Integrating Trigonometric Functions with Pythagorean Identities

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Homework Help Overview

The discussion revolves around integrating the function \(\int \frac{1}{d^2 + z^2}dz\) using trigonometric identities and relationships. The original poster references Pythagorean identities and a triangle setup to understand the integration process.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relevance of the triangle and trigonometric relationships to the integration problem. There are attempts to clarify how to approach the integral and the connection to the given variables.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and the integration process. Some guidance has been offered regarding the use of a right triangle and trigonometric substitutions, but no consensus has been reached on the best approach.

Contextual Notes

There is mention of a larger problem context that is not fully detailed, which may affect the understanding of the integral. Participants express confusion over the original problem's setup and the relationship between the variables involved.

geft
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Given

[tex]R = \sqrt{d^2 + z^2}[/tex]

and

[tex]\cos \alpha = \frac{d}{R}[/tex]

find

[tex]\int \frac{1}{d^2 + z^2}dz[/tex]

The answer given is

[tex]\frac{1}{d}\tan^{-1}(\frac{z}{d})[/tex]

I only know that I have to convert the denominator to secant squared, which integrates to a tangent. However, I don't know how the tangent inverse manages to get there, replacing the angle.
 
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You're question doesn't make too much sense. I mean if you're asked to find the antiderivative of a function (1/(d^2 + z^2)), what do R and cos alpha have to do with it ?
 
Without them, I don't think you'd able to tell which variable is on which side of the triangle, because the answer is in trigonometric form. Also, the question is part of a bigger problem, so please disregard the choice of variables used.
 
What triangle ? Why don't you post your problem's text along with your attempts to solve it ?
 
geft said:
Given

[tex]R = \sqrt{d^2 + z^2}[/tex]

and

[tex]\cos \alpha = \frac{d}{R}[/tex]

find

[tex]\int \frac{1}{d^2 + z^2}dz[/tex]

The answer given is

[tex]\frac{1}{d}\tan^{-1}(\frac{z}{d})[/tex]

I only know that I have to convert the denominator to secant squared, which integrates to a tangent. However, I don't know how the tangent inverse manages to get there, replacing the angle.

Use a right triangle to get the relationships in your trig substitution. In my triangle, the acute angle a (for alpha) is to the left, and the right angle is at the right. The base of the triangle is d and the altitude is z. The hypotenuse is sqrt(d^2 + z^2). You don't need R.

From this triangle, tan(a) = z/d, so d*tan(a) = z, and d*sec^2(a)da = dz. From the triangle you can also see that sec(a) = sart(d^2 + z^2)/d.

Make the appropriate substutions in your original integral, and find the antiderivative. When you are done, make sure you undo your substitution to get an antiderivative in terms of z.
 
@dextercioby: It's not exactly a problem, but a guided solution. The book just sort of skips right to the answer without working out the integration, which got me confused.

@Mark44: Many thanks!
 

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