2
$\begingroup$

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:

Specifically, I am in the last lecture, doing the exercises.

But the professor, as far I remember, has not talked about these situations.

$\endgroup$
4
  • $\begingroup$ What happens when you differentiate $\sec{x}$ with respect to $x$? Remember that $\int f(x) dx = F(x) \iff F'(x) = f(x)$ (that, informally speaking, one should redirect oneself to the Fundamental theorem of calculus) $\endgroup$
    – Miguelgondu
    May 25, 2015 at 23:23
  • $\begingroup$ anyone got a geometric version? $\endgroup$
    – MichaelChirico
    May 25, 2015 at 23:43
  • $\begingroup$ @Miguelgondu Your statement is quite formal seeing as the symbol $\int$ without indices is literally defined as taking a primitive. $\endgroup$
    – GPerez
    May 26, 2015 at 0:13
  • $\begingroup$ @GPerez The $c$'s missing, though, :) $\endgroup$
    – Miguelgondu
    May 26, 2015 at 0:32

4 Answers 4

Reset to default
7
$\begingroup$

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,

$$ \int \sec(x)\tan(x) dx = -\int \frac{1}{u^2} du= \frac{1}{u} + C = \frac{1}{\cos(x)} +C = \sec(x) + C$$

$\endgroup$
6
$\begingroup$

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 $u = \cos x$.

$\endgroup$
6
  • $\begingroup$ I'm sorry, but I don't understand your question. $\endgroup$
    – Aaron Maroja
    May 25, 2015 at 23:32
  • $\begingroup$ Substitution, take $u = \cos x$. I'll edit it. $\endgroup$
    – Aaron Maroja
    May 25, 2015 at 23:34
  • 7
    $\begingroup$ I really don't recommend using $\cos^{-1}(x)$ notation to denote $\cos(x)^{-1}$. $\endgroup$
    – Cameron Williams
    May 25, 2015 at 23:34
  • 1
    $\begingroup$ @CameronWilliams Not to confuse with $\arccos x$, right? $\endgroup$
    – Aaron Maroja
    May 25, 2015 at 23:39
  • $\begingroup$ @AaronMaroja Yeah. I avoid using $\operatorname{trig}^{-1}$ notation myself but a lot of people don't. $\endgroup$
    – Cameron Williams
    May 25, 2015 at 23:41
3
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.