Help with an intermediate integral

In summary, the person is seeking help with evaluating the integral ## \int_{-a}^{a}\frac{1}{(z^2+a^2)(z^2+2a^2)^{1/2}} dz ##, which they have been attempting to solve without the use of tables or computer aids. They have tried using integration by parts and changing variables, but have been unsuccessful in finding a solution. They are seeking assistance in evaluating this integral.
  • #1

Homework Statement


I have been trying to evaluate an integral that has come up in the process of me solving a different problem, but am completely stuck. As I have confirmed with Wolfram Alpha that the integral once solved yields the correct solution to my problem. However, I am trying to better my ability to solve integrals without the help of tables or computer aids, so I am trying to find the correct solution to this one. Also, because my issue deals more with the math rather than the physics of the problem, I have included it here among the Math section. If this was inappropriate to put this here or if this is too little to prove that I have done enough work, please let me know and I will take this post down and include the full problem in the Physics section.[/B]

The integral is ## \int_{-a}^{a}\frac{1}{(z^2+a^2)(z^2+2a^2)^{1/2}} dz ##

The problem here simply is that every time I attempt this integral, the math becomes a "runaway train". I can use any help or hints with regards to how to fully evaluate this integral.

Homework Equations



## \int_{-a}^{a}\frac{1}{(z^2+a^2)(z^2+2a^2)^{1/2}} dz ##

The Attempt at a Solution


[/B]
I started off by using integration by parts and designating ##dv=[z^2+a^2]^{-1}## and ##u=[z^2+2a^2]^{-1/2}##. This led to the result of ##\frac{1}{a}(arctan(\frac{z}{a}))[z^2+2a^2]^{-1/2}\bigg|_{-a}^a +\int_{-a}^a\frac{z}{a}(arctan(\frac{z}{a}))[z^2+2a^2]^{-3/2} dz ##. When I try to evaluate the integral in this difference, that's when I go into a long never ending train of Integration by Parts. Any help with this integration would be must appreciated.
 
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  • #2
FallenLeibniz said:

Homework Statement


I have been trying to evaluate an integral that has come up in the process of me solving a different problem, but am completely stuck. As I have confirmed with Wolfram Alpha that the integral once solved yields the correct solution to my problem. However, I am trying to better my ability to solve integrals without the help of tables or computer aids, so I am trying to find the correct solution to this one. Also, because my issue deals more with the math rather than the physics of the problem, I have included it here among the Math section. If this was inappropriate to put this here or if this is too little to prove that I have done enough work, please let me know and I will take this post down and include the full problem in the Physics section.[/B]

The integral is ## \int_{-a}^{a}\frac{1}{(z^2+a^2)(z^2+2a^2)^{1/2}} dz ##

The problem here simply is that every time I attempt this integral, the math becomes a "runaway train". I can use any help or hints with regards to how to fully evaluate this integral.

Homework Equations



## \int_{-a}^{a}\frac{1}{(z^2+a^2)(z^2+2a^2)^{1/2}} dz ##

The Attempt at a Solution


[/B]
I started off by using integration by parts and designating ##dv=[z^2+a^2]^{-1}## and ##u=[z^2+2a^2]^{-1/2}##. This led to the result of ##\frac{1}{a}(arctan(\frac{z}{a}))[z^2+2a^2]^{-1/2}\bigg|_{-a}^a +\int_{-a}^a\frac{z}{a}(arctan(\frac{z}{a}))[z^2+2a^2]^{-3/2} dz ##. When I try to evaluate the integral in this difference, that's when I go into a long never ending train of Integration by Parts. Any help with this integration would be must appreciated.

Preliminary simplification is always worthwhile. If you call your integral ##F(a)##, changing variables to ##z = ay## gives
[tex] F(a) = \frac{1}{a^2} F(1), \\
\text{where } \;\; F(1) = \int_{-1}^1 \frac{dy}{(y^2+1)\sqrt{y^2+2}} = 2 \int_0^1 \frac{dy}{(y^2+1)\sqrt{y^2+2}} [/tex]
Submit this last integration to Wolfram Alpha (a free on-line version of Mathematical lite) to see what you get.

Alternatively, try the change of variable ##u = y/\sqrt{y^2+2}##.
 

1. What is an intermediate integral?

An intermediate integral refers to an integral that falls in between a basic integral and an advanced integral. It typically involves more complex functions and techniques compared to a basic integral, but may not require the use of advanced methods such as integration by parts or substitution.

2. How do I solve an intermediate integral?

Solving an intermediate integral requires a good understanding of basic integration techniques such as power rule, u-substitution, and trigonometric identities. It may also involve more advanced methods such as partial fractions and integration by parts. Practice and familiarity with different types of integrals will also help in solving intermediate integrals.

3. Can I use a calculator to solve intermediate integrals?

While some calculators may have a built-in function to solve integrals, it is important to note that these functions may not always be accurate and may not work for more complex integrals. It is always best to solve intermediate integrals by hand to ensure accuracy and a better understanding of the underlying concepts.

4. How do I know if I have solved an intermediate integral correctly?

There are a few ways to check if you have solved an intermediate integral correctly. One way is to differentiate your answer and see if it matches the original function. Another way is to use a graphing calculator to graph both the original function and your answer and see if they are the same. You can also try plugging in different values for the variable and see if the results match.

5. Are there any tips for solving intermediate integrals?

Yes, there are a few tips that can help with solving intermediate integrals. First, make sure to review and understand the basic integration techniques before attempting intermediate integrals. It is also helpful to break down the integral into smaller parts and use substitution to simplify the problem. Additionally, practicing with different types of integrals and seeking help from a tutor or teacher can also improve your skills in solving intermediate integrals.

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