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

  • Thread starter Thread starter DryRun
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary

Homework Help Overview

The discussion revolves around the integral of sinh(x) multiplied by ln(cosh^2(x)) with respect to x. Participants are exploring integration techniques, particularly integration by parts and substitution methods.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss using integration by parts, with one suggesting a substitution of cosh(x) to simplify the integral. There are attempts to clarify the differentiation of ln(cosh^2(x)) and its implications for the integration process.

Discussion Status

Some participants have offered guidance on the differentiation and integration techniques, while others express uncertainty about the steps involved. There is acknowledgment of a simpler approach that emerged from the discussion, but no consensus on a final method has been reached.

Contextual Notes

There are indications of confusion regarding the differentiation of ln(cosh^2(x)) and the setup for integration by parts. Participants are navigating through various interpretations of the integral and the implications of their chosen methods.

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.

[itex]\displaystyle \frac{d}{dx}\ln(\cosh^2(x))= \text{sech}^2(x)\cdot(2\cosh(x)\sinh(x))[/itex]
[itex]\displaystyle =2\,\text{sech}(x) \sinh(x)[/itex]​
Of course that could have been arrived at more easily by noticing that [itex]\displaystyle \ln(\cosh^2(x))=2\ln(\cosh(x))\,.[/itex]
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: [itex]\displaystyle dv=\sinh(x)\,dx\,.[/itex]

This should result in [itex]\displaystyle\int v\,du=2\int\sinh(x)\,dx\,.[/itex]
 
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).
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
7
Views
2K
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 1 ·
Replies
1
Views
1K