Dealing with the $dx$ that is in the denominator of a definite integral.

Question

I’m not sure if I should specify between a definite or an indefinite integral here but that’s the instance I’m specifically interested in.

I was thinking about an integration rule that states that if we have $\sin x$ raised to an odd power we let $u=\cos x$ and respectively with $\cos x$. I’m aware that this isn’t something that always works out but I tried to play with it.

Consider $$I=\int_{0}^{\frac{\pi}{4}}\tan x dx = \int_{0}^{\frac{\pi}{4}} \frac{\sin x}{\cos x}dx $$

This works out easily by letting $u=\cos x$ but letting $u=\sin x$

$$\int_{0}^{\frac{\sqrt 2}{2}} \frac{u}{du} $$

Can this be worked out? A comment here said we could maybe say $\frac{1}{dx}= \infty$

Do we just say that this is a bad substitution and move on? Is the answer to that integral different to what it would’ve been by doing the normal substitution?

Answer 1

$\int_0^{\pi/4} \frac{\sin{x}}{\cos{x}}dx$

Note $u=\sin{x}$ does not imply $du=\cos{x}$, but rather $du=\cos{x}dx$.That in turn implies $dx=\frac{du}{\sqrt{1-u^2}}$

So $\frac{\sin{x} dx}{\cos{x}}=\frac{u}{du/dx}dx=\frac{u}{du}dx^2=\frac{u}{du}\frac{du^2}{1-u^2}=\frac{udu}{1-u^2}$

$\int\frac{udu}{1-u^2}=(-1/2)\ln{(1-u^2)}+C=-\ln{(\cos{x})}+C $

When messing with differential equations you might actually find something like $du$ in the denominator, but you'll have to cancel it out to get a meaningful integrand. There might be more flexibility with numerical methods.