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 $$\int_0^{9.124 * 10^6} \sqrt{1 - 96.24*t^2 / 8.98755 * 10^{16}} \, dt$$ = $$\int_0^{9.124 * 10^6} \sqrt{1 - 1.071 * 10^ {-15} *t^2} \, dt$$. 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.