A Function to Integrate

[Septim]

Which method do you use in integrating \int\((1-\beta^{2})sin(\varphi)d\varphi/(1-\beta^{2}(sin(\varphi)^{2})^{3/2} This integral is from Berkeley Physics Course Volume 2. Thanks in advance.

[LCKurtz]

What method would I use? Maple of course. Here's a link:

http://math.asu.edu/~kurtz/pix/integrated.pdf

[Septim]

Is it possible to do it by hand ? Because in the book it says you can try this as an exercise on integration, then I think it is possible to do it by hand. If so, can you please provide me the steps?

[LCKurtz]

Is it possible to do it by hand ? Because in the book it says you can try this as an exercise on integration, then I think it is possible to do it by hand. If so, can you please provide me the steps?

You might try changing the sin2(φ) in the denominator to 1 - cos2(φ) then letting u = cos(φ) and see what happens. That's where I would start if I were to try working it by hand, which I'm not.

[Septim]

I have tried it but I am still stuck. Is it the right way ?

[KingNothing]

I don't think anyone here will know the 'right' way off the top of their head. You might just have to go through and try all the methods you can think of. If you do get 'stuck', we can't help at all unless you post your work.

[coelho]

You might try changing the sin2(φ) in the denominator to 1 - cos2(φ) then letting u = cos(φ) and see what happens. That's where I would start if I were to try working it by hand, which I'm not.

This is the way to go. Also, for making the calculations a bit shorter, write:

a^2=(1-\beta^2)

If this substituitions were done right, the integral becomes:

\int \frac{-a^2 du}{(a^2+\beta^2 u^2)^{3/2}}

This can be integrated using the following trigonometric substituition:

\frac{\beta}{a} u = tan \theta

And this substituition leads to a very simple integral. (Just remember that 1+tan^2(x)=sec^2(x))

[Septim]

Thank you, I did not think of substituting
a^2=(1-\beta^2)
. By the way when I tried with different softwares the results were different. Maple, Microsoft Mathemetics etc. What may be the reason behind this?

[coelho]

Are you sure the "different" results arent different ways of writing the same thing?

If you were sure, could you post them all?

[Septim]

I am not exactly sure, but here is the integration done by Maple 14 (Hve a look at the attachment). By the way Microsoft Mathematics was unable to evaluate the integral this time.

[coelho]

Man... Maple is at the version number 14 and it still didnt learn how to simplify expressions? :frown:
I used to get this problem when i used Maple V R4 (long long ago), and rarely could use it to help me with the integrations. The only help i got from it was to know if the integral was possible to calculate or not. But the "hard work" had to be done by myself alone.

Anyway... i think this result Maple gave is some unsimplified version of the answer, which is:

\int \frac { \left( 1-\beta^2 \right) sin(\varphi) d\varphi }{ \left( 1-\beta^2 sin^2(\varphi) \right)^{3/2} } = \frac{-cos(\varphi)}{ \sqrt{1-\beta^2 sin^2(\varphi)} }

I didnt verify if this answer is equivalent to the one given by Maple, but this one can be shown to be right by derivating it.

[g_edgar]

if I say to simplify, Maple it does

[Septim]

Yes, it simplifies the expression when you click simplify in the context menu. However it does not simplify the expression: (1+\beta^2cos(\phi)^2-\beta^2) to (1-\beta^2sin(\phi)^2) Why does it not simplify such a simple expression?

[LCKurtz]

I have gotten so used to having Maple fail to simplify complicated expressions that I didn't even ask it to try on this problem. It certainly brings up the question of why Maple wouldn't automatically give the simplified answer instead of the multiline mess it gives if you don't ask.

Maybe it's Maple's version of "Don't ask, don't tell". :grumpy: