I am confused why we can introduce differentials into an integral when performing an integration by substitution.
Consider the integral $$\int \frac{1}{ x \sqrt{1-x} } dx.$$ We can perform the substitutions $$x=\sin^2u,$$ $$dx=2\sin u \cos{u} du ,$$ on the integral to give $$\int \frac{1}{ \sin^2u \sqrt{1-\sin^2u} } 2\sin u \cos u du .$$
Why can you treat $dx$ as a differential?
From my understanding the integration sign $\int dx$ works as as if it is an operator, just like how $\frac {d}{dx}$ works as an operator and not as a fraction. Which means $\int$ and $dx$ should not be interpreted separately.
But at the same time treating $dx$ as a differential always work out fine so there must be some validness in treating it as an differential.
the $\frac {d}{dx}$ is just a notation of the differential
$$\frac {d}{dx} f(x) = \lim_{\delta x\to0}{\frac {f(x+\delta x)-f(x)}{\delta x}} $$ or we can say that :
$$\frac {d}{dx} f(x) = \lim_{\delta x\to0} {\frac {\delta f(x)}{\delta x}} $$
with $$ df(x) =\lim_{\delta x\to0} f(x+\delta x)-f(x)= \lim_{\delta x\to0} \delta f(x)$$ so the notation $\frac {d}{dx}$ is actually a fraction and not just a notation of an Operator.
by writing this: $$ g(x)=\frac {d f(x)}{dx} $$
if you know the function $g(x)$ and you looking for $f(x)$ we need firstly to know this $d f(x)$
we must multiply both side with $dx$ and get that : $$d f(x) = g(x) \space dx $$
so Now the time of the Inverse Operator of the " $d $ " the sum of Riemann is the same idea
$$\sum d f(x) = \sum g(x) \space dx = f(x) $$ it's not correct exactly because the sum must be continued, then we can renotated the continued sum as an Integral $\int$ :
$$\int d f(x) = \int g(x) \space dx = f(x) $$
so it's not just a symbole