Usage of variable in integration

In summary, the conversation discusses the application of a change of variable ##f=\sin(\theta)\sin(u)## in solving the differential equation $$\frac{df}{d\theta}=-\frac{\sqrt{\sin(\theta)^2-f^2}}{\theta^2}##. The group discusses whether the letters ##f## and ##\theta## should be considered independent or not. It is noted that if f is a function of ##\theta##, the derivative ##\frac{df}{d\theta}## will not be zero in general. The conversation also mentions an example where the change of variable does not satisfy the differential equation.
  • #1
jk22
729
24
When solving differential equations the following scripture can arise, for example:

$$\int \frac{df}{\sqrt{\sin(\theta)^2-f^2}}$$

If the change of variable ##f=\sin(\theta)\sin(u)##

Is performed, do the letters ##f,\theta## shall be considered independent or is

$$df=\cos(\theta)\sin(u)d\theta+\sin(\theta)\cos(u)du$$ ?
 
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  • #2
In that integral, ##\theta## is treated as a constant, so the change of variable formula is ##df = \sin\theta\ \cos u\,du##
 
  • #3
What is the application? does θ depend on f ? Suppose θ is independent of f - then what would you expect to be?
 
  • #4
I took an example supposing $$\int\frac{df}{\sqrt{\sin(\theta)^2-f^2}}=\frac{1}{\theta}$$

It comes from the differential equation $$\frac{df}{d\theta}=-\frac{\sqrt{\sin(\theta)^2-f^2}}{\theta^2}$$

If we put ##df=\sin\theta\cos udu## we can integrate to ##u=\frac{1}{\theta}\Rightarrow f=\sin\theta\sin(\frac{1}{\theta})##

But it does not satisfy the differential equation. Why is this ?
 
  • #5
jk22 said:
If the change of variable
##f=\sin(\theta)\sin(u)## Is performed, do the letters ##f,\theta## shall be considered independent or is
$$df=\cos(\theta)\sin(u)d\theta+\sin(\theta)\cos(u)du$$ ?
It would have been helpful to include the differential equation in post #1.
jk22 said:
It comes from the differential equation $$\frac{df}{d\theta}=-\frac{\sqrt{\sin(\theta)^2-f^2}}{\theta^2}$$
If f is a function of ##\theta##, the ##\frac{df}{d\theta}## won't be zero, in general.
 

Related to Usage of variable in integration

1. What is the purpose of using variables in integration?

Variables are used in integration to represent changing quantities or values in a mathematical equation. They allow us to solve for unknown values and make the integration process more flexible and efficient.

2. How do variables affect the accuracy of an integration?

The use of variables in integration can greatly affect the accuracy of the results. By choosing the appropriate variables, we can simplify the integration process and reduce the chances of making errors. However, using the wrong variables or too many variables can lead to incorrect results.

3. Can variables be used in both definite and indefinite integrals?

Yes, variables can be used in both definite and indefinite integrals. In definite integrals, variables are used to represent the limits of integration, while in indefinite integrals, they are used to represent the unknown constant of integration.

4. How do we choose the appropriate variables for integration?

The choice of variables for integration depends on the specific problem at hand. Generally, we want to choose variables that will simplify the equation and make it easier to integrate. It is also important to choose variables that are consistent with the limits of integration.

5. Can variables be used in integration for multivariable functions?

Yes, variables can be used in integration for multivariable functions. In this case, we would use multiple variables to represent the different independent variables in the function. This allows us to integrate over multiple dimensions and solve for the appropriate values.

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