Evaluating an integral with ${\mathrm{d}t}$ in the numerator, $\int \frac{\mathrm{d}t}{\cos(t)^2}$. How do I solve an integral with a differential on top? E.g.: given this integral to evaluate: $$\;\int \frac{dt}{(\cos(t))^2}\;\;?$$
What does it even mean when there's a differential?
 A: Ever since the time of Leibniz, the notation $\int f(t)\,dt$ has been used to denote the set of all functions whose derivative is $f(t)$.
The notation
$$\int \frac{dt}{(\cos t)^2}$$
is a space-saving version of
$$\int \frac{1}{(\cos t)^2}\,dt.$$  
We can save even more space by writing 
$$\int \frac{dt}{\cos^2 t}.$$
A: Differential notation is that - a notation. As such, as long as you're being unambiguous you can do just about whatever you want. If you want to get even fancier about it, recall that under Riemannian integration $\int f(x) dx$ is a limit of $\Sigma f(x) \delta x$. So while the $dx$ is no longer representing a real quantity, the notation comes out of something that was, and if you can rearrange the original expression in a particular way that doesn't really change the result, then you can probably get away with doing something similar to the integral.
For example, I've seen $\int dx f(x)$ used, in contexts where it's clear that the $f(x)$ is still part of the integral, and similarly in this case it's not a huge sin to write $\int 1/f(x) dx$ as $\int dx/f(x)$.
A: To address your notational question, the term $\,dt\,$ in an integral tells us the variable with respect to which we are integrating. Integrals should always include $d\alpha$, where $\alpha$ should be replaced by whatever variable with respect to which we are integrating. 
Writing $\,dt$ in the numerator is simply space-saving, and means no more and no less than the following:
$$\int \frac {dt}{(\cos t)^2} = \underbrace{\int \frac 1{\cos^2(t)}\,dt}_{\large = \int \frac 1{\left(\cos(t)\right)^2}\,dt}$$
And integrating: $$\int \frac 1{\cos^2(t)}\,dt = \int \sec^2(t) \,dt = \tan(t) + C$$
It always helps to "double check" an integral, if you have doubts about $\int \sec^2(t)\,dt$:
$$y = \tan(t) + C \iff$$ $$\dfrac{dy}{dt}\left(\tan(t) + C\right) = \sec^2(t) + 0 \iff dy = \sec^2(t) \,dt$$
A: Since the derivative of $\tan(t)$ is $\dfrac{1}{(\cos(t))^2}$, the integral is equal to $\tan(t)+C$.
