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Integral of Sech^5x*Cschx dx

  • Thread starter Crush1986
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  • #1
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Homework Statement



integral of Sech^5(x)*Csch(x) dx

Homework Equations



I think Coth^2(x)-1 = Csch^2(x) may help

The Attempt at a Solution



I tried a few things. The latest being breaking the problem up and doing some re-working.

int(Sech^2(x)*Sech^3(x)*Csch(x) dx) I then multiplied the whole expression by Sinh^3(x)/Sinh^3(x) to get
int(Sech^2(x)*tanh^3(x)*Csch^4(x) dx)
I then got rid of the Csch^4 by breaking it up into two (Coth(x)^2-1)'s and multiplying the tanh^3(x) term through to end up with
int(Sech^2(x)*tanh(x)*(coth^2(x)+tanh^2(x) dx)

I think I just made a huge mess and probably some mistakes...

I was doing really well through two books worth of problems until this one :s, one of the very last ones lol.

If I could get some guidance or a little hint to move me toward a better understanding of the technique needed here I'd greatly appreciate it.

Thank you.
 

Answers and Replies

  • #2
AlephZero
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A hint: how would you do ##\int \sin^5x \cos x \,dx##? The easy way is make a substitution.....
 
  • #3
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A hint: how would you do ##\int \sin^5x \cos x \,dx##? The easy way is make a substitution.....

That one I see easy. U substitution. U would be sinx and du would be cosx. So it'd be 1/6sin^6(x)+C. I don't see an obvious sub here though.

I was trying to create one with tanh(x) and sech(x) functions, but think I was going down a wrong path.

Am I missing some connection here? It seems like sech and csch just don't have an obvious sub to me.

The 1/coshx and 1/sinhx I do see. But I get like an inverse du. I don't know how to deal with that.
 
  • #4
vela
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Try ##u = \sinh x##.
 
  • #5
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Try ##u = \sinh x##.

I tried this last night, I must be missing the obvious.

So ok, u= sinhx, du would be coshx dx

I then have the integral of 1/u but I don't see how to incorporate the du?

I'm guessing there is just some reworking of the du to make this work that I'm not seeing. I have a 1/cosh^5(x) in the problem and a du = to cosh(x) dx.
 
  • #6
vela
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Oh, sorry, I meant ##u = \cosh x##.

EDIT: Actually, I think you can get it to work either way.

If you use ##u = \sinh x##, you get ##du = \cosh x\,dx##, so ##dx = \text{sech } x\, du##. So
$$\int \text{sech}^5 x \text{ csch } x \,dx = \int \text{sech}^6 x\ u^{-1}\,du.$$ Now you just need to write ##\text{sech}^6 x## in terms of ##u##.
 
Last edited:
  • #7
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Oh, sorry, I meant ##u = \cosh x##.

I'm Sure I tried this too.

So U = cosh(x) du = sinh(x) dx

Gives me integral of 1/u^5 and again I'm not sure what do do with the du. So in the integral I have a 1/sinh(x) and the du is equal to sinh(x) dx. I just don't see how you work the du in there.
 
  • #8
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Oh, sorry, I meant ##u = \cosh x##.

EDIT: Actually, I think you can get it to work either way.

If you use ##u = \sinh x##, you get ##du = \cosh x\,dx##, so ##dx = \text{sech } x\, du##. So
$$\int \text{sech}^5 x \text{ csch } x \,dx = \int \text{sech}^6 x\ u^{-1}\,du.$$ Now you just need to write ##\text{sech}^6 x## in terms of ##u##.

Ohhhhhh I see that now... Ok going to really study this and try to find similar problems to work on too.
 
  • #9
AlephZero
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My idea was
$$\begin{aligned} & \int \text{sech}^5 x\, \text{csch}\, x\, dx\\
=& \int \dfrac{dx}{\cosh^5 x \sinh x}\\
=& \int \dfrac{\sinh x\,dx}{\cosh^5 x \sinh^2 x}\end{aligned}$$
Then let ##u = \cosh x## and you get to something you an integrate using partial fractions.
 
  • #10
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Well, I thought I could take it from there, I'm still unable to express the whole thing in terms on only U and no X's... I am just not getting this one problem lol. the -1's are just throwing me off when I'm trying to do both terms. I can express one just fine but when it comes to the other I'm just NOT seeing it...

:( sorry, I just don't get this one.
 
  • #11
Dick
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Well, I thought I could take it from there, I'm still unable to express the whole thing in terms on only U and no X's... I am just not getting this one problem lol. the -1's are just throwing me off when I'm trying to do both terms. I can express one just fine but when it comes to the other I'm just NOT seeing it...

:( sorry, I just don't get this one.
I would try following AlephZero's lead. I find it easier to think about sinh and cosh than about sech and csch.
 
  • #12
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I would try following AlephZero's lead. I find it easier to think about sinh and cosh than about sech and csch.

I do too. We haven't gotten to partial fractions though. So I should probably solve it by U-substitution. I just can't get an integral that is completely in terms of U and in a U*du form of any type. It just doesn't come to me.
 
  • #13
vela
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Are you sure the problem isn't to integrate sech5 x tanh x? I don't think you can do this with just a simple substitution.
 
  • #14
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Are you sure the problem isn't to integrate sech5 x tanh x? I don't think you can do this with just a simple substitution.

I'm thinking the instructor skipped this problem in this section for that reason now. He wants us to master U substitution. He said he'd pick problems from these sections, hopefully he doesn't do this one. I found the answer online and it is a monster kind of...

The instructions in the book do say to use one of the identities but yah.. It looked way more involved than that.
 
  • #15
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Are you sure the problem isn't to integrate sech5 x tanh x? I don't think you can do this with just a simple substitution.

No it's Cschx I found a way to do it with 2 subs and starting off like alpha said. I did need to do some partial fractions.
 
  • #16
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I got 1/4sech^4(x) - 1/2 sech^2(x) + 1/2(ln(tanh^2(x)) + C


That was with starting at

Sinh(x)/(cosh^5(x)sinh^2(x)) u= coshx du= sinhx

So I had 1/(u^5(u^2-1)) du then subbed v for u^2 so dv =2u du

Got 1/v^3(v-1) did partial fractions.
 
Last edited:
  • #17
vela
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Looks like you have a sign error somewhere. The sech^2 term should be positive as well.
 

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