- #1
dodo
- 695
- 2
The stupid question of the day.
Is it fair to say that\frac{du}{dx} = \frac 1 {dx/du}<br />since this comes (I think) from the chain rule,<br /> \frac{dx}{du} \frac{du}{dx} = \frac{dx}{dx} = 1<br />
Which means that, when integrating by substitution, I can choose to do either of<br /> \int f(u) du = \int f(x) \frac {du}{dx} dx = \int \frac {f(x)} {dx/du} dx<br />depending on which derivative I happen to have at hand.
(Just checking; you can't be too careful when treating differentials as it they were fractions, which they aren't.)
The matter came out while studying Weierstrass' substitution, where<br /> \begin{align*}<br /> \int \frac 1 {\sin x} dx &= \int \frac {\frac{1+t^2}{2t}} {\frac{1+t^2}{2}}<br /> = \int \frac 1 t dt &\mbox{with }t=\tan \frac x 2<br /> \end{align*}<br />because \frac {dt}{dx} is easier to figure out than \frac {dx}{dt}.
Any more detail (particularly from the viewpoint of analysis) is welcome. Right now I feel like I'm just shuffling symbols without really knowing what am I doing.
Is it fair to say that\frac{du}{dx} = \frac 1 {dx/du}<br />since this comes (I think) from the chain rule,<br /> \frac{dx}{du} \frac{du}{dx} = \frac{dx}{dx} = 1<br />
Which means that, when integrating by substitution, I can choose to do either of<br /> \int f(u) du = \int f(x) \frac {du}{dx} dx = \int \frac {f(x)} {dx/du} dx<br />depending on which derivative I happen to have at hand.
(Just checking; you can't be too careful when treating differentials as it they were fractions, which they aren't.)
The matter came out while studying Weierstrass' substitution, where<br /> \begin{align*}<br /> \int \frac 1 {\sin x} dx &= \int \frac {\frac{1+t^2}{2t}} {\frac{1+t^2}{2}}<br /> = \int \frac 1 t dt &\mbox{with }t=\tan \frac x 2<br /> \end{align*}<br />because \frac {dt}{dx} is easier to figure out than \frac {dx}{dt}.
Any more detail (particularly from the viewpoint of analysis) is welcome. Right now I feel like I'm just shuffling symbols without really knowing what am I doing.