Why the anti derivative of $\sec(x) \cdot \tan(x)$ is $\sec(x)$? I have discovered that $$\sec(x) = \frac{1}{\cos(x)}$$ but I do not understand why the indefinite integral of $\sec(x) \cdot \tan(x)$ is $\sec(x)$.
I am watching the following videos:


*

*https://www.khanacademy.org/math/integral-calculus/indefinite-definite-integrals/indefinite_integrals/. 


Specifically, I am in the last lecture, doing the exercises.
But the professor, as far I remember, has not talked about these situations.
 A: Lets consider the indefinite integral,
$$ \int \sec(x)\tan(x) dx = \int \frac{\sin(x)}{\cos^2(x)} dx$$
We can then perform a $u$ substitution with $u=\cos(x)$ and $du = -\sin(x) dx $ obtaining,
$$ \int \sec(x)\tan(x) dx  = -\int \frac{1}{u^2} du= \frac{1}{u} + C = \frac{1}{\cos(x)} +C = \sec(x) + C$$
A: Hint: Here's something for you to think about. 
By the Chain Rule  $$\frac{d}{dx} \left(\cos (x)\right)^{-1} = (-1) (\cos (x))^{-2} \dot \, (-\sin (x)) = \frac{1}{\cos x} \dot\, \frac{\sin x}{\cos x}$$
Edit: In order to find $\int \sec x \tan x dx $. Write it as $$\frac{\sin x}{\cos^2 x}$$
and make the substitution $u = \cos x$.  
A: The expression $\sec x\tan x$ can be written
$$
\frac{1}{\cos x}\frac{\sin x}{\cos x}=\frac{\sin x}{\cos^2 x}
=-\frac{-\sin x}{\cos^2 x}=-\frac{f'(x)}{f(x)^2}
$$
where $f(x)=\cos x$. Consider, for a generic differentiable function $f$,
$$
g(x)=\frac{1}{f(x)}.
$$
By the chain rule
$$
g'(x)=-\frac{f'(x)}{f(x)^2}.
$$
In the special case of $f(x)=\cos x$, we see that
$$
g(x)=\frac{1}{\cos x}=\sec x
$$
There's really nothing more than this; it is what it is.
A: Here's a more elementary way of seeing it from the derivative of $\sec{\theta}$:
$\frac{\sec(\theta+\delta)-\sec\theta}{\delta}=\frac{\frac{1}{\cos(\theta+\delta)}-\frac{1}{\cos\theta}}{\delta}=\frac{cos\theta-\cos(\theta+\delta)}{\delta\cos\theta\cos(\theta+\delta)}$,
So
$\frac{d}{d\theta}(\sec{\theta})=\lim\limits_{\delta\rightarrow 0}\frac{\sec(\theta+\delta)-\sec\theta}{\delta}=\lim\limits_{\delta\rightarrow 0}-\frac{\cos(\theta+\delta)-\cos\theta}{\delta}\frac{1}{\cos\theta}\frac{1}{\cos(\theta+\delta)}$.
The first term is the (opposite of) the derivative of $\cos\theta$; the other two simply go to $\cos\theta$.
Since $\frac{d}{d\theta}(\cos\theta)=-\sin\theta$, we get the result.
For a nice geometric proof of the derivative of $\sin\theta$ which can easily be adjusted to prove the cosine derivative, see here.
