Evaluating Integrals Using Trigonometric Function Substitutions Question

  • Thread starter GreenPrint
  • Start date
  • #1
1,196
0
Hi,

I just had this idea pop into my head... Can you use a trig sub with a reference triangle who has sides equal to zero? or more like a value close to zero such as dx or da or something?

For example

integral 1/sqrt(9+dx^2) (dx)^2

would have a reference triangle were the hypotenuse is sqrt(9+dx^2)
one of the legs would be 3
the other leg would be dx

you then could express dx/sqrt(9+dx^2) as either csc(theta) or sec(theta) depending on what leg you set 3 be equal to...

integral 1/sqrt(9+dx^2) (dx)^2 = integral csc(theta) dx
 
Last edited:

Answers and Replies

  • #2
1,196
0
I really wounder what can be done if one sets the side of the length of the triangle to dx and try and make a sub some how if that makes any sense
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,833
956
Hi,

I just had this idea pop into my head... Can you use a trig sub with a reference triangle who has sides equal to zero? or more like a value close to zero such as dx or da or something?

For example

integral 1/sqrt(9+dx^2) (dx)^2
This is an invalid integral. You cannot treat the "dx" as if it were a variable.
An integral in x must be of the form [itex]\int f(x)dx[/itex].

would have a reference triangle were the hypotenuse is sqrt(9+dx^2)
one of the legs would be 3
the other leg would be dx

you then could express dx/sqrt(9+dx^2) as either csc(theta) or sec(theta) depending on what leg you set 3 be equal to...

integral 1/sqrt(9+dx^2) (dx)^2 = integral csc(theta) dx
 

Related Threads on Evaluating Integrals Using Trigonometric Function Substitutions Question

  • Last Post
Replies
5
Views
2K
Replies
8
Views
2K
Replies
2
Views
1K
Replies
8
Views
1K
Replies
3
Views
3K
Replies
16
Views
788
Replies
34
Views
3K
Replies
16
Views
670
Replies
4
Views
1K
Top