Integration (1 Viewer)

Timothy.Siu · May 20, 2009

Petyo said:
No. that was the full question.
btw can you explain a bit more, what does " n=/=0" mean?

n=/= 0 means n cant equal 0

study-freak · May 20, 2009

Petyo said:
No. that was the full question.
btw can you explain a bit more, what does " n=/=0" mean?

if you see the working first part, towards the end, you can see n in the demominator.
Thus n cannot be 0.

eldore44 · May 20, 2009

tan(pi/2) can only be dealt with when you use tan^-1(tan(pi/2))

I was trying to use the first U(0) we get and minus the second U(0) we get.
ie $\frac{1}{2}\int_{0}^{\pi} \frac{2}{5-4cosx} dx = \frac{1}{5}\int_{\pi}^{0}\frac{4cosx}{5-4cosx}$
and then mess around with that.(I dont like latex)

lolokay · May 20, 2009

Petyo said:
No. that was the full question.
btw can you explain a bit more, what does " n=/=0" mean?

when doing the working you get to:
$-1/2 \int_{0}^{\pi }cos(nx)dx\\ = -1/2n [sin(nx)]|_{0}^{\pi }\\ but\ if\ n\ =0,\ you're\ dividing\ by\ zero,\ so\ n\neq 0\\ if\ n=0,\ cos(nx)=cos(0)=1\\ -1/2 \int_{0}^{\pi }cos(nx)dx\\ = -1/2 \int_{0}^{\pi }1dx\\ = -1/2[x]|_{0}^{\pi }\\ = -\pi/2\\ \neq 0$

Petyo · May 20, 2009

study-freak said:
if you see the working first part, towards the end, you can see n in the demominator.
Thus n cannot be 0.

Yea, true. Thanks

Petyo · May 20, 2009

If n=0 isn't it true that U (n) becomes U (0) = S 1/ (5-4cosx) dx [ from x =0 to x= pi] ?

Trebla · May 20, 2009

Drongoski said:
Great effort study-freak!

Per Timothu.Siu's suggestion, for U(0):

$\100dpi Let \ \ tan\frac{x}{2}\ =\ t \Rightarrow cosx\ =\ \frac{1-t^{2}}{1+t^{2}};\ \ dx\ =\ \frac{2dt}{1+t^{2}}\\ \\ U_{0}\ =\ \int_{0}^{\pi }\frac{1}{5-4cosx}dx\ =\ \int_{0}^{\pi } \frac{1}{5\ -\ 4(\frac{1-t^{2}}{1+t^{2}})}\times \frac{2dt}{1+t^{2}}\\ \\ =\ \ \int_{0}^{\pi }\frac{2}{1+9t^{2}}dt\ \ =\ \ \frac{2}{3}tan^{-1}(3t)]^{\pi }_{0}\ \ =\ \ \left [ \frac{2}{3}tan^{-1}(3tan\frac{x}{2}) \right ]^{\pi }_{0}\\ \\ =\ \ \frac{2}{3}[tan^{-1}(3\ tan\frac{\pi }{2})\ -\ tan^{-1}(3tan\ 0)]\ \ =\ \ \frac{2}{3}[\frac{\pi }{2}\ -\ 0]\ \ =\ \ \frac{\pi }{3}$

Drongoski, you forgot to change the limits as you made the substitution...

What can be done, is to find:
$\int_0^a \dfrac{1}{5-4\cos x}\,dx$
using a t-substitution and let a approach $\pi$

So your integral becomes:
$\dfrac{2}{3}\left[\tan^{-1}(3t)\right]_0^{\tan\frac{a}{2}}=\dfrac{2}{3}\tan^{-1}(3\tan\frac{a}{2})$
As a approaches $\pi$ , this expression approaches $\dfrac{\pi}{3}$

eldore44 · May 20, 2009

Petyo said:
If n=0 isn't it true that U (n) becomes U (0) = S 1/ (5-4cosx) dx [ from x =0 to x= pi] ?

yes.

study-freak · May 20, 2009

Trebla said:
Drongoski, you forgot to change the limits as you made the substitution...

What can be done, is to find:
$\int_0^a \dfrac{1}{5-4\cos x}\,dx$
using a t-substitution and let a approach $\pi$

So your integral becomes:
$\dfrac{2}{3}\left[\tan^{-1}(3t)\right]_0^{\tan\frac{a}{2}}=\dfrac{2}{3}\tan^{-1}(3\tan\frac{a}{2})$
As a approaches $\pi$ , this expression approaches $\dfrac{\pi}{3}[/B]$

How did you know this?

Timothy.Siu · May 20, 2009

Trebla said:
Drongoski, you forgot to change the limits as you made the substitution...

What can be done, is to find:
$\int_0^a \dfrac{1}{5-4\cos x}\,dx$
using a t-substitution and let a approach $\pi$

So your integral becomes:
$\dfrac{2}{3}\left[\tan^{-1}(3t)\right]_0^{\tan\frac{a}{2}}=\dfrac{2}{3}\tan^{-1}(3\tan\frac{a}{2})$
As a approaches $\pi$ , this expression approaches $\dfrac{\pi}{3}$

i dont think he forgot, although i guess the working out is incorrect with the limits.

he just intended to substitute t=tan x/2 back in after he integrated it so he wouldn't have to change the limits.

Trebla · May 21, 2009

study-freak said:
How did you know this?

You know that for y = tan^-1x, as x approaches infinity, y approaches π/2.
Now with tan^-1(3tan a/2)
as a-->π, tan a/2 --> ∞
and
3tan a/2 --> ∞
hence tan^-1(3tan a/2) --> π/2
so (2/3) tan^-1(3tan a/2) --> π/3

study-freak · May 21, 2009

Trebla said:
You know that for y = tan^-1x, as x approaches infinity, y approaches π/2.
Now with tan^-1(3tan a/2)
as a-->π, tan a/2 --> ∞
and
3tan a/2 --> ∞
hence tan^-1(3tan a/2) --> π/2
so (2/3) tan^-1(3tan a/2) --> π/3

I don't think this works because I remember doing one integration problem in two different methods and one method gave answers like Ctan^(-1) (ktan(pi/2)) where C and k were integral, C,k>1 while the other method gave out an actual real number like pi/3 or something like that.
When I tried using your method to evaluate the answer from the 1st method, I failed to produce the same answer as the one I obtained using the 2nd method (which was the t method). I was sure that there was no error with other calculations and I also had others to check it. I don't remember the question so I can't post it though.

Trebla · May 21, 2009

study-freak said:
I don't think this works because I remember doing one integration problem in two different methods and one method gave answers like Ctan^(-1) (ktan(pi/2)) where C and k were integral, C,k>1 while the other method gave out an actual real number like pi/3 or something like that.
When I tried using your method to evaluate the answer from the 1st method, I failed to produce the same answer as the one I obtained using the 2nd method (which was the t method). I was sure that there was no error with other calculations and I also had others to check it. I don't remember the question so I can't post it though.

I'm pretty sure that it should work in this case. In Ctan^(-1) (ktan(pi/2)), the answer (in the limit) would be Cpi/2...The method I posted above was the evaluation step of the t-substitution.

Petyo · May 21, 2009

Trebla said:
Drongoski, you forgot to change the limits as you made the substitution...

What can be done, is to find:
$\int_0^a \dfrac{1}{5-4\cos x}\,dx$
using a t-substitution and let a approach $\pi$

So your integral becomes:
$\dfrac{2}{3}\left[\tan^{-1}(3t)\right]_0^{\tan\frac{a}{2}}=\dfrac{2}{3}\tan^{-1}(3\tan\frac{a}{2})$
As a approaches $\pi$ , this expression approaches $\dfrac{\pi}{3}$

Even if the integral seems to approach (pi/3) when x approaches pi as you've stated above, pi/3 still can't be regarded as an answer can it?
In fact I find it hard to accept that tan (pi/2) really exists and can be used during the calculations.

Trebla · May 21, 2009

Yes, it can be regarded as an answer even though it just converges to the value (but doesn't equal until the "point of infinity"), so the limit exists. These are called improper integrals.

In fact, the definition of a definite integral also involves limits. i.e. the value you get when integrating something actually gives you a limit of infinitely small summations of rectangles that form an area.

The reason that limits were used is so that we can get close to the solution that contains the problematic tan(pi/2) without actually reaching the undefined expression. Since the limit converges, a finite value can be obtained...

Petyo · May 22, 2009

Thanks for your explanation.
btw how did you get to know all these things such as the "improper integrals"?

Trebla · May 23, 2009

If you intend to pursue a science degree at university, you'll cover them in more detail in first year maths units.

Petyo · May 31, 2009

my teacher said we can have tan (pi/2) = infinity (or "undefined" but "infinity" is preferred in this circumstance") and the result was pi/3

tommykins · May 31, 2009

Petyo said:
Thanks for your explanation.
btw how did you get to know all these things such as the "improper integrals"?

It is covered in the first session of mathematics for Engineering as well.

Basically, first year math in Uni defines nearly everything from scratch to prevent mathematical notation 'abuse'.

ie. cis@ is not used, neither is your tan(pi/2) = infinity, a more suitable statement is as x approaches pi/2, tan(x) approaches infinity

For further reading on Trebla's post - Riemann integral - Wikipedia, the free encyclopedia

Basically we calculate the area of 'n' amounts of rectangles from a function (ie. from point x to point x - a where a is the 'length' of the rectangle), the height can be determined by evaluating the function at that point. By taking the HIGHEST possible value of the rectangles, as well as the LOWEST - we apply a limit as the 'n' amount of rectangles approaches infinity, we can find the area under the curve.

I learnt it this year and it's quite genius.

Integration (1 Viewer)

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