Find dx in a Complex Integration Problem: Anti-Derivative of Cos(2x)

In summary, to find the anti-derivative of (cos(2x)), the conversation suggests using u-du and u = 2x, du = 2dx, and dx = 1/2 du. For a more general case, u = f(x), du = f'(x) dx, and dx = du/f'(x) can be used. To find the anti-derivative of cos(cos(x)) ⋅ sin(x), the suggestion is to substitute u = cos(x).
  • #1
Whatupdoc
99
0
how would i find the anti-derv. of (cos(2x)). I am really confused with sub.

ok i'll use u-du

u = 2x
du = 2 dx
dx = 1/2 du

k so... cos(u)*dx

since the dx is there, you plug in the dx that i found right? so... sin(2x)1/2

ok so that's a simple integration problem, can someone give me a harder problem and show me how to find "dx", that's what I am really confused about.
 
Physics news on Phys.org
  • #2
So if in general:
u = f(x)
du = f'(x) dx
dx = du/f'(x)
Find the anti-derivative of:
cos(cos(x)) ⋅ sin(x)
and use:
u = cos(x)
as a substitution.
 
  • #3


To find dx in a complex integration problem, we need to use substitution. In this case, we are given the function cos(2x) and we want to find its anti-derivative.

First, we let u = 2x and du = 2dx. This is because the derivative of 2x is 2, which is the coefficient of x in our function.

Next, we solve for dx by dividing both sides by 2: dx = 1/2 du.

Now, we can substitute this value for dx in our original function: cos(2x)dx = cos(u)(1/2 du).

From here, we can integrate the function as usual, treating u as our variable and using the power rule. The result will be the anti-derivative of cos(2x) in terms of u.

Finally, we can substitute back in for u to get the anti-derivative in terms of x. In this case, the anti-derivative will be sin(2x) + C.

To demonstrate this process with a harder problem, let's find the anti-derivative of e^(3x).

Again, we start by letting u = 3x and du = 3dx. Solving for dx, we get dx = 1/3 du.

Substituting this value for dx in our original function, we get e^(3x)dx = e^u(1/3 du).

Integrating this function with respect to u, we get the anti-derivative 1/3 e^u + C.

Finally, substituting back in for u, we get the anti-derivative in terms of x: 1/3 e^(3x) + C.

In summary, to find dx in a complex integration problem, we use substitution to solve for dx in terms of du. Then, we can integrate the function with respect to du and substitute back in for our original variable to get the final anti-derivative.
 

Related to Find dx in a Complex Integration Problem: Anti-Derivative of Cos(2x)

1. What is the anti-derivative of Cos(2x)?

The anti-derivative of Cos(2x) is 1/2 Sin(2x) + C, where C is the constant of integration.

2. How do you approach finding dx in a complex integration problem?

To find dx in a complex integration problem, you can use substitution, integration by parts, or other techniques depending on the complexity of the problem. It is important to have a good understanding of basic integration rules and techniques.

3. Why is finding dx important in complex integration problems?

Finding dx is important because it allows us to solve for the integral and evaluate the anti-derivative. This helps us find the area under a curve, volume of a solid, or other quantities that can be represented by integrals.

4. Can you provide an example of finding dx in a complex integration problem?

Sure, for example, if we have the integral of 2x Cos(3x^2) dx, we can use the substitution u = 3x^2, du = 6x dx to rewrite the integral as 2/3 Cos(u) du. Then we can use the anti-derivative of Cos(u) to solve for the integral.

5. Are there any tips for finding dx in a complex integration problem?

One tip is to try different techniques and see which one works best for the given problem. It is also helpful to simplify the integrand as much as possible before attempting to find dx. Additionally, checking your answer using differentiation can help ensure the correctness of your solution.

Similar threads

  • Calculus and Beyond Homework Help
Replies
15
Views
850
  • Introductory Physics Homework Help
Replies
3
Views
769
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
28
Views
528
  • Calculus
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
558
  • Introductory Physics Homework Help
Replies
5
Views
686
  • Introductory Physics Homework Help
Replies
4
Views
5K
Back
Top