Evaluate the integral? $\int \frac{ln tanh (\frac{x}{2})}{{cosh}^{2} x} d x$ $int ln tanh(x/2)/cosh^2x dx$

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Evaluate the integral? $\int \frac{ln tanh (\frac{x}{2})}{{cosh}^{2} x} d x$ $int ln tanh(x/2)/cosh^2x dx$

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Answer 1 · 2017-07-24T20:03:36

I got

$= tanh (x) ln (tanh (\frac{x}{2})) - 2 arctan (tanh (\frac{x}{2})) + C$ $= tanh(x)ln(tanh(x/2)) - 2arctan(tanh(x/2)) + C$ , $x > 0$ $" "x > 0$

DISCLAIMER: VERY LONG ANSWER!

Well, I don't work with hyperbolics much, but I do know that:

the derivative of $tanh (x)$ $tanh(x)$ is still ${sech}^{2} (x)$ $sech^2(x)$ .
we still have $cosh (x) = \frac{1}{sech (x)}$ $cosh(x) = 1/(sech(x))$ .
we however, have that $- {sinh}^{2} (x) + {cosh}^{2} (x) = 1$ $-sinh^2(x) + cosh^2(x) = 1$ , so that ${tanh}^{2} (x) + {sech}^{2} (x) = 1$ $tanh^2(x) + sech^2(x) = 1$ and $1 + {csch}^{2} (x) = {coth}^{2} (x)$ $1 + csch^2(x) = coth^2(x)$ .

So, we have:

$= \int {sech}^{2} (x) ln (tanh (\frac{x}{2})) d x$ $= int sech^2(x)ln (tanh(x/2))dx$

Now let's try an integration by parts. Let:

$u = ln (tanh (\frac{x}{2}))$ $u = ln(tanh(x/2))$
$d v = {sech}^{2} (x) d x$ $dv = sech^2(x)dx$
$d u = \frac{{sech}^{2} (\frac{x}{2})}{2 tanh (\frac{x}{2})} d x$ $du = (sech^2(x/2))/(2tanh(x/2))dx$
$v = tanh (x)$ $v = tanh(x)$

$u v - \int v d u$ $uv - intvdu$

$= tanh (x) ln (tanh (\frac{x}{2})) - \int \frac{tanh (x) {sech}^{2} (\frac{x}{2})}{2 tanh (\frac{x}{2})} d x$ $= tanh(x)ln(tanh(x/2)) - int (tanh(x)sech^2(x/2))/(2tanh(x/2))dx$

With the above $sech$ $sech$ identity, this integral becomes:

$\int \frac{tanh (x) {sech}^{2} (\frac{x}{2})}{2 tanh (\frac{x}{2})} d x$ $int (tanh(x)sech^2(x/2))/(2tanh(x/2))dx$

$= int (tanh(x))/(2tanh(x/2))dx - int (tanh(x) tanh^cancel(2)(x/2))/(2cancel(tanh(x/2)))dx$

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 $(1)$ , let $u = tanh(x/2)$ . Then, $du = 1/2 sech^2(x/2)dx = 1/2(1 - tanh^2(x/2))dx$ , so:

$2int 1/(1 + tanh^2(x/2))dx$

$= 2 int 1/((1 + u^2)(1 - u^2))du$

$= 2 int 1/((1 + u^2)(1+u)(1-u))du = 2 int (Ax + B)/(1 + u^2) + C/(1 + u) + D/(1 - u)du$

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 $(1)$ :

$(1) = 2 cdot [ 1/2 arctan(tanh(x/2)) + 1/4 ln |1 + tanh(x/2)| - 1/4 ln|1 - tanh(x/2)|]$

$= ul(arctan(tanh(x/2)) + 1/2 ln |(1 + tanh(x/2))/(1 - tanh(x/2))|)$

Now, for $(2)$ , we obtain the same thing, except in terms of $coth(x/2)$ and we account for the opposite sign out front:

$(2) = -arctan(coth(x/2)) - 1/2 ln |(1 + coth(x/2))/(1 - coth(x/2))|$

$= arctan(tanh(x/2)) - 1/2 ln |(tanh(x/2) + 1)/(tanh(x/2) - 1)|$

$= ul(arctan(tanh(x/2)) - 1/2 ln |(1 + tanh(x/2))/(1 - tanh(x/2))|)$

In adding $(1)$ and $(2)$ together, the $ln$ terms cancel to give:

$(1) + (2) = 2arctan(tanh(x/2))$

Therefore, the overall integral (the answer!) is:

$color(blue)(barul(|stackrel(" ")(" "tanh(x)ln(tanh(x/2)) - 2arctan(tanh(x/2)) + C" ")|))$

Answer 2 · 2017-07-24T21:18:19

$int \ ln tanh(x/2)/cosh^2x \ dx = tanh(x) \ lntanh(x/2) - 2 arctan(tanh(x/2)) + C$

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 $J=int \ tanh(2u)(sech^2u/tanh u) \ du$ , then. Using the hyperbolic identity:

$tanh(2A) = (2 tanh(A))/{1 + tanh^2A)$

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 $J$ :

$J = 2 \ int \ 1/(1+v^2) \ dv$
$\ \ = 2 arctan(v) + C'$

Inserting this result into [A] we get:

$I = tanh(2u) \ lntanhu - 2 arctan(v) + C$

And restoring the substitutions we get:

$I = tanh(2x/2) \ lntanh(x/2) - 2 arctan(tanhu) + C$
$\ \ = tanh(x) \ lntanh(x/2) - 2 arctan(tanh(x/2)) + C$

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