Integral that I have been trying to solve

[yanic]

Hello,

Please check the attached file.
It's an integral that I have been trying to solve for a while.

I hope that I have been enough specific.


Regards

 

[mfb]

Do you know substitution?
sin(x)=u is a possible approach, and that method might work as well.

The first part is easy to integrate, just ignore it for now and focus on the fraction.

As this is a homework-like question, I moved your thread to our homework section.

 

[D H]

Do you know substitution?
sin(x)=u is a possible approach, and that method might work as well.

Substitution won't help much here. This is not integrable in the elementary functions. Ignoring the first part, which is easy, the result is a constant times the difference between an elliptical integral of the first kind and an elliptical integral of the second kind.

 

[D H]

Hint: Rewrite the numerator $-\sin^2(x)$ as $c-(\sin^2(x)+c)$.

 

[yanic]

Hint: Rewrite the numerator $-\sin^2(x)$ as $c-(\sin^2(x)+c)$.

Hi,
The numerator is a square root. I can't see how rewriting the numerator in such a way
will lead me to the expected result.
Please can you give further indications?

 

[yanic]

Substitution won't help much here. This is not integrable in the elementary functions. Ignoring the first part, which is easy, the result is a constant times the difference between an elliptical integral of the first kind and an elliptical integral of the second kind.

Hello,
I totally agree with you. Substitution in such a way won't work and the equation has nothing to do
with homework. I am trying to figure out a Mathematical model and I need to solve this part of the entire expression.

Regard

 

[D H]

Hi,
The numerator is a square root. I can't see how rewriting the numerator in such a way
will lead me to the expected result.
Please can you give further indications?

The *denominator* is a square root. The *numerator* is -sin(x)².

 

[yanic]

The *denominator* is a square root. The *numerator* is -sin(x)².

Sorry
Yeah I see. I have tried but that won't help either.
DId you get any result with that substitution?
The problem is the last term of the integration by part
that always contains at the denominator √([sin]^{2}(x)+c)

 

[D H]

Sorry
Yeah I see. I have tried but that won't help either.
DId you get any result with that substitution?
The problem is the last term of the integration by part
that always contains at the denominator √([sin]^{2}(x)+c)

Did you read my first post? You can try all the substitutions you want and you will *never* be able to find an analytic solution to this problem in the elementary functions. This problem is not integrable in the elementary functions. Fortunately, there are two special functions that are very applicable to this problem, the elliptical integrals of the first and second kind. The standard definitions of these functions are

$$\begin{aligned} F(\phi,k) &= \int_0^{\phi} \frac{d\theta}{\sqrt{1-k^2\sin^2(\theta)}} \\ E(\phi,k) &= \int_0^{\phi} \sqrt{1-k^2\sin^2(\theta)} \, d\theta \end{aligned}$$
Here, k is the "elliptic modulus" and must lie between 0 and 1. The restriction is reduced if we substitute k² with m:

$$\begin{aligned} F(\phi;m) &= \int_0^{\phi} \frac{d\theta}{\sqrt{1-m\sin^2(\theta)}} \\ E(\phi;m) &= \int_0^{\phi} \sqrt{1-m\sin^2(\theta)} \, d\theta \end{aligned}$$
Your integral, with appropriate modifications, can be made to look very much like these latter forms.

So what's the point? Simple: You can find mathematical tables the list these elliptical integrals and software that implements them.

 

[jackmell]

And what exactly you're going to do with it once you get it yanac? Now I don't want to get in D.H. way since I think he's doing a great job explaining to me and you how to solve this thing but seems to me there is more to it for example what happens when the expression in the root becomes negative or how exactly can I use those special functions to actually solve an integral and they do agree with numerical calculation right? Might want to check. Even if $1-m\sin^2(t)<0$? I'm not rightly sure. Just saying that's all.