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Integral of sech^3 (x)

  1. Dec 9, 2011 #1

    sharks

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    The problem statement, all variables and given/known data
    Integral of sech^3 (x)


    The attempt at a solution

    sech^3 (x) = (sech x)(sech^2 (x)) = (sech x)(1-tanh^2 (x))
    And then i expanded by multiplying:

    (sech x)(1-tanh^2 (x)) = (sech x) - (sech x)(tanh^2 (x))
    The integral of sech x can be found.
    But what about the integral of (sech x)(tanh^2 (x))?
    I've tried various substitutions, but failed. Let t = tanh x and also let t = sech x.
    I also tried this way:
    (sech x)(tanh^2 (x)) = (sech x)(1 - sech^2 (x)) = sech x - sech^3 (x)
    But then i end up with the original problem again.
     
  2. jcsd
  3. Dec 9, 2011 #2

    dextercioby

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    [tex] \int \frac{d(\sinh x)}{(1+\sinh^2 x)^2} = \int \frac{du}{(1+u^2)^2} [/tex]

    Can you solve this ?
     
  4. Dec 9, 2011 #3

    LCKurtz

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    sech(x)tanh2x = [STRIKE]sech(x)[sech(x)tanh(x)]. Let u=sech(x), what is du[/STRIKE]?

    [Edit:] Nevermind. I hurriedly wrote what I wanted it to be instead of what it is :mad:
     
    Last edited: Dec 9, 2011
  5. Dec 9, 2011 #4
    try approaching it from integrating by parts:

    ∫sec^3(x) dx

    u= sec(x) dv=sec^2(x) dx
    du=sec(x)tan(x) v=tan(x)


    ∫sec^3(x) dx = sec(x)*tan(x)-∫sec(x)*tan^2(x) dx
    ∫sec^3(x) dx = secx*tanx-∫sec^3(x)-sec(x) dx
    ∫sec^3(x) dx = secx*tanx-∫sec^3(x) dx+∫sec(x) dx

    Can you figure out the rest from here?
     
  6. Dec 9, 2011 #5

    sharks

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    Unfortunately, i must be getting stupider... I can't understand either of you.

    @dextercioby In your example, if by letting u = sinh x, then du/dx = cosh x, so... the R.H.S. of your equation should be something different. The numerator sinh x can't be absorbed cleanly into du.

    @LCKurtz How can sech(x)tanh^2(x) = sech(x)[sech(x)tanh(x)??
    If i start with the R.H.S of this equation, it means sech^2(x)tanh(x).

    I think you mean, by using the hyperbolic identity: 1 - tanh^2(x) = sech^2(x)
    I would then get tanh^2(x) = 1 - sech^2(x)

    This would become upon replacing in sech(x)tanh^2(x):
    sech(x)[1 - sech^2(x)]

    I don't know where this is going...

    @McAfee This problem is not about finding the integral of sec^3(x) but integral of sech^3(x). Thanks though.
     
  7. Dec 9, 2011 #6

    vela

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    Integration by parts is probably the easiest way to do the problem.

    Let u = sinh x. Then du = cosh x dx, so dx = du/cosh x, so your integral becomes
    [tex]\int\text{sech}^3 x\,dx = \int\frac{dx}{\cosh^3 x} = \int\frac{du}{\cosh^4 x}[/tex]Finish changing variables from x to u.
     
  8. Dec 9, 2011 #7

    sharks

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    OK, i was able to get dextercioby's R.H.S. equation.

    Now, i'm unsure about the method of integration. Here are the 2 methods that i'm thinking of. I don't know if both methods are valid in this case though.

    Method 1:
    Integral of (1 + u^2)^(-2).du which after integration would give: -1/[2u(1 + u^2)] + C

    Method 2:
    Do partial fractions first and then integrate. This would give:
    (Au + B)/(1 + u^2) + (Cu + D)/[(1 + u^2)^(2)]
    I hope the format above is correct? I used (Au + B) and (Cu + D) as the numerators since the denominator is quadratic. Otherwise i would have put just A and B instead.
     
  9. Dec 9, 2011 #8

    vela

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    The first method is wrong. With partial fractions, you'll find A=B=C=0 and D=1. It's already in its simplest form.

    I'm not sure what dextercioby had in mind. I usually go the other way. Use the trig substitution to go from the u-form to the trig form, and then use another technique to evaluate the trig integral.
     
  10. Dec 9, 2011 #9

    LCKurtz

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    It isn't. Just a careless error.
     
  11. Dec 9, 2011 #10

    SammyS

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    Use the power reduction formula for [itex]\displaystyle\int\text{sech}^m(x)dx\,.[/itex] In this case m=3.

    You get [itex]\displaystyle\int\text{sech}^3(x)dx=\frac{1}{2} \tanh(x) \text{sech}(x)+\frac{1}{2} \int \text{sech}(x)\,dx\,.[/itex]

    To get this result you can do integration by parts with [itex]dv=\text{sech}^2(x)\,dx[/itex] and [itex]u=\text{sech}(x)[/itex] along with some hyperbolic function identities & algebra.
     
  12. Dec 9, 2011 #11

    sharks

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    OK, this is getting immensely complicated! I'm still at undergraduate level and haven't yet done anything pertaining to the power reduction formulae.

    So, here is the answer from my notes:
    http://s2.ipicture.ru/uploads/20111210/gBo5g6t3.jpg

    Hopefully, this will help all of us to trace a better and simpler route to solve this problem.
     
  13. Dec 9, 2011 #12

    vela

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    It only seems immensely complicated because you keep ignoring the suggestion to use integration by parts.
     
  14. Dec 11, 2011 #13

    sharks

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    I took a break, and i apologize for being a little bit frustrated with this problem.

    From this point: [tex]\int \frac{d(\sinh x)}{(1+\sinh^2 x)^2} = \int \frac{du}{(1+u^2)^2}[/tex]
    I expressed it in partial fractions (can confirm the results that A=B=C=0 and D=1), and got [tex]\int \frac{du}{(1+u^2)^2}[/tex]
    But now how to proceed next?
    Could you please explain how to go about it using your method?
     
    Last edited: Dec 11, 2011
  15. Dec 11, 2011 #14
    Solve [tex]\int \frac{du}{(1+u^2)}[/tex] where [tex]t=\frac{1}{(1+u^2)}[/tex] and [tex]{dv}={du}[/tex] (Integration by parts)
     
  16. Dec 12, 2011 #15

    sharks

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    Hi Karamata

    OK, to solve the problem you typed above, here is what i have so far:

    [tex]\frac{u}{(1+u^2)} + 2\int\frac{u^2}{(1+u^2)^2}[/tex]

    Now, to evaluate [tex]\int\frac{u^2}{(1+u^2)^2}[/tex]

    If i break the integral using partial fractions, i believe i will get the same answer, as it's already in its simplest form.

    So, i integrate by parts again.
     
    Last edited: Dec 12, 2011
  17. Dec 12, 2011 #16
    [tex]\frac{u^2}{(1+u^2)^2}=\frac{u^2+1-1}{(1+u^2)^2}=\frac{u^2+1}{(1+u^2)^2}-\frac{1}{(1+u^2)^2}[/tex].....
     
  18. Dec 12, 2011 #17

    sharks

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    To continue from above:

    [tex]\int\frac{1 + u^2}{(1 + u^2)^2} = \int\frac{1}{1 + u^2} = \arctan u[/tex]

    But [tex]\int\frac{1}{(1 + u^2)^2}[/tex] is a bit tricky. I cannot simplify it further using partial fractions.

    I suppose i'll have to use substitution.

    Let [tex]u = tan \theta[/tex] Not sure though.
     
    Last edited: Dec 12, 2011
  19. Dec 12, 2011 #18
    [tex]\int\frac{1}{1 + u^2}= \arctan u[/tex]

    I don't get it.

    [tex]\int {\frac{du}{1+u^2}}=\frac{u}{u^2+1}+2\left( \arctan {u} + C -\int \frac{du}{(1+u^2)^2}\right)[/tex]
    Or [tex]\arctan {u}=\frac{u}{u^2+1}+2\left( \arctan {u} + C -\int \frac{du}{(1+u^2)^2}\right)[/tex]
    And now, [tex]\int \frac{du}{(1+u^2)^2}=[/tex]
     
    Last edited: Dec 13, 2011
  20. Dec 12, 2011 #19

    sharks

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    I had made a mistake when evaluating.
    [tex]\int\frac{1}{1 + u^2}= \arctan u[/tex]
    I edited my post as you were replying. Thank you for pointing it out.

    In evaluating the integral below, surely, it can't be as simple as this (but i'll take a chance as i can't see what substitution to make):
    [tex]\int \frac{du}{(1+u^2)^2} = \arctan^2 u[/tex]
    It's probably wrong though.

    If i use the substitution:
    Let [tex]u = \tan \theta[/tex]

    Then the integral evolves into:
    [tex]\int \frac{du}{sec^4 \theta}[/tex]
    But
    [tex]\frac{du}{d\theta} = sec^2 \theta[/tex]
    So, the integral becomes
    [tex]\int \frac{d\theta}{sec^2 \theta} = \int cos^2 \theta.d\theta[/tex]

    Next, i use double angle formula to convert the above integral into:
    [tex]\int \frac{1}{2}.d\theta + \int \frac{cos 2\theta}{2}.d\theta[/tex]
    Then, after integration, i get:
    [tex] \frac{\theta}{2} + \frac{sin 2\theta}{4}[/tex]
    From
    [tex]u = \tan \theta[/tex]
    I get
    [tex]\theta = \arctan u[/tex] which i substitute into the integration results, and i get:
    [tex]\frac{\arctan u}{2} + \frac{sin (2\arctan u)}{4}[/tex]
     
    Last edited: Dec 12, 2011
  21. Dec 12, 2011 #20
    [tex]\arctan {u}=\frac{u}{u^2+1}+2\left( \arctan {u} + C -\int \frac{du}{(1+u^2)^2}\right)[/tex]
    And now, [tex]\int \frac{du}{(1+u^2)^2}=[/tex]

    [tex]\hbox{TUT}=\hbox{BRB}+2\left(\hbox{TUT}-\int \frac{du}{(1+u^2)^2}\right)[/tex]
    [tex]\int \frac{du}{(1+u^2)^2}=\frac{1}{2}(\hbox{BRB}+\hbox{TUT})[/tex]


    This isn't true.
     
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