Integrating Trigonometric Functions with Pythagorean Identities

In summary: I didn't realize that the triangle was related to the substitution. I thought it was just a random given information. Now it makes more sense. In summary, to find the antiderivative of the function 1/(d^2 + z^2), given R = sqrt(d^2 + z^2) and cos alpha = d/R, use a right triangle to get the relationships in the trig substitution. From the triangle, tan alpha = z/d, so d*tan alpha = z and d*sec^2 alpha da = dz. Make the appropriate substitutions in the original integral and find the antiderivative. The final answer is 1/d * tan^-1 (z/d), where z and d are the sides
  • #1
geft
148
0
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.
 
Physics news on Phys.org
  • #2
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 ?
 
  • #3
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.
 
  • #4
What triangle ? Why don't you post your problem's text along with your attempts to solve it ?
 
  • #5
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.
 
  • #6
@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!
 

FAQ: Integrating Trigonometric Functions with Pythagorean Identities

What is trigonometry integration?

Trigonometry integration is a mathematical process used to find the area under a curve of a trigonometric function. It involves using trigonometric identities and techniques to solve integrals.

Why is trigonometry integration important?

Trigonometry integration is important because it allows us to solve problems related to curved shapes, such as finding the area of a circle or the volume of a sphere.

What are some common trigonometric identities used in integration?

Some common trigonometric identities used in integration include the Pythagorean identities, the double angle identities, and the power reducing identities.

What are the different techniques used in trigonometry integration?

The different techniques used in trigonometry integration include substitution, integration by parts, and using trigonometric identities to simplify the integral.

What are some real-life applications of trigonometry integration?

Trigonometry integration is used in many fields, such as physics, engineering, and architecture, to solve problems related to curved shapes and motion. It is also used in calculating areas and volumes in real-world scenarios.

Back
Top