Trig Substitution vs. Regular Substitution for Indefinite Integrals

  • Thread starter Thread starter sapiental
  • Start date Start date
  • Tags Tags
    Confusion Trig
Click For Summary

Homework Help Overview

The discussion revolves around the evaluation of the indefinite integral of the function x(sqrt(4-x^2))dx, focusing on the appropriateness of using trigonometric substitution versus regular substitution.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the use of trigonometric substitution (x = 2sin(θ)) and its corresponding differential, while others consider a regular substitution (u = 4 - x^2) that seems simpler. There is a question about the validity of both approaches and whether they yield the same result.

Discussion Status

The conversation reflects differing opinions on the best substitution method to use. Some participants express confusion regarding the necessity of trigonometric substitution, while others affirm that both methods can be valid. Guidance has been offered to check if both substitutions lead to the same answer.

Contextual Notes

There is mention of a communication from a professor suggesting that the regular substitution is acceptable, which adds to the uncertainty in the discussion. Participants are also considering the implications of having x in front of the square root in the integral.

sapiental
Messages
110
Reaction score
0
Hi

If I have the indefinite integral x(sqrt(4-x^2))dx

must I replace x with 2sint and then dx = 2cosdt

or can I just do regular substitution with u = 4-x^2 since xdx occurs in the integral already.

when I solve it this way I get -1/3(4-x^2)^3/2 + C

it just seems a lot more work to do it the trigonometric substitituon way.

Any help is much appreciated. Thanks!
 
Physics news on Phys.org
you have to use the substitution [tex]x = 2\sin \theta[/tex].

Thus [tex]dx = 2\cos \theta d\theta[/tex].

At the end you have to solve for [tex]\theta[/tex] to convert the integral back in terms of x. you know that [tex]\sin \theta = \frac{x}{2}[/tex], so use a right triangle to express this relationship.
 
Last edited:
hey,

are you positive? My professor Just emailed me saying that the substitution u = 4-x^2 is acceptable.

Did you miss my x infront of the sqrt(4-x^2)dx ?

Really confused now..
 
There is nothing wrong with your own substitution.
You might check, though, that the two substitutions yield the same answer.
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 105 ·
4
Replies
105
Views
11K
Replies
4
Views
3K