Where am i doing it wrong(integration)?

  • Thread starter vkash
  • Start date
I don't know what it should be called. "Typesetting"? "Markup"? "Mark-up"? The computer language used to create the document, anyhow.
  • #1
vkash
318
1
Determine the value of integral from -pie to pie [2x(1+sin(x)]/[1+cos(x)*cos(x)]

sorry i can't provide it's answer because i am not able to tex tags i tried a lot but every time i am getting error. that doesn't print things as it should. see this(next line) where am i wrong
[tex]\frac{2x(1+sinx)}{1+cos^2x}[\tex]. did you see it's effect.?
My steps of solving question are.
(1) change x by -x(property)
(2)put cos(x)=t upper and lower limmit will became equal so it's answer should zero but it's answer is not zero.
It's answer is pie square. i have reached to pie square. How. After first step the integration can also be written as limit 0 to pie(do also twice of it). After that i solve it ans reached to answer pie square.
Afterall my question is why it' answer came zero if i put cos(x)=t in first case.

where can i found the pdf or htm file containing the way of using tex or any other such things.
 
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  • #2
The second tex marker should have a forward slash (/tex), not a back slash(\tex). I just took what you typed and changed the back slash. Then it works:

[tex]\frac{2x(1+sinx)}{1+cos^2x}[/tex].
 
  • #3
phyzguy said:
The second tex marker should have a forward slash (/tex), not a back slash(\tex). I just took what you typed and changed the back slash. Then it works:

[tex]\frac{2x(1+sinx)}{1+cos^2x}[/tex].

https://www.physicsforums.com/showthread.php?t=386951 I see all this from that pdf.

After all leave this tex itex issue and try to answer.
 
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  • #4
vkash said:
https://www.physicsforums.com/showthread.php?t=386951 I see all this from that pdf.
Near the top of the PDF there is an error in with the closing tex tag that is shown. I have reported this error, so it should be changed soon.
vkash said:
After all leave this tex itex issue and try to answer.

This is your integral:
[tex]\int_{-\pi}^{\pi}\frac{2x(1 + sin(x))~dx}{1 + cos^2(x)}[/tex]

By the way, the name of this Greek letter is "pi" not "pie."

Regarding the integral, I wouldn't use that substitution. Instead, I would rewrite the denominator, replacing cos2(x) by 1 - sin2(x). That would make the integrand
[tex]\frac{2x(1 + sin(x))}{-sin^2(x)}[/tex]

At this point you could split the integral into two integrals:
[tex]-2\int x~csc^2(x)dx - 2\int x csc(x)dx[/tex]

Integration by parts might work on these, but I haven't done the work, so don't know for sure.
 
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  • #5
Mark44 said:
Near the top of the PDF there is an error in with the closing tex tag that is shown. I have reported this error, so it should be changed soon.This is your integral:
[tex]\int_{-\pi}^{\pi}\frac{2x(1 + sin(x))~dx}{1 + cos^2(x)}[/tex]

By the way, the name of this Greek letter is "pi" not "pie."

Regarding the integral, I wouldn't use that substitution. Instead, I would rewrite the denominator, replacing cos2(x) by 1 - sin2(x). That would make the integrand
[tex]\frac{2x(1 + sin(x))}{-sin^2(x)}[/tex]

At this point you could split the integral into two integrals:
[tex]-2\int x~csc^2(x)dx - 2\int x csc(x)dx[/tex]

Integration by parts might work on these, but I haven't done the work, so don't know for sure.
great answer mark
but you forget to answer the point i raise.
I have also solved this question. But my question is that, that if you substitute cos(x) by t then upper and lower limits became equal in that case it's answer should zero(upper and lower limit are same). but it is not so.
thanks for reporting error.
 
  • #6
vkash said:
great answer mark
but you forget to answer the point i raise.
I have also solved this question. But my question is that, that if you substitute cos(x) by t then upper and lower limits became equal in that case it's answer should zero(upper and lower limit are same). but it is not so.
thanks for reporting error.

By the way: in tex you should use "\sin", etc., not "sin", etc. Here is the difference: [itex] \sin[/itex] (using "\sin") vs. [itex] sin [/itex] (using "sin"). This also applies to cos, tan, sec, csc, cotan, ln, log, exp, cosh, sinh, tanh and some others.

RGV
 
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  • #7
I haven't done the calculations, but the antiderivative of the integrand may not be expressible in terms of elementary functions. I might be wrong, though, so the OP is urged to try if the substitution [itex] t=\tan \frac{x}{2} [/itex] might solve

[tex]\int {x}\frac{1+\sin x}{2-\sin^2{x}} dx [/tex]
 
  • #8
If t = cos(x), then x = cos-1(t), provided that x [itex]\in[/itex] the interval [0, [itex]\pi[/itex]]. To change the limits of integration, they need to be in this restricted domain, and -[itex]\pi[/itex] isn't in it. That's the best explanation I can come up with.
 
  • #9
Ray Vickson said:
By the way: in tex you should use "\sin", etc., not "sin", etc. Here is the difference: [itex] \sin[/itex] (using "\sin") vs. [itex] sin [/itex] (using "sin"). This also applies to cos, tan, sec, csc, cotan, ln, log, exp, cosh, sinh, tanh and some others.
IMO, that's a matter of style. I don't see any advantage in using \sin etc. in LaTeX expressions over not using it.

Also, I try to use as little LaTeX as I can get away with. Pages with large amounts of LaTeX take an inordinate amount of time to render in my browser.
 
  • #10
Mark44 said:
IMO, that's a matter of style. I don't see any advantage in using \sin etc. in LaTeX expressions over not using it.

Also, I try to use as little LaTeX as I can get away with. Pages with large amounts of LaTeX take an inordinate amount of time to render in my browser.

Well, using \sin instead of sin looks much better when typeset in a document; it may not matter so much in a homework forum. However, pointing out the distinction to the OP cannot be harmful and might even be helpful. Certainly, when submitting papers to journals the publisher would likely insist upon correct usage.

RGV
 
  • #11
Mark44 said:
If t = cos(x), then x = cos-1(t), provided that x [itex]\in[/itex] the interval [0, [itex]\pi[/itex]]. To change the limits of integration, they need to be in this restricted domain, and -[itex]\pi[/itex] isn't in it. That's the best explanation I can come up with.

I like this explanation.:smile:

Your explanation arises another question that is
why 0 to [itex]\pi[/itex] why not -[itex]\pi[/itex] to 0 is taken as range of cos-1 function.
When i have read inverse trigonometric equation chapter it was written that, we take it from 0 to pi to make it a function. else it will not fit in definition of function.
There it was for just to make a function but that point is affecting integration's limit so i think i need to know more deeply.
 
  • #12
They (I don't know who) had to restrict the domain of the cos-1 function, and the interval [0, [itex]\pi[/itex]] was a reasonable choice. Any interval on which the function was one-to-one would have worked.
 

1. What is integration and why is it important in science?

Integration is the process of combining or merging different parts or components into a whole. In science, it is important because it allows us to understand complex systems and phenomena by analyzing their individual parts and how they interact with each other.

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Some common mistakes include not considering all relevant variables, using incorrect or outdated data, and not properly accounting for sources of error or uncertainty.

3. How can I improve my integration skills?

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