Choosing the Right u to Integration by Substitution

In summary, when solving integrals by substitution, the key is to have a good understanding of which integrals can be solved in terms of elementary functions. Look for patterns and factorizations, and pay attention to what types of terms will result from taking derivatives. In the example given, the substitution u=1+z^3 was chosen because it simplifies the integral and makes it easier to solve. Experience is key in identifying the right substitution.
  • #1
sporus
21
0
i don;t have a specific homework question. i have a sort of conceptual question instead

when integrating by substitution, how do i know what to choose as u?

for example


integral of z^2 / (1 + z^3)^(1/3) dz

i am suppposed to choose u as 1+z^3. any other value for u won't give me the right answer. so my question is how do i know what to choose as u? i could have chosen something else but how do i choose the right one?
 
Physics news on Phys.org
  • #2
sporus said:
any other value for u won't give me the right answer.

It's not that other choices won't give the right answer, it's that other choices might not lead to integrals that you can do. The trick to identifying substitutions is having a firm grasp of what integrals can be solved in terms of elementary functions. You then look for substitutions that put the integral in question into one of those forms. There aren't really a set of rules to follow, it's something that comes about from experience.

However, your example is of the form

[tex]\int f(z) g(z) dz,[/tex]

where [tex]g(z)[/tex] can be observed to be proportional to [tex]f'(z)[/tex]. This type of factorization is one of the first things you look for.
 
  • #3
right, this is definitely something that comes with experience. You'll notice patterns that will help you choose u. For this example, since you know you will get (something)x^2 when you take the derivative of (something)x^3, you know that the term with the x^3 will want to include u. You include the 1+ because it turns a binomial into one term, which is much easier to integrate.

Most Calc 1 problems use simple patterns like these... Generally when you have any polynomial in parentheses raised to some power, you will want to see if you can set what's inside the parentheses as u first.
 
  • #4
ok, thanks
 

Related to Choosing the Right u to Integration by Substitution

1. What is integration by substitution?

Integration by substitution is a method of solving integrals in calculus. It involves replacing a variable in the integrand with a new variable, in order to simplify the integral and make it easier to solve.

2. How do you know when to use integration by substitution?

Integration by substitution is typically used when the integrand contains a complicated function or expression. It can also be used when the integrand contains a product of functions or a chain of functions.

3. What is the general process for integration by substitution?

The general process for integration by substitution includes: 1) identifying a substitution variable, 2) calculating the derivative of the substitution variable, 3) substituting the original function and its derivative into the integral, 4) simplifying the integral, and 5) solving for the original variable.

4. Are there any specific techniques for choosing a substitution variable?

Yes, there are a few techniques that can help in choosing a substitution variable. These include looking for a function and its derivative within the integrand, using trigonometric identities, and choosing a variable that will cancel out with another term in the integrand.

5. Are there any tips for solving difficult integrals using integration by substitution?

One tip for solving difficult integrals using integration by substitution is to practice and familiarize oneself with different substitution techniques. It can also be helpful to simplify the integrand as much as possible before choosing a substitution variable. Additionally, it is important to check the final answer by differentiating it to ensure accuracy.

Similar threads

  • Calculus and Beyond Homework Help
Replies
12
Views
1K
Replies
1
Views
646
  • Calculus and Beyond Homework Help
Replies
4
Views
233
  • Calculus and Beyond Homework Help
Replies
16
Views
1K
  • Calculus and Beyond Homework Help
Replies
27
Views
2K
Replies
23
Views
1K
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
700
  • Calculus and Beyond Homework Help
Replies
13
Views
2K
Back
Top