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.

$\endgroup$
  • $\begingroup$ 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 '15 at 23:23
  • $\begingroup$ @Miguelgondu To be honest, I don't know. I just know that $sec(x) = \frac{1}{cos(x)}$. $\endgroup$ – nbro May 25 '15 at 23:25
  • $\begingroup$ anyone got a geometric version? $\endgroup$ – MichaelChirico May 25 '15 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 '15 at 0:13
  • $\begingroup$ @GPerez The $c$'s missing, though, :) $\endgroup$ – Miguelgondu May 26 '15 at 0:32
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 $u = \cos x$.

$\endgroup$
  • $\begingroup$ I didn't think to go backwards to confirm it. The problem is that I needed to find the original function $sec(x)$, and I am not seeing how one usually proceeds by integrating such a function $sec(x) \cdot tan(x)$ $\endgroup$ – nbro May 25 '15 at 23:31
  • $\begingroup$ I'm sorry, but I don't understand your question. $\endgroup$ – Aaron Maroja May 25 '15 at 23:32
  • $\begingroup$ My question is: if I didn't have the undefined integral $sec(x)$ (which is actually what I needed to find), how would you proceed to find the antiderivate of a function similar to $sec(x) \cdot tan(x)$? $\endgroup$ – nbro May 25 '15 at 23:33
  • 7
    $\begingroup$ I really don't recommend using $\cos^{-1}(x)$ notation to denote $\cos(x)^{-1}$. $\endgroup$ – Cameron Williams May 25 '15 at 23:34
  • 1
    $\begingroup$ @CameronWilliams Not to confuse with $\arccos x$, right? $\endgroup$ – Aaron Maroja May 25 '15 at 23:39
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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.