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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.

Taenyfan
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  • $1-x>0\implies x<1$ But $x$ can be $<-1$ right? Better use $\sqrt{1-x}=y$ – lab bhattacharjee Jul 05 '18 at 08:30
  • @labbhattacharjee Hi, I was using a suggested substitution given in this post: https://math.stackexchange.com/questions/2841480/how-to-integrate-x-sqrt1-x-1. Anyway, the exact substitution isn't really the problem I am trying to understand. But thanks! – Taenyfan Jul 05 '18 at 08:32
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    The $\mathrm d x$ term in an integral should remind you of the $\Delta x$ in Riemann sums, which is really an increase of the integration variable and can be approximated by the differential. – Bernard Jul 05 '18 at 08:41
  • This is just a handy recipe. (Of course there is a proof behind it, using the chain rule.) – Christian Blatter Jul 05 '18 at 08:44

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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

El-Mo
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