Method for finding non-obvious substitution in integration

In summary, to find the integral of sec(x), a clever substitution is used to eliminate the x variable and be left with u and du. One method involves using partial fractions and another involves a more obvious choice for the expression of 1. However, the key is to look for a way to manipulate the integrand into an equivalent but more complicated fraction to see if the resulting numerator is the derivative of the resulting denominator, making the problem easier to solve. This method can be applied to other situations where a substitution term is not immediately obvious.
  • #1
DocZaius
365
11
To find the integral of the sec(x), you have to substitute a term that is not immediately obvious.

[tex]\int sec(x) dx = \int sec(x) \frac{sec(x)+tan(x)}{sec(x)+tan(x)} dx[/tex]

[tex]u= sec(x)+tan(x)[/tex]
[tex]du= (sec(x)tan(x)+sec^{2}(x))dx[/tex]

[tex]\int sec(x) \frac{sec(x)+tan(x)}{sec(x)+tan(x)} dx = \int \frac{sec^{2}(x)+sec(x)tan(x)dx}{sec(x)+tan(x)} dx = \int \frac{du}{u} = ln |u| + C = ln |sec(x)+tan(x)| + C[/tex]

My question is: What method is used to find this clever substitution term? All the online resources tell me what the term is, but not how they arrived at it. I understand that the goal is to eliminate the x variable and be left with u and du, but for some reason when I try to "reverse engineer" the process, I cannot arrive at u. I ask for a method that would work with other situations where a substitution term is not immediately obvious.

Thanks!
 
Last edited:
Physics news on Phys.org
  • #2
I'm not so sure there is an answer to that question. I think there is really no "method" for coming up with such a clever choice for an expression of 1 but that it was just that...a very clever choice that happens to work!
 
  • #3
yep, calculus itself originated with trial and error, then come up with proofs to describe
 
  • #4
Here's another method to do this integral which is a bit more obvious:
[tex]\int sec(x) dx =\int \frac{dx}{cos(x)} =\int \frac{cos(x) dx}{cos^{2}(x)}=\int \frac{cos(x) dx}{1-sin^{2}(x)}=\int \frac{d(sin(x))}{1-sin^{2}(x)}[/tex]

Now let's use partial fractions on [tex]\frac{1}{1-sin^{2}(x)}[/tex], by letting [tex]\frac{1}{1-sin^{2}(x)}=\frac{A}{1+sin(x)}+\frac{B}{1-sin(x)}[/tex].

[tex]A(1-sin(x))+B(1+sin(x))=1[/tex], so [tex](A+B)+(B-A)sin(x)[/tex] and [tex]A=B=\frac{1}{2}[/tex].

Thus we have [tex]\int sec(x) dx =\int (\frac{1}{2}\frac{d(sin(x))}{1+sin(x)}+\frac{1}{2}\frac{d(sin(x))}{1-sin(x)})=\int (\frac{1}{2}\frac{d(1+sin(x))}{1+sin(x)}-\frac{1}{2}\frac{d(1-sin(x))}{1-sin(x)})[/tex].

From there the problem is trivial.
 
  • #5
Or substitute x = pi/2 - t so that you get the integral of 1/sin(t) and then a method analogous to what Lugita15 used is even simpler:

1/sin(t) = 1/[2 sin(t/2) cos(t/2)] =

[cos^2(t/2) + sin^2(t/2)]/[2 sin(t/2)cos(t/2)] =

1/2 [cot(t/2) + tan(t/2)]

which is trivial to integrate.
 
  • #6
This is probably one of the sneakiest tricks I encountered when doing integrals (apparently wikipedia thinks integrating sec^3(x) is tricky - they devoted an entire article to that integral - but this feels way trickier). The particular multiplication by a factor that is equal to 1 does not seem to be the important point here. When I encountered this integral after having seen the trick once, I forgot the exact factor that was needed. But I remembered that really one of the main reasons for turning an integrand into an equivalent but more complicated fraction is to see if you can get the resulting numerator to be the derivative of the resulting denominator. If you succeed in doing this, you know that a particular antiderivative will basically be the natural log of that resulting denominator. I think keeping this general idea in mind is more important than its usefulness in this particular example. For instance, there are many integrals that seem to require the use of partial fractions, but if you can manipulate the integrand in the manner just described, you can make the problem a lot easier (especially if you hate dealing with partial fractions like I do).
 

1. What is a "Method for finding non-obvious substitution in integration"?

The Method for finding non-obvious substitution in integration is a mathematical technique used to simplify and solve complex integrals by making a substitution of variables. It involves identifying a new variable that can be substituted into the integral to make it easier to solve.

2. Why is it important to find non-obvious substitutions in integration?

Finding non-obvious substitutions in integration can greatly simplify the process of solving integrals, especially for complex functions. It allows for a more efficient and accurate solution to be obtained, which is important in many areas of science and engineering.

3. How is the method for finding non-obvious substitution in integration different from regular substitution?

The method for finding non-obvious substitution in integration differs from regular substitution in that it involves identifying a new variable that may not immediately seem related to the original integral, but can be substituted in a way that simplifies the integral. Regular substitution involves replacing a variable with a known value or function.

4. What are some common techniques for finding non-obvious substitutions in integration?

Some common techniques for finding non-obvious substitutions in integration include using trigonometric identities, completing the square, and using u-substitution. These techniques can help to identify a new variable that can be substituted into the integral to make it easier to solve.

5. How can the method for finding non-obvious substitution in integration be applied in real-world situations?

The method for finding non-obvious substitution in integration can be applied in many real-world situations, such as in physics, engineering, and economics. It can be used to solve a variety of integrals that arise in these fields, making it a valuable tool for scientists and researchers.

Similar threads

  • Calculus
Replies
4
Views
804
  • Calculus
Replies
6
Views
1K
Replies
1
Views
945
Replies
2
Views
920
Replies
2
Views
829
Replies
13
Views
2K
Replies
6
Views
2K
Replies
14
Views
1K
Replies
2
Views
1K
Back
Top