# Integral of sinhx.lncosh^2(x).dx

1. Dec 8, 2011

### sharks

The problem statement, all variables and given/known data
Integral of sinhx.lncosh^2(x).dx

The attempt at a solution
I used integration by parts.
Let U = lncoshx^2(x)
Let dV = sinhx

So, dU =sech^2(x)/(2coshx.sinhx) = 1/(2cosh^3(x).sinhx) [i hope this line is correct?]
And, V = coshx

When using the formula for integration by parts, the integral of V.dU proves to be difficult. I think i should use substitution to simply it somehow, but not sure how.

2. Dec 8, 2011

### dextercioby

Substitute $\cosh x = t$.

3. Dec 8, 2011

### sharks

So, i used the substitution suggested above into V.dU, and i get:

Integral of V.dU = Integral of 1/[2(t^4 - t^2)] w.r.t.t

I could use this equivalent (re-arranged) expression: Integral of 1/[2(t^4 - t^2)] w.r.t.t = Integral of 1/[(2t^2).(t^2 - 1)] w.r.t.t

But then i would probably have to integrate this by parts again. I guess i'm stuck once more.

Last edited: Dec 8, 2011
4. Dec 8, 2011

### SammyS

Staff Emeritus
That line is incorrect.

$\displaystyle \frac{d}{dx}\ln(\cosh^2(x))= \text{sech}^2(x)\cdot(2\cosh(x)\sinh(x))$
$\displaystyle =2\,\text{sech}(x) \sinh(x)$​
Of course that could have been arrived at more easily by noticing that $\displaystyle \ln(\cosh^2(x))=2\ln(\cosh(x))\,.$
Your integration by pars should work fine, except,
dv should be: $\displaystyle dv=\sinh(x)\,dx\,.$

This should result in $\displaystyle\int v\,du=2\int\sinh(x)\,dx\,.$

5. Dec 9, 2011

### sharks

I finally got it. Thanks SammyS for the hint about 2ln(cosh x). I used that and it all got much simpler, like you said.

6. Dec 9, 2011

### dextercioby

I mean without part integration at the beginning. First substitute, then the new integral should be ~ int ln t dt. Now you can use part integration to compute the antiderivative of the logarithm (which is also tabluated).