How to find antiderivative of product

Click For Summary
SUMMARY

The discussion focuses on finding the antiderivative of the expression x^5 + tan(2x)sec(2x)dx. Participants emphasize the necessity of using u-substitution twice to simplify the integration process. One user suggests changing the trigonometric functions to sine and cosine before applying substitutions, specifically letting u = cos(x) and du = -sin(x). The conversation highlights the importance of analyzing the equation thoroughly before attempting manipulation to avoid unnecessary complexity.

PREREQUISITES
  • Understanding of antiderivatives and integration techniques
  • Familiarity with u-substitution in calculus
  • Knowledge of trigonometric identities, particularly secant and tangent functions
  • Basic skills in manipulating trigonometric expressions
NEXT STEPS
  • Study the process of u-substitution in calculus
  • Learn about trigonometric identities and their applications in integration
  • Practice finding antiderivatives of composite functions
  • Explore advanced integration techniques, such as integration by parts
USEFUL FOR

Students and educators in calculus, particularly those focusing on integration techniques and antiderivatives. This discussion is beneficial for anyone looking to enhance their understanding of trigonometric integration methods.

donjt81
Messages
70
Reaction score
0
problem: find the anti derivative of x^5 + tan(2x)sec(2x)dx

how do you find the anti derivative of the second half of that problem tan(2x)sec(2x)
 
Physics news on Phys.org
donjt81 said:
problem: find the anti derivative of x^5 + tan(2x)sec(2x)dx

how do you find the anti derivative of the second half of that problem tan(2x)sec(2x)

Does the function y=sec(u)tan(u) look familiar at all? Perhaps as the derivative of some common elementary function?
 
yeah, you are going to have to use u-substitution twice. I don't know if I am correct or not, but I changed everything to sin and cos before anything and manipulated it that way. I then let u = cosx and du = -sinx. For the second u-substitution, I let v=u²-1 and (1/2)dv = udu. I hope that helps and does not confuse you and more.
 
prace said:
yeah, you are going to have to use u-substitution twice. I don't know if I am correct or not, but I changed everything to sin and cos before anything and manipulated it that way. I then let u = cosx and du = -sinx. For the second u-substitution, I let v=u²-1 and (1/2)dv = udu. I hope that helps and does not confuse you and more.

What are you doing? What is the derivative of secant?
 
oh yeah... I see that the deriveitive of secx = secx tanx... Hmm... looks like I went a long.. long long long way around it.. haha. Sorry about that donjt81, I hope I didn't lead you too far the wrong way. I think it comes out to the same answer. Does it? Well, I guess we can both learn a lesson, or at least myself. And that is to really look at what's in front of you with the equation before just jumping into it. I just jumped right in and started to manipulate it, when I could have just taken a minute to think about what was really there and solved it much quicker and easier.
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
19
Views
2K
Replies
4
Views
2K
  • · Replies 48 ·
2
Replies
48
Views
5K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
2
Views
1K
Replies
15
Views
3K