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| mgralaska |
 Posted: Sat Feb 09, 2008 4:51 pm Post subject: antiderivate of csc^3 x |
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Forum Freshman
Joined: 09 Feb 2008
Posts: 5
Location: Alaska
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Hey everybody,
I was just searching the Internet, trying to find an answer to the indefinite integral of cosecant cubed of x and I came across this forum (although I couldn't find an answer here). This looks like a really neat site.
So does anybody know the steps involved to getting the correct answer for this problem? (I can look up the answer in my book; I just don't know how to get there, and it has me stumped.)
Thanks. |
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| serpicojr |
| Posted: Sat Feb 09, 2008 5:09 pm Post subject: |
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Forum Ph.D.
Joined: 17 Jul 2007
Posts: 871
Location: JRZ
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| I'm going to let Demen Tolden, our resident integration expert-in-training, handle this one. |
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| Demen Tolden |
| Posted: Sat Feb 09, 2008 10:18 pm Post subject: |
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Forum Senior
Joined: 19 Sep 2007
Posts: 387
Location: St. Paul, Minnesota, USA
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Thanks! I am so happy that I get to use my new found integration skills to help someone else!
I'm working on it now though, just give me some time. 
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| Demen Tolden |
| Posted: Sat Feb 09, 2008 10:54 pm Post subject: |
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Forum Senior
Joined: 19 Sep 2007
Posts: 387
Location: St. Paul, Minnesota, USA
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Ok, this problem is done almost exactly like you would do
∫ sec3 dx
Many calculus students seem to be using the same book as I am using, James Stewart's Calculus Early Transendentals. If this is the case there is an example that the book does in section 7.2 Trignometric Integrals, Example 8. The strategy, at least how I see it, is to set up your integral like this:
A = ∫ csc3 dx
and then evaluate your integral so you also have to subract A on the right side, like so:
Integration by parts gives:
u = csc x
dv = csc2dx
du = -csc x cot x
v = -cot x
A = -cot x csc x - ∫ csc x cot2x
Now you may notice that if you use the trig identity
cot2x + 1 = csc2x
you can change this equation to this:
A = -cot x csc x - ∫ csc x ( csc2 x - 1)dx
At this point I get excited and think "That's it! There is the other A I need contained in that second integrand!"
A = -cot x csc x - ∫ (csc3 - csc x)dx
A = -cot x csc x - A + ∫ csc x dx
2A = -cot x csc x + ln (cscx - cot x)
A = (1/2)(-cot x csc x + ln (csc x - cot x) )
I am glad to be of assitance sir.
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| mgralaska |
| Posted: Sun Feb 10, 2008 4:29 pm Post subject: |
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Forum Freshman
Joined: 09 Feb 2008
Posts: 5
Location: Alaska
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Thanks so much! For some reason I thought I had to only use a u and du substitution, so didn't even try using integration by parts. That works much easier.
If you are willing (and want) to try and solve another one, I've got another for you: integral of du / (u sqrt (5-u^2)). It is in the chapter where I am supposed to be using a trigonometric substitution to get rid of the integral, but I can't seem to do that because of the 5. I tried doing various other substitutions, but can't find any that work. |
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| Demen Tolden |
| Posted: Sun Feb 10, 2008 9:49 pm Post subject: |
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Forum Senior
Joined: 19 Sep 2007
Posts: 387
Location: St. Paul, Minnesota, USA
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You are going to be running into problems like this very often. They will be in the form of
∫ dx / (9-x2)
or just
∫ (9 - x2)1/2
or
∫ dx / (x2 + 6)
ect..
The pattern here is that the part of the integral that is difficult is a polynomial. The trick is to convert the polynomial into something that you can take the square root of such as sec2 or sin2. If you have to deal with something such as
∫ dx / ( x ( 5 - x2)1/2 )
You should find a substitution for x2 where you can factor out the 5. In this case we could try to turn
5 - x2
into
5(1 - sin2 Θ)
with just the right substitution for x. What I usually do is write:
x2 = 5 sin2 Θ
then solve for x:
x = 51/2 sin Θ
then we can find dx by differentiating since we have x alone
dx = 51/2 cos Θ dΘ
And now with this information, the integral becomes much easier to solve:
∫ 51/2 cos Θ (51/2 sin Θ 51/2 cos Θ)-1 dΘ
∫ 5-1/2 sec Θ dΘ
5-1/2 ln (sec Θ + tan Θ)
and then you can just convert back to x since you know that sin Θ = x / (51/2)
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| mgralaska |
| Posted: Tue Feb 12, 2008 3:20 am Post subject: |
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Posts: 5
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| Demen Tolden |
| Posted: Tue Feb 12, 2008 9:55 am Post subject: |
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Forum Senior
Joined: 19 Sep 2007
Posts: 387
Location: St. Paul, Minnesota, USA
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| Quote: |
∫ 51/2 cos Θ (51/2 sin Θ 51/2 cos Θ)-1 dΘ
∫ 5-1/2 sec Θ dΘ
5-1/2 ln (sec Θ + tan Θ)
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Oops. I think that should have been:
∫ 5-1/2 csc Θ
5-1/2 ln (csc Θ - cot Θ)
but the idea is still correct.
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| n77ler |
| Posted: Thu Mar 27, 2008 2:16 pm Post subject: |
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Joined: 27 Mar 2008
Posts: 6
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| Your answer is wrong. You don't have the constant of integration |
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| serpicojr |
| Posted: Thu Mar 27, 2008 2:31 pm Post subject: |
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| Selene |
| Posted: Thu Mar 27, 2008 3:10 pm Post subject: |
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Forum Ph.D.
Joined: 04 Feb 2008
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Location: I live in Bertrand Russells teapot!
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Demen
I think you are very brainy!!
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| n77ler |
| Posted: Tue Apr 01, 2008 1:42 pm Post subject: |
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Forum Freshman
Joined: 27 Mar 2008
Posts: 6
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| well you lose a lot of marks without it, sorry for pointing it out |
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| serpicojr |
| Posted: Tue Apr 01, 2008 4:53 pm Post subject: |
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| I wasn't objecting to you pointing out that the constant of integration was missing. Reminding people that antiderivatives are not unique is a worthwhile thing! Rather, I was reacting to you calling the answer "wrong" due this minor error. It's like you throwing your gin and tonic at me because I forgot the swizzle stick. |
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| JaneBennet |
| Posted: Mon Apr 07, 2008 6:43 am Post subject: |
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Forum Junior
Joined: 06 Apr 2008
Posts: 257
Location: UK
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| n77ler wrote: |
| well you lose a lot of marks without it, sorry for pointing it out |
Why should you lose “a lot of” marks just for not remembering the arbitrary constant? One or two marks maybe, but surely not “a lot of”? If you’ve just done a difficult integration problem faultlessly except for the arbitrary constant at the end, surely you would expect to get at least 8 or 9 marks out of 10 for your hard work – but getting say 5 or 6 out of 10 just because you forgot a minor detail seems contrary to common sense!
The point about integration questions is to test how well students understand integration and can apply integration techniques (substitution, integration by parts, etc) – not how anal-retentive they are over minor details.
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