# Integral of Sech^5x*Cschx dx

## 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.

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

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.

vela
Staff Emeritus
Homework Helper
Try ##u = \sinh x##.

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.

vela
Staff Emeritus
Homework Helper
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##.

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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.

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.

AlephZero
Homework Helper
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.

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.

Dick
Homework Helper
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.

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.

vela
Staff Emeritus
Homework Helper
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.

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.

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.

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.

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vela
Staff Emeritus