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

[morrobay]

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

 

[Gib Z]

Try a simple substitution, x=\frac{\cos u}{\sqrt{a}}

 

[morrobay]

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

If you calculate the velocity as a function of time v(t) of C in the frame of A, then between two coordinate times t_0 and t_1 in A, the elapsed time on C would be \int_{t_0}^{t_1} \sqrt{1 - v(t)^2 / c^2} \, dt. 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 \int_0^{9.124 * 10^6} \sqrt{1 - 96.24*t^2 / 8.98755 * 10^{16}} \, dt = \int_0^{9.124 * 10^6} \sqrt{1 - 1.071 * 10^<br /> {-15} *t^2} \, dt. 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

 

[HallsofIvy]

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 \sqrt{1- x^2}dx. With x= sin(u), that \sqrt{1- x^2}= \sqrt{1- sin^2(u)}= cos(u) and dx= cos(u)du. \int \sqrt{1- x^2} dx= \int cos^2(u)du which can be integrated using the trig identity "cos²(u)= (1/2)(1+ cos(2u))".

If you have integrand \sqrt{a^2- x^2} just factor out the "a²": a\sqrt{1- x^2/a^2} and obvious substitution is x/a= sin(u).

 

[Gib Z]

\int \sqrt{ 1- ax^2} dx

Make the substitution from the previous post.

Then dx = - \frac{\sin u}{\sqrt{a}}.

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

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

-\frac{1}{\sqrt{a}} \int \sin^2 u du

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

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

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

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

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