dt`

`\ \ \ = int_(-pi/2)^(pi/2) \ ((1 + e^t) cost)/(1+e^t) \ dt`

`:. 2I = int_(-pi/2)^(pi/2) \ cost \ dt`

And this is now a trivial integral:

`\ \ \ \ \ 2I = [sint]_(-pi/2)^(pi/2)`

`:. 2I = sin(pi/2) -sin(-pi/2)`

`:. 2I = 1 - (-1)`

`:. 2I = 2`

Hence:

`I = int_(-pi/2)^(pi/2) \ cosx/(e^x+1) \ dx = 1`