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

  • Thread starter Thread starter DryRun
  • Start date Start date
  • Tags Tags
    Integral
DryRun
Gold Member
Messages
837
Reaction score
4
Homework Statement
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.
 
Physics news on Phys.org
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:
sharks said:
Homework Statement
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?]
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))\,.
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.

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\,.
 
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.
 
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).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top