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The Integral of 5x*csc(7x)*cot(7x)

  1. Sep 8, 2009 #1
    1. The problem statement, all variables and given/known data
    Integral of 5x*csc(7x)*cot(7x)

    2. Relevant equations
    Integration by parts
    Trigonometric Derivatives (csc(u) = -csc(u)cot(u))
    Substitution

    3. The attempt at a solution

    *Note, I apologize beforehand for the lack of knowledge on how to display mathematical equations nicely on PF *

    5*int[x*csc(7x)*cot(7x)]

    I am preparing for the 7x substitution here:
    5/7*int[7x*csc(7x)*cot(7x)]

    let z = 7x
    dz = 7dx

    5/7 * 1/7 *int[z*csc(z)*cot(z)]
    5/49 * int[z*csc(z)*cot(z)]

    Preparing for the csc(z) substitution here:
    -1* 5/49 * int[-1*z*csc(z)*cot(z)]

    let w = csc(z)
    dw = -csc(z)cot(z)

    arccsc(w) = z

    -1*5/49*int[arccsc(w)]

    Integration by parts

    u = arccsc(w) dv=dw
    du = -1/|w|*sqrt(w^2-1) v = w

    -1*5/49*[w*arccsc(w) - int[ -1/|w|*sqrt(w^2-1) * w ] ]

    -1*5/49*[w*arccsc(w) + int[w/|w|*sqrt(w^2-1)]

    In between these two steps I may have made an error since I canceled the absolute value w on the denominator with the w on the numerator.

    -1*5/49*[w*arccsc(w) + int[1/sqrt(w^2-1)]

    Solved for answer in two different ways (both wrong)
    arccosh way -
    -1*5/49*[w*arccsc(w) + arccosh(w)]
    -1*5/49*[csc(z)*arccsc(csc(z)) + arccosh(csc(z))]
    -1*5/49*[csc(7x)*arccsc(csc(7x)) + arccosh(csc(7x))]

    trig substitution way -
    -1*5/49*[w*arccsc(w) + int[1/sqrt(w^2-1)]

    let w = 1*sec(theta)
    dw = sec(theta)tan(theta)

    -1*5/49*[w*arccsc(w) + int[sec(theta)tan(theta)/sqrt(sec(theta)^2-1)]
    -1*5/49*[w*arccsc(w) + int[sec(theta)tan(theta)/tan(theta)]
    -1*5/49*[w*arccsc(w) + int[sec(theta)]
    -1*5/49*[w*arccsc(w) + ln|sec(theta)+tan(theta)|]

    Building the right triangle

    Since w/1 = sec(theta)

    w is the hyp, 1 is the adj side, & sqrt(w^2-1) is then the opposite side

    w -> /| <- sqrt(w^2-1)
    / |
    ---
    ^ (1)

    tan(theta) = sqrt(w^2-1)
    sec(theta) = w

    -1*5/49*[w*arccsc(w) + ln|w+sqrt(w^2-1)|]

    Final answer:
    -1*5/49*[csc(7x)*arccsc(csc(7x)) + ln|csc(7x)+sqrt(csc(7x)^2-1)|]


    So, I'm unsure as to why both of these answers were marked wrong (I've submitted them into webassign). Any help would be appreciated!

    http://img168.imageshack.us/img168/3822/mathwoesonwebassign5xcs.th.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 8, 2009 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    This looks like a problem to me; trig functions are periodic, and assign the same value to more than one input (eg. sin(0)=sin(pi)=0 ), arccsc(w) will return the so-called principle value only, and there is no guarantee that that value is equal to 'z' (it could be z+2pi or z+4pi...etc.)

    Instead, try applying integration by parts directly to [itex]\int -z\csc z\cot z dz[/itex] by using [itex]u=z[/itex] and [itex]dv=-\csc z\cot z dz[/itex]
     
    Last edited: Sep 8, 2009
  4. Sep 8, 2009 #3
    So:

    5*int[x*csc(7x)*cot(7x)]

    5/7*int[7x*csc(7x)*cot(7x)]

    let z = 7x
    dz = 7dx

    5/7 * 1/7 *int[z*csc(z)*cot(z)]
    5/49 * int[z*csc(z)*cot(z)]

    -5/49 * int[z*-csc(z)*cot(z)]

    ====== (Working the problem from your advice now) ======

    u = z dv = -csc(z)cot(z)
    du = dz v = csc(z)

    -5/49 *(z*csc(z) - int[csc(z)])

    -5/49 *(z*csc(z) - ln|csc(z) - cot(z)|)

    Plugging in for z.

    -5/49 *(7x*csc(7x) - ln|csc(7x)-cot(7x)|) + C

    Should be my final answer.... (Can anyone verify that please?)
     
    Last edited: Sep 8, 2009
  5. Sep 8, 2009 #4

    gabbagabbahey

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    Homework Helper
    Gold Member

    Looks good to me!:approve:
     
  6. Sep 8, 2009 #5
    Yup seems right :)

    Stay Up
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