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Homework Help: Trig. Integration - Please Help?

  1. Aug 6, 2006 #1
    http://img57.imageshack.us/img57/5647/integraljr9.jpg [Broken]

    Can anybody please help me explain how to do this step-by-step by using the techniques of integration?

    Thank you. :smile:
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Aug 6, 2006 #2

    VietDao29

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    There should be a chaper in your textbook about trigonometric integral. You can take a look at it.
    Generally, to integrate:
    [tex]\int \tan ^ {2n}x dx[/tex], or [tex]\int \cot ^ {2n}x dx[/tex], we can do as follow:
    Say, we want to integrate:
    [tex]\int \tan ^ 6 x dx = \int ( \tan ^ 6 x + \tan ^ 4 x - \tan ^ 4 x - \tan ^ 2 x + \tan ^ 2 x + 1 - 1) dx[/tex]
    [tex]= \int \tan ^ 4 x (\tan ^ 2 x + 1) dx - \int \tan ^ 2 x (\tan ^ 2 x + 1) dx + \int \sec ^ 2 x dx - \int dx[/tex]
    [tex]= \int \tan ^ 4 x \sec ^ 2 x dx - \int \tan ^ 2 x \sec ^ 2 x dx + \int \sec ^ 2 x dx - \int dx[/tex]
    [tex]= \int \tan ^ 4 x d(\tan x) - \int \tan ^ 2 x d(\tan x) + \int \sec ^ 2 x dx - \int dx[/tex]
    [tex]= \frac{\tan ^ 5 x}{5} - \frac{\tan ^ 3 x}{3} + \tan x - x + C[/tex].
    Ok, can you go from here? :)
     
  4. Aug 6, 2006 #3
    For the problem I posted, don't I have to switch "cot" to "cos/sin" (ratio identity) and then use the half-angle/double identities?
     
  5. Aug 6, 2006 #4

    VietDao29

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    If you change to sin, and cos, and work from there, I think you will be messed up. You'll get some expression quite 'ugly'. You can try to see if it works. :)
    Anyway, have you tried the way I suggested? Did you get the answer?
    In the example, I used the identity: sec2x = tan2x + 1.
    To tackle the problem you asked, you should note that: csc2x = cot2x + 1.
     
  6. Aug 6, 2006 #5
    I really appreciate your help, but I'm still kind of lost. I understand how to do the ones with a "6" exponential, but usually the ones with a "4" exponential confuses me. The integrals with a trig. function alone with an "even" exponential (e.g. Tan^[x]) is what gets me. However, I understand the integrals with two trig. functions (e.g. Tan^3[x]Sec^4[x]). For the integral with a single trig. function and a "4" exponential, are you supposed to split it into two (ex. Cot^2[x]Cot^2[x])?

    I tried doing my posted problem and I got:
    1/3csc^3[x] - cscx + C
     
    Last edited: Aug 6, 2006
  7. Aug 7, 2006 #6

    VietDao29

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    You can add, and then subtract cot2x, and 1 respectively. Like this:
    [tex]\cot ^ 4 x = \cot ^ 4 x + \cot ^ 2 x - \cot ^ 2 x - 1 + 1[/tex]
    Then we'll try to factor the expression, and we have:
    [tex]\cot ^ 4 x = \cot ^ 4 x + \cot ^ 2 x - \cot ^ 2 x - 1 + 1 = \cot ^ 2 x (\cot ^ 2 x + 1) - (\cot ^ 2 x + 1) + 1 = \cot ^ 2 x \csc ^ 2 x - \csc x + \fbox{1}[/tex].
    The main aim is to factor it so that csc2x appear, and we can then use the substitution:
    [tex]u = \cot x \Rightarrow du = - \csc ^ 2 x[/tex]
    It's the same as my example above. :)
    Be careful with the signs. And you are forgetting an x there. After factoring, you'll be left with a 1 (see the boxed part above). The answer should be:
    something + x + C
    Can you go from here? :)
     
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