Trigonometric Substitution Problem

In summary, the conversation discusses a problem involving an integral with a variable coefficient. The solution involves using the substitution sin(theta) = 2x and drawing a right triangle to simplify the equation. The conversation also includes a joke about banking.
  • #1
Lancelot59
646
1
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!
 
Physics news on Phys.org
  • #2
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).
 
  • #3
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!
 
  • #4
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!
 
  • #5
I hit the minus sign by accident. :p
 
  • #6
I hope you don't get a job at the bank I go to :yuck:

(Just kidding!)
 
  • #7
Mark44 said:
I hope you don't get a job at the bank I go to :yuck:

(Just kidding!)

Eww...banking.:yuck:
 

1. What is trigonometric substitution?

Trigonometric substitution is a method used to simplify integrals involving expressions with radicals. It involves replacing the variable with a trigonometric function to make the integral easier to solve.

2. When should I use trigonometric substitution?

Trigonometric substitution is useful when the integral contains expressions with radicals, or when the integrand can be rewritten in terms of trigonometric functions.

3. How do I choose which trigonometric function to substitute?

The choice of trigonometric function to substitute depends on the form of the integral. The most commonly used substitutions are for integrals involving expressions with the form a^2 - x^2, a^2 + x^2, and x^2 - a^2.

4. What are the steps for solving a trigonometric substitution problem?

The steps for solving a trigonometric substitution problem are:
1. Identify the form of the integral and choose an appropriate trigonometric substitution.
2. Substitute the variable with the chosen trigonometric function and perform any necessary algebraic manipulations.
3. Use trigonometric identities to rewrite the integral in terms of a single trigonometric function.
4. Solve the resulting integral using basic integration techniques.

5. Are there any common mistakes to avoid when using trigonometric substitution?

One common mistake is forgetting to change the limits of integration when substituting the variable. It is also important to be careful when performing algebraic manipulations to avoid introducing errors. Additionally, it is important to choose the appropriate trigonometric substitution for the specific form of the integral.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
103
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
547
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
863
  • Calculus and Beyond Homework Help
Replies
34
Views
5K
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
362
  • Calculus and Beyond Homework Help
Replies
1
Views
819
  • Calculus and Beyond Homework Help
Replies
2
Views
641
Back
Top