Trigonometric Substitution Problem

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Homework Help Overview

The discussion revolves around a trigonometric substitution problem involving the integral \(\int{\sqrt{1-4x^{2}}}dx\). Participants explore methods to simplify the integral using trigonometric identities and geometric interpretations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts a substitution with \(x=\sin(\theta)\) but encounters difficulties due to the coefficient in front of the variable. They seek guidance on simplifying the integral further. Other participants suggest using a different substitution, \( \sin(\theta) = 2x\), and discuss the geometric interpretation of the problem using right triangles.

Discussion Status

Participants are actively engaging with the problem, offering different substitution methods and clarifying the geometric relationships involved. There is a mix of interpretations regarding the setup of the triangle and the application of the Pythagorean theorem, indicating a productive exploration of the topic.

Contextual Notes

Some participants express confusion over the correct interpretation of the triangle sides and the hypotenuse in relation to the integral, highlighting the need for clarity in the geometric approach. There is also a light-hearted exchange among participants, indicating a casual atmosphere in the discussion.

Lancelot59
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This problem looks relatively simple, but the coefficient in front of the variable is causing issues:

[tex]\int{\sqrt{1-4x^{2}}}dx[/tex]

So I started like this:

[tex]x=sin(\theta)[/tex]
[tex]dx=cos(\theta)d\theta[/tex]
[tex]\int{\sqrt{1-4sin^{2}(\theta)}cos(\theta)d\theta}[/tex]

Normally you can remove the constant from the root and go about the rest of the integral, but I'm stumped on this. How can I break this down into something more manageable? I've tried several identities but nothing has worked yet.

A pointer in the right direction would be great. Thanks in advance!
 
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Use the substitution sin(theta) = 2x. Rather than trying to remember which substitution to use, I draw a right triangle, labellings its sides according to the quantities in the radical. My triangle has a hypotenuse of 1, opposite side of 2x, and adjacent side of sqrt(1 - 4x^2).
 
So you just treat the root as the Pythagorean theorem? Makes sense.

To make sure I understand this fully, if my equation was [tex]\int{\sqrt{1+16x^{2}}}dx[/tex]

I would say that one of triangle's sides was 1, and the other was 4x? Making the hypotenuse the square root of 1-16x^2?

Okay, so going back to the original I can say that sin(theta) = 2x/1, and then isolate x to get x=sin(theta)/2. Putting that in it squares and cancels out the 4. Wonderful, thanks for the help!
 
Lancelot59 said:
So you just treat the root as the Pythagorean theorem? Makes sense.

To make sure I understand this fully, if my equation was [tex]\int{\sqrt{1+16x^{2}}}dx[/tex]

I would say that one of triangle's sides was 1, and the other was 4x?
Yes.
Lancelot59 said:
Making the hypotenuse the square root of 1-16x^2?
No, the hypotenuse would be sqrt(1 + 16x^2).
Lancelot59 said:
Okay, so going back to the original I can say that sin(theta) = 2x/1, and then isolate x to get x=sin(theta)/2. Putting that in it squares and cancels out the 4. Wonderful, thanks for the help!
 
I hit the minus sign by accident. :p
 
I hope you don't get a job at the bank I go to

(Just kidding!)
 
Mark44 said:
I hope you don't get a job at the bank I go to

(Just kidding!)

Eww...banking.
 

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