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integration problem

  • 17-03-2007 08:51PM
    #1
    balzarywex
    Closed Accounts Posts: 96 ✭✭


    does anyone have any tips for for finding the integral of (sec x)^4 dx?
    iv looked thorugh the table of integrals but no inspiration as of yet! any help would be greatly appreciated!


Comments

  • #2
    JC 2K3
    Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    hmmm

    try writing it as ((sec x)^2)((sec x)^2) dx and do integration by parts.

    (integ of (sec x)^2 dx = tanx + c)


  • #3
    Useless Fecker
    Registered Users, Registered Users 2 Posts: 18 ✭Useless Fecker


    Is the answer 2?


  • #4
    Kevster
    Closed Accounts Posts: 2,980 ✭✭✭Kevster


    Sec(x) is 1/Cos(x)

    Therefore, Sec(x)^4 is 1/Cos(x)^4

    1/Cos(x)^4 is then Cos(x)^-4

    I think you can proceed much easier from that point.

    ...although, JC 2K3's way is obviously much much simpler!


  • #5
    JC 2K3
    Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    ((sec x)^2)((sec x)^2) dx

    u = sec(x)^2 = cos(x)^-2
    du = -2cos(x)^-3(-sinx).dx = 2sin(x)/cos(x)^3.dx = 2tan(x)/cos(x)^2.dx
    dv=sec(x)^2.dx
    v = tan(x)

    uv = tan(x)/cos(x)^2 = tan(x)sec(x)^2

    integ(v.du) = 2(integ(tan(x)^2/cos(x)^2.dx))
    = 2(integ((tan(x)^2)(sec(x)^2).dx))

    w = tan(x)^2
    dw = 2sec(x)^2(tan(x))dx
    2(integ((tan(x)^2)(sec(x)^2).dx)) = integ(w^(1/2).dw)
    =2w^(3/2)/3
    2tan(x)^3/3

    uv - integ(v.du) = tan(x)/cos(x)^2 - 2tan(x)^3/3
    tan(x)sec(x)^2 - 2tan(x)^3/3
    (3tan(x)sec(x)^2 - 2tan(x)^3)/3
    (1/3)(3tan(x)sec(x)^2 - 2tan(x)^3)


    Hopefully you can follow that, I think it's right. I'm only doing my LC, so perhaps there's a quicker way to do it, I dunno, but hope I helped.


  • #6
    Kevster
    Closed Accounts Posts: 2,980 ✭✭✭Kevster


    Phwoarrr..... I need a longer sheet with the way I'm doing it. I'm managing to do nested 'Integral by parts' if you get me. So far, I've nested it thrice.

    ...make that fourth. I think I'll give-up.


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  • #7
    Crumbs
    Registered Users, Registered Users 2 Posts: 1,080 ✭✭✭Crumbs


    That integration by parts method looks right but I think there's a quicker way.

    Use the identity: sec^2(x) = 1 + tan^2(x)

    Int [sec^4(x) dx] = Int [(1 + tan^2(x)).sec^2(x) dx]

    Substitution:
    u = tan x
    du = sec^2(x).dx

    Int [1 + u^2 du]

    = u + (u^3)/3 + c

    = tan x + (tan^3(x))/3 + c


  • #8
    JC 2K3
    Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    Ah, damn you!

    If you replace the sec(x)^2 with (1+tan(x)^2) in my answer you get the same answer:

    (1/3)(3tan(x)sec(x)^2 - 2tan(x)^3)
    = (1/3)(3tan(x)(1+tan(x)^2) - 2tan(x)^3)
    = (1/3)(3tan(x) + 3tan(x)^3 - 2tan(x)^3)
    = (1/3)(3tan(x) + tan(x)^3)
    = tan(x) + (tan(x)^3)/3


  • #9
    balzarywex
    Closed Accounts Posts: 96 ✭✭balzarywex


    you boardsies are legends for help! too bad i forgot i posted up here and got the answer two days ago (the 1/Cosx way!) :D cheers!


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