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Integration (missing link)

  • Thread starter Kawakaze
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  • #1
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Homework Statement



Determine

[tex]\int sec(2x) dx[/tex] for interval [tex]\frac{3\pi}{8}[/tex] - [tex]\frac{\pi}{2}[/tex]


The Attempt at a Solution



I get

[tex]\frac{1}{\pi/2} ln (sec(\frac{\pi}{2}x+tan(\frac{\pi}{2}x))-\frac{1}{3\pi/8} ln (sec(\frac{3\pi}{8}x+tan(\frac{3\pi}{8}x))[/tex]

I am unsure how to simplify this further, wolphram alpha gives a much more elegant

http://www.wolframalpha.com/input/?....**Integral.rangestart-.*Integral.rangeend---
 

Answers and Replies

  • #2
dextercioby
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Well, the hyperbolic functions enter this as follows

[tex] \int \frac{dx}{\cos 2x} = \frac{1}{2} \int \frac{d(2x)}{\cos 2x} = \frac{1}{2} \int \frac{\cos 2x \ d(2x)}{\cos^2 2x} = \frac{1}{2} \int \frac{d(\sin 2x)}{1-\sin^2 2x} [/tex]

Ok, but now you know that

[tex] (\coth^{-1} x)' = \frac{1}{1-x^2} [/tex]

So that's how you get the argcotanh function in the final answer.
 
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  • #3
33,503
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Homework Statement



Determine

[tex]\int sec(2x) dx[/tex] for interval [tex]\frac{3\pi}{8}[/tex] - [tex]\frac{\pi}{2}[/tex]


The Attempt at a Solution



I get

[tex]\frac{1}{\pi/2} ln (sec(\frac{\pi}{2}x+tan(\frac{\pi}{2}x))-\frac{1}{3\pi/8} ln (sec(\frac{3\pi}{8}x+tan(\frac{3\pi}{8}x))[/tex]

I am unsure how to simplify this further, wolphram alpha gives a much more elegant

http://www.wolframalpha.com/input/?....**Integral.rangestart-.*Integral.rangeend---
For your antiderivative you should have gotten (1/2)ln|sec(2x) + tan(2x)|. You can check this by differentiating to get your original integrand, sec(2x).

To evaluate your definite integral, evaluate (1/2)ln|sec(2x) + tan(2x)|at pi/2 and 3pi/8. The answer you showed above should not still have x in it.
 
  • #4
tiny-tim
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Hi Kawakaze! :smile:

Alternative formula which may shorten the calculation slightly:

sec2x + tan2x = (1 + sin2x)/cos2x

= (1 + cos(π/2 - 2x))/sin(π/2 - 2x)

= 2cos2(π/4 - x)/2sin(π/4 - x)cos(π/4 - x) = cot(π/4 - x) :wink:


Another alternative formula:

d/dx tanh-1(cotx)

= -cosec2x/(1 - cot2x)

= 1/(cos2x - sin2x)

= 1/cos2x = sec2x,

so ∫sec2x dx = tanh-1(cotx) + C :smile:

(only meaningful for cotx < 1, ie tan x > 1, otherwise have to use tanh-1(tanx))
 
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  • #5
tiny-tim
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To match your elegant wolfram alpha result (1/2 coth-1(√2), = 1/2 tanh-1(1/√2)) :wink:
I am unsure how to simplify this further, wolphram alpha gives a much more elegant

http://www.wolframalpha.com/input/?....**Integral.rangestart-.*Integral.rangeend---
tanh-1(cotx) = 1/2 tanh-1(sin2x)

because sin2x = 2tanx/(1 + tan2x) = 2cotx/(cot2x + 1),

so if u = tanh-1(cotx), then cotx = tanhu, so sin2x = 2tanhu/(tanh2u + 1) = tanh2u


We can get this directly from

d/dx tanh-1(sin2x)

= 2cos2x/(1 - sin22x)

= 2sec2x,

so ∫sec2x dx = 1/2 tanh-1(sin2x) + C :smile:

the advantage of this over the tanh-1(cotx) solution is that the function tanh-1 is only defined between ±1, so we needed to say tanh-1(cotx) for cotx < 1 and tanh-1(tanx) for cotx > 1 … but |sin2x| is never > 1, so the single formula 1/2 tanh-1(sin2x) is valid (except for x = any odd multiple of π/4, which is also when sec2x = ∞)
 
  • #6
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Hi guys, thanks for the responses. Again, sorry but over my head! I did type the equation in wrong to begin with. The hyperbolic functions you talk about are not in this chapter of my course book, so I assume out of scope for this assessment, its my first week of noob calculus.

Standard integral is -
[tex]\frac{1}{a} ln (sec(ax)+tan(ax))[/tex]
(when sec(ax)+tan(ax)>0)

Are the [tex]\pi/2[/tex] & [tex]3\pi/8[/tex] to be plugged in as the constant or for x?
 
  • #7
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For a definite integral, you don't need to add the constant of integration. Substitute the two limits of integration into x.
 
  • #8
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Ok thanks! =) But that does raise more questions, the questions states "There is no need to evaluate your answer numerically"

Now what could that possibly mean? If i plug in the values then I am answering it numerically. I should have mentioned this sooner. It also says evaluate my answer with mathcad, and mathcad gives a numerical answer! =S
 
  • #9
dextercioby
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Then you either haven't posted the complete text of the problem, or the author of it is plain stupid. A definite integral will always have a numerical value...
 
  • #10
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Yeah I was wondering the same, i thought I just didnt get it. Here is the entire question -

Evaluate

[tex]\int sec(2x) dx[/tex]

(There is no need to evaluate your answer numerically.)
Verify your answer using Mathcad.

I cant get the right integration symbol, at the top its pi/2 and and at the bottom its 3pi/8

I thought it would be F(b) - F(a), where F is the indefinite integral with b or a plugged in.b being pi/2 and a being 3pi/8
 
  • #11
dextercioby
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A retard conceived the problem.

As for the LaTex issue, click here to find the code:

[tex] \int_{\frac{3\pi}{8}}^{\frac{\pi}{2}} \sec 2x \ dx = \mbox{Numerical value with or without mathcad} [/tex]

:D
 
  • #12
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A retard wrote the problem.
Haha that made my day, what a grin I have on my face. Thanks for the tex help too. Im gonna see if I cant find something in the errata, i doubt it but, yea...
 
  • #13
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nope nothing in the errata, could they mean , show the working but not the answer??? =S
 
  • #14
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Ok thanks! =) But that does raise more questions, the questions states "There is no need to evaluate your answer numerically"

Now what could that possibly mean? If i plug in the values then I am answering it numerically. I should have mentioned this sooner. It also says evaluate my answer with mathcad, and mathcad gives a numerical answer! =S
That means leave your answer in its exact form, in terms of ln(sec(3pi/4) + tan(3pi/4)) and ln(sec(pi) + tan(pi)) or whatever, without using a calculator to get an approximation of this stuff.

Then you either haven't posted the complete text of the problem, or the author of it is plain stupid. A definite integral will always have a numerical value...
A definite integral represents a number, but that number can be exact (such as [itex]\sqrt{2}[/itex]) or approximate (such as 1.414). The author wants the exact answer, not an approximation.
 
  • #15
dextercioby
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A definite integral represents a number, but that number can be exact (such as [itex]\sqrt{2}[/itex]) or approximate (such as 1.414). The author wants the exact answer, not an approximation.
Don't wanna be a jerk, but nowhere in the problem is there a mention of <approximate> values. The problem states: "Evaluate A (definite integral). (There is no need to evaluate your answer numerically.) [...]".

But evaluating (at least correctly, according to the commonly accepted mathematics) a definite integral will always yield a number, so his remark b/w the brackets is plain stupid.

That's the point I was trying to make.
 
  • #16
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Ok, lets have one last stab at this. Its only 3 marks and its 1.20am so i know where my priorities are at the minute. (:

[tex]\int_{\frac{3\pi}{8}}^{\frac{\pi}{2}} \sec 2x \ dx[/tex] = F(b) - F(a)

[tex] = (ln(sec(\frac{\pi}{2})+tan(\frac{\pi}{2})) - (ln(sec(\frac{3\pi}{8})+tan(\frac{3\pi}{8}))[/tex]
 
  • #17
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Don't wanna be a jerk, but nowhere in the problem is there a mention of <approximate> values. The problem states: "Evaluate A (definite integral). (There is no need to evaluate your answer numerically.) [...]".

But evaluating (at least correctly, according to the commonly accepted mathematics) a definite integral will always yield a number, so his remark b/w the brackets is plain stupid.

That's the point I was trying to make.
Evaluating the answer numerically and approximating it by the use of a calculator or computer mean the same thing.
 
  • #18
dextercioby
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Referring to Post 16 of the thread:

Hmm, but aren't sec(pi/2) and tan(pi/2) undefined in the reals ?? There must be a trick that you're missing, probably from the (2x) in the argument of the integrated function...;)
 
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  • #19
dextercioby
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Evaluating the answer numerically and approximating it by the use of a calculator or computer mean the same thing.
It's not the point I was making. I was addressing the semantics or logics behind the question, you were telling me that writing somewhere on a piece of paper sqrt(2) or 1.414 means the same thing.
 
  • #20
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It's not the point I was making. I was addressing the semantics or logics behind the question, you were telling me that writing somewhere on a piece of paper sqrt(2) or 1.414 means the same thing.
Not at all, and that's not what I said.

Suppose you do some integration that results in [itex]\sqrt{2}[/itex]. That is the exact answer. If you use some sort of device to evaluate this numerically, you will get 1.4142135623730950488016887242097 (maybe with more digits, maybe with fewer digits). That is an approximation.

If the instructions for the problem say that you do not have to evaluate your answer numerically or that you should not evaluate it numerically, then the right answer is [itex]\sqrt{2}[/itex], not 1.4142135623730950488016887242097.
 
  • #21
dextercioby
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I don't follow you at all. Evaluating numerically an integral simply means putting a number on the RHS of the following line

Integral =

If on the RHS you put instead of a number the following text: <A blue-eyed lion which hasn't eaten people lately>, you don't evaluate the integral numerically anymore. :))

In other words, putting sqrt(2) in the RHS of Integral = means evaluating the integral numerically, rightfully or not, depending of course if sqrt(2) is the correct NUMERICAL value of the integral.

I drop out now, I don't think we're on the same page...
 
  • #22
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The word "numerically" implies approximation. In the examples I used, we evaluated the integral to get [itex]\sqrt{2}[/itex]. If you leave the answer as a radical, you aren't evaluating the integral in the usual sense of "numerically". We evaluated it numerically to get 1.414.
 
  • #23
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Ok, lets have one last stab at this. Its only 3 marks and its 1.20am so i know where my priorities are at the minute. (:

[tex]\int_{\frac{3\pi}{8}}^{\frac{\pi}{2}} \sec 2x \ dx[/tex] = F(b) - F(a)

[tex] = \frac{1}{2}(ln(sec(\pi)+tan(\pi)) - \frac{1}{2}(ln(sec(\frac{3\pi}{4})+tan(\frac{3\pi}{4}))[/tex]
Uhm guys? That did make interesting reading but you kinda missed my reply tucked away at the end of page one :D

I think one thing is clear, if 2 people such as yourselves cant agree on the meaning of a question, then the question is written badly.
 
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  • #24
tiny-tim
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Hi Kawakaze! :smile:
Ok thanks! =) But that does raise more questions, the questions states "There is no need to evaluate your answer numerically"

Now what could that possibly mean? If i plug in the values then I am answering it numerically. I should have mentioned this sooner. It also says evaluate my answer with mathcad, and mathcad gives a numerical answer! =S
Mark44 :smile: is right …

they obviously mean you to evaluate sec and tan of π and 3π/4 because they're easy, but to leave any square-roots as they are

and they certainly don't expect you to evaluate the ln :wink:

(though if you had something like ln(√3), I'm sure they'd expect you to simplify it, to 1/2 ln(3))
 
  • #25
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Ok basically keep it tidy then? Like when you work with trig, and leave the root(3)/2 s as they are.

I submitted the paper today, so we shall see. Thanks again! I cant wait to see my result.
 

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