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Integral of Sech^5x*Cschx dx

  1. Jan 18, 2014 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

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

    3. 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.
     
  2. jcsd
  3. Jan 18, 2014 #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.....
     
  4. Jan 18, 2014 #3

    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.
     
  5. Jan 18, 2014 #4

    vela

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    Try ##u = \sinh x##.
     
  6. Jan 18, 2014 #5

    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.
     
  7. Jan 18, 2014 #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: Jan 18, 2014
  8. Jan 18, 2014 #7

    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.
     
  9. Jan 18, 2014 #8

    Ohhhhhh I see that now... Ok going to really study this and try to find similar problems to work on too.
     
  10. Jan 18, 2014 #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.
     
  11. Jan 18, 2014 #10
    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.
     
  12. Jan 18, 2014 #11

    Dick

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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.
     
  13. Jan 18, 2014 #12

    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.
     
  14. Jan 18, 2014 #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.
     
  15. Jan 19, 2014 #14

    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.
     
  16. Jan 19, 2014 #15

    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.
     
  17. Jan 19, 2014 #16
    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: Jan 19, 2014
  18. Jan 19, 2014 #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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