Integral help from S.R. and G.R. physicsforums

  • Thread starter morrobay
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In summary, the integral helps calculate the elapsed time on an accelerating clock. The substitution x= sin(u) makes it easy to integrate.
  • #1
morrobay
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Would someone evaluate this integral:
See Physics Forums -Special and General Relativity above.
On page one with title: Elapsed time on accelerating clock. by morrobay.
In particular reply # 4 by JesseM .
I seem to be missing something with The Integrator reference.
And one semester calculus , (Thomas ) is not enough.
thanks
 
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  • #2
Try a simple substitution, [tex]x=\frac{\cos u}{\sqrt{a}}[/tex]
 
  • #3
Integral help from S.R. and G.R physicsforums

JesseM said:
If you calculate the velocity as a function of time v(t) of C in the frame of A, then between two coordinate times [tex]t_0[/tex] and [tex]t_1[/tex] in A, the elapsed time on C would be [tex]\int_{t_0}^{t_1} \sqrt{1 - v(t)^2 / c^2} \, dt[/tex]. If the elapsed time in A's frame is 105.6 days = 9.124 * 10^6 s, and v(t) in A is 9.81*t m/s (so v(t)^2 = 96.24*t^2 m^2/s^2), then to calculate the elapsed time you'd evaluate the integral [tex]\int_0^{9.124 * 10^6} \sqrt{1 - 96.24*t^2 / 8.98755 * 10^{16}} \, dt[/tex] = [tex]\int_0^{9.124 * 10^6} \sqrt{1 - 1.071 * 10^
{-15} *t^2} \, dt[/tex]. You can enter Sqrt[1 - a x^2] into the integrator to see how to evaluate this integral.

If that substitution is simple for you or anyone else I would like to see it
 
  • #4
Then look in any introductory calculus book!

Since cos2(u)= 1- sin2(u), setting x= sin(u) is a "natural" choice for any integrand of the form [itex]\sqrt{1- x^2}dx[/itex]. With x= sin(u), that [itex]\sqrt{1- x^2}= \sqrt{1- sin^2(u)}= cos(u)[/itex] and dx= cos(u)du. [itex]\int \sqrt{1- x^2} dx= \int cos^2(u)du[/itex] which can be integrated using the trig identity "cos2(u)= (1/2)(1+ cos(2u))".

If you have integrand [itex]\sqrt{a^2- x^2}[/itex] just factor out the "a2": [itex]a\sqrt{1- x^2/a^2}[/itex] and obvious substitution is x/a= sin(u).
 
  • #5
[tex] \int \sqrt{ 1- ax^2} dx[/tex]

Make the substitution from the previous post.

Then [tex]dx = - \frac{\sin u}{\sqrt{a}}[/tex].

The integral is then transformed into;
[tex]\int \sqrt{ 1- a \cdot \frac{\cos^2 u}{a} } ( - \frac{\sin u}{\sqrt{a}}) du[/tex].

Constants may be taken out of an integral, and [tex] 1 - \cos^2 u = \sin^2 u[/tex] by the Pythagorean Identity.

[tex]-\frac{1}{\sqrt{a}} \int \sin^2 u du[/tex]

[tex]\sin^2 u = \frac{1 - \cos (2u)}{2} [/tex] which is verifiable by the double angle identity; [itex]\cos (2t) = \cos^2 t - \sin^2 t[/itex].

[tex]-\frac{1}{\sqrt{a}} \int \left( \frac{1}{2} - \frac{2 \cos (2u)}{4} du \right)[/tex]

Split the integral and in the second one, let w= 2u, so dw= 2 du.
[tex] = -\frac{1}{\sqrt{a}} \left( \frac{u}{2} - \frac{\sin (2u)}{4} \right)[/tex]

where [tex]x= \frac{\cos u}{\sqrt{a}}[/tex], or [tex] u = \arccos \left( \sqrt{a}u\right)[/tex]

EDIT: Just saw halls post, should have chosen his substitution because the signs work out more nicely, but this still works out fine.
 
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