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Trig. Integration - Please Help?

  • Thread starter p4nda
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  • #1
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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:
 
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Answers and Replies

  • #2
VietDao29
Homework Helper
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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? :)
 
  • #3
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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?
 
  • #4
VietDao29
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p4nda said:
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?
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.
 
  • #5
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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
 
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  • #6
VietDao29
Homework Helper
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p4nda said:
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])?
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. :)
I tried doing my posted problem and I got:
1/3csc^3[x] - cscx + C
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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