Evaluate the integral? # int ln tanh(x/2)/cosh^2x dx #
2 Answers
I got
#= tanh(x)ln(tanh(x/2)) - 2arctan(tanh(x/2)) + C# ,#" "x > 0#
DISCLAIMER: VERY LONG ANSWER!
Well, I don't work with hyperbolics much, but I do know that:
- the derivative of
#tanh(x)# is still#sech^2(x)# . - we still have
#cosh(x) = 1/(sech(x))# . - we however, have that
#-sinh^2(x) + cosh^2(x) = 1# , so that#tanh^2(x) + sech^2(x) = 1# and#1 + csch^2(x) = coth^2(x)# .
So, we have:
#= int sech^2(x)ln (tanh(x/2))dx#
Now let's try an integration by parts. Let:
#u = ln(tanh(x/2))#
#dv = sech^2(x)dx#
#du = (sech^2(x/2))/(2tanh(x/2))dx#
#v = tanh(x)#
#uv - intvdu#
#= tanh(x)ln(tanh(x/2)) - int (tanh(x)sech^2(x/2))/(2tanh(x/2))dx#
With the above
#int (tanh(x)sech^2(x/2))/(2tanh(x/2))dx#
#= int (tanh(x))/(2tanh(x/2))dx - int (tanh(x) tanh^cancel(2)(x/2))/(2cancel(tanh(x/2)))dx#
I had to look these identities up to verify them though:
#sinh(u pm v) = sinhu coshv pm coshu sinhv# #cosh(u pm v) = coshu coshv pm sinhu sinhv#
Thus,
#tanh(x) = tanh(x/2 + x/2) = sinh(x/2 + x/2)/(cosh(x/2 + x/2))#
#= (2sinh(x/2)cosh(x/2))/(cosh^2(x/2) + sinh^2(x/2)) xx (cosh^2(x/2))/(cosh^2(x/2))#
#= (2tanh(x/2))/(1 + tanh^2(x/2))#
Therefore, the integral becomes:
#= int cancel((2tanh(x/2)))/((1 + tanh^2(x/2))cancel(2tanh(x/2)))dx - cancel(1/2)int (cancel(2)tanh(x/2)tanh(x/2))/(1 + tanh^2(x/2)) xx (1//tanh^2(x/2))/(1//tanh^2(x/2))dx#
#= ul(int 1/(1 + tanh^2(x/2))dx - int 1/(coth^2(x/2) + 1)dx)# #" "bb((1) + ( 2))#
For
#2int 1/(1 + tanh^2(x/2))dx#
#= 2 int 1/((1 + u^2)(1 - u^2))du#
#= 2 int 1/((1 + u^2)(1+u)(1-u))du = 2 int (Ax + B)/(1 + u^2) + C/(1 + u) + D/(1 - u)du#
Through getting common denominators and using partial fraction decomposition, we obtain:
#A = 0, B = 1/2, C = 1/4, D = -1/4#
This gives for
#(1) = 2 cdot [ 1/2 arctan(tanh(x/2)) + 1/4 ln |1 + tanh(x/2)| - 1/4 ln|1 - tanh(x/2)|]#
#= ul(arctan(tanh(x/2)) + 1/2 ln |(1 + tanh(x/2))/(1 - tanh(x/2))|)#
Now, for
#(2) = -arctan(coth(x/2)) - 1/2 ln |(1 + coth(x/2))/(1 - coth(x/2))|#
#= arctan(tanh(x/2)) - 1/2 ln |(tanh(x/2) + 1)/(tanh(x/2) - 1)|#
#= ul(arctan(tanh(x/2)) - 1/2 ln |(1 + tanh(x/2))/(1 - tanh(x/2))|)#
In adding
#(1) + (2) = 2arctan(tanh(x/2))#
Therefore, the overall integral (the answer!) is:
#color(blue)(barul(|stackrel(" ")(" "tanh(x)ln(tanh(x/2)) - 2arctan(tanh(x/2)) + C" ")|))#
# int \ ln tanh(x/2)/cosh^2x \ dx = tanh(x) \ lntanh(x/2) - 2 arctan(tanh(x/2)) + C #
Explanation:
We seek:
# I = int \ ln tanh(x/2)/cosh^2x \ dx #
I will take a very similar approach to that of Truong-Son N. but I will simplify the expression in the integrand via an initial substitution. We can remove the half-angle via a substitution of the form:
# u=x/2 => (du)/dx = 1/2 # and#x=2u#
So then substituting into the integral, it becomes:
# I = int \ ln tanhu/cosh^2 (2u) \ 2 \ du #
# :. I/2 = int \ ln tanhu \ sech^2 (2u) \ du #
Then an application of Integration By Parts (IBP) will rmove the logarithm from the integrand:
Let
# { (U,=lntanh u, => , U',=sech^2u/tanh u), (V',=sech^2 (2u), =>, V,=1/2tanh(2u) ) :}#
Then plugging into the IBP formula:
# int \ (U)(V') \ dx = (U)(V) - int \ (V)(U') \ dx #
Gives us
# int \ (lntanhu)(sech^2(2u)) \ du = (lntanh)(1/2tanh(2u)) - int \ (1/2tanh(2u))(sech^2u/tanh u) \ du #
# :. I/2 = 1/2 \ tanh(2u) \ lntanhu - 1/2 J#
# :. \ \ I = tanh(2u) \ lntanh u- J# ..... [A]
Where
# tanh(2A) = (2 tanh(A))/{1 + tanh^2A) #
We have:
# J = int \ ((2 tanh(u))/(1 + tanh^2u))(sech^2u/tanh u) \ du #
# \ \ = 2 \ int \ ( sech^2u )/( 1 + tanh^2u ) \ du #
And with this second integral, a simple substitution of the form:
# v = tanhu => (dv)/(du) = sech^2u #
Gives us after substituting into
# J = 2 \ int \ 1/(1+v^2) \ dv #
# \ \ = 2 arctan(v) + C' #
Inserting this result into [A] we get:
# I = tanh(2u) \ lntanhu - 2 arctan(v) + C #
And restoring the substitutions we get:
# I = tanh(2x/2) \ lntanh(x/2) - 2 arctan(tanhu) + C #
# \ \ = tanh(x) \ lntanh(x/2) - 2 arctan(tanh(x/2)) + C #
