Assistance with a very tough integral

In summary, the conversation was about evaluating the integral given in the Homework Equations section. The Attempt at a Solution involved integrating by parts and setting c=a+α, but the person was unsure of how to proceed. Other strategies proposed were working backwards from the solution and using a trig substitution. The person also mentioned additional restrictions on the variables involved.
  • #1
TheFerruccio
220
0

Homework Statement


Evaluate the integral.

Homework Equations


$$\int{\arccos{\frac{a}{a+\alpha}}\sqrt{\frac{(a+\alpha)^2}{(a+\alpha)^2-x^2}}d(a+\alpha)}$$

For reference, this is the solution, but I do not know how to get here:

$$\frac{a}{2}\ln{\frac{\xi+1}{\xi-1}} -\frac{x}{2}\ln{\frac{\xi+\frac{x}{a}}{\xi-\frac{x}{a}}}$$
where
$$\xi^2=\frac{(a+\alpha)^2-x^2}{(a+\alpha)^2-a^2}$$

The Attempt at a Solution


First step for me was to integrate by parts. I set $$c=a+\alpha$$ and integrated using c as my working variable.

After integrating by parts once, I end up with:

$$\left[\arccos{\frac{a}{c}}\sqrt{c^2-x^2}\right]_{boundary}-\int{\frac{a}{c}\sqrt{\frac{c^2-x^2}{c^2-a^2}}dc}$$

I am not sure what to do here. I was thinking of trying some sort of u substitution, maybe having $$u=\sqrt{\frac{c^2-a^2}{c^2-x^2}}$$.

There are additional restrictions on how these all relate to each other. For instance:

c > 0
a > 0
α > 0
c=a+α

Perhaps they can assist me with further limiting the scope of this integral and making it evaluate. Right now, if I try to put this integral into Mathematica, I end up with a statement which includes the Appell Hypergeometric function.

It should be noted that I also do not know what the limits of integration would be. Perhaps having the solution would give some insight into what the limits of integration are, but I do not see it. I have generated a stack of paper over the course of a week trying to figure out this integral, and I am getting absolutely nowhere. Some further assistance would be fantastic.

As a side note: Please tell me if my questions are being somehow vague on these forums. I would greatly like to improve the clarity of my questions for others. Given that every thread I have created in the past several months on this forum has been completely devoid of replies, I have to wonder whether I am fundamentally missing some key piece of information in my descriptions that results in scaring at the potential help away.
 
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  • #2
Have you tried working backward from the answer to see if you can see how to proceed?
 
  • #3
For this integral -- ##\int \frac{a}{c}\sqrt{\frac{c^2-x^2}{c^2-a^2}}dc## -- I'd be more inclined to try a trig substitution.
 
  • #4
Both of these are very good ideas. I will see if I can do these. I went to the professor and he said that it's probably in a table somewhere, so I do not think he did the algebra either. After seeing a square root of squares, I did default to thinking it must be some kind of triangle equality I could set up.
 
  • #5
Any time you have a sum or difference of squares, or the square root of a sum or difference of squares, trig substitution is a good strategy. Keep in mind here that x is kind of a red herring - the variable of integration is c.
 

What is an integral?

An integral is a mathematical concept that represents the area under a curve on a graph. It is used to find the exact value of an unknown quantity.

Why is it important to find assistance with a tough integral?

Integrals can be challenging and time-consuming to solve, especially when they are complex. Seeking assistance can help save time and ensure accuracy in solving the integral.

What are some techniques for solving tough integrals?

Some techniques for solving tough integrals include substitution, integration by parts, and partial fractions. It is also important to have a good understanding of basic integration rules and properties.

How can I check if my solution to a tough integral is correct?

One way to check if your solution is correct is by using a graphing calculator or software to graph the original function and compare it to the graph of your solution. Another way is to differentiate your solution and see if it gives you the original function.

Where can I find reliable assistance with a tough integral?

You can seek assistance from a math tutor, a study group, online forums, or educational websites. It is important to make sure the source is reputable and qualified to provide assistance with integrals.

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