High School Usage of variable in integration

[jk22]

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

$$df=\cos(\theta)\sin(u)d\theta+\sin(\theta)\cos(u)du$$ ?

 

[andrewkirk]

In that integral, $\theta$ is treated as a constant, so the change of variable formula is $df = \sin\theta\ \cos u\,du$

 

[scottdave]

What is the application? does θ depend on f ? Suppose θ is independent of f - then what would you expect dθ to be?

 

[jk22]

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 ?

 

[Mark44]

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.

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.