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

#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#

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 #