Evaluate the integral? : int_(-pi/2)^(pi/2) cosx/(e^x+1) dx ∫π2−π2cosxex+1dx

2 Answers
Cem Sentin
Nov 15, 2017

int_(-pi/2)^(pi/2) cosx/(e^x+1)*dx=1∫π2−π2cosxex+1⋅dx=1

Explanation:

I=int_(-pi/2)^(pi/2) cosx/(e^x+1)*dxI=∫π2−π2cosxex+1⋅dx

After using x=-yx=−y and dx=-dydx=−dy transforms,

I=int_(pi/2)^(-pi/2) cos(-y)/(e^(-y)+1)*(-dy)I=∫−π2π2cos(−y)e−y+1⋅(−dy)

=int_(-pi/2)^(pi/2) cos(-y)/(e^(-y)+1)*dy=∫π2−π2cos(−y)e−y+1⋅dy

=int_(-pi/2)^(pi/2) cosy/(e^(-y)+1)*dy=∫π2−π2cosye−y+1⋅dy

=int_(-pi/2)^(pi/2) (e^y*cosy)/(e^y+1)*dy=∫π2−π2ey⋅cosyey+1⋅dy

=int_(-pi/2)^(pi/2) (e^x*cosx)/(e^x+1)*dx=∫π2−π2ex⋅cosxex+1⋅dx

After collecting 2 integrals,

2I=int_(-pi/2)^(pi/2) ((e^x+1)*cosx)/(e^x+1)*dx2I=∫π2−π2(ex+1)⋅cosxex+1⋅dx

=int_(-pi/2)^(pi/2) cosx*dx=∫π2−π2cosx⋅dx

=int_0^(pi/2) 2cosx*dx=∫π202cosx⋅dx

=[2sinx]_0^(pi/2)[2sinx]π20

=22

Hence I=int_(-pi/2)^(pi/2) cosx/(e^x+1)*dx=1I=∫π2−π2cosxex+1⋅dx=1

Steve M
Nov 16, 2017

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

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