Evaluate the integral? : # int_(-pi/2)^(pi/2) cosx/(e^x+1) dx #
2 Answers
Explanation:
After using
After collecting 2 integrals,
=
=
Hence
# 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 #