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I was wondering why can't you apply the following when dealing with $\sec (x)$

$1$- convert $$\sec (x) = \frac{1}{\cos (x)}$$ 2- Integrate $$\int \frac{1}{\cos (x)} d(\cos x)= \ln \left\lvert \cos x \right\rvert.$$

I know the standard process of finding the integral of $\sec (x)$. I know that the aforementioned steps are incorrect. But, I want a mathematical reason as to why we don't proceed in that way.

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    $\begingroup$ $\int \frac{1}{\cos x}\, d(\cos x)=\ln|\cos x|+C\neq \int\frac{1}{\cos x}\, dx$. $\endgroup$ – user236182 Feb 26 '16 at 1:25
  • $\begingroup$ Just check of the meaning of your question is same after the edits. $\endgroup$ – user230452 Feb 26 '16 at 2:09
  • $\begingroup$ The comment by @user236182 pretty much answered the question. :) $\endgroup$ – Asker123 Feb 26 '16 at 2:28
  • $\begingroup$ $d(\cos x) = -\sin x + C \neq dx$ $\endgroup$ – user230452 Feb 26 '16 at 2:46
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Tangent half-angle substitution is also a simple way to solve the problem $t=\tan(\frac x2)$, $dx=2\frac{dt}{1+t^2}$, $\cos(t)=\frac{1-t^2}{1+t^2}$. Using all of that $$I=\int \frac{dx}{\cos(x)}=\int \frac{1+t^2}{1-t^2}\frac{2}{1+t^2}dt=2\int \frac{dt}{1-t^2}=\int \Big(\frac 1{1+t}+\frac 1{1-t}\Big)\,dt$$ $$I=\log\Big(\frac {1+t}{1-t}\Big)=\log\Big(\frac {1+\tan \left(\frac{x}{2}\right)}{1-\tan \left(\frac{x}{2}\right)}\Big)=\log\Big(\frac {\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)} {\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)}\Big)$$

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Note,

$$\frac{d}{dx} \sec (x)=\sec (x) \tan (x)$$

And,

$$\frac{d}{dx} \tan x=\sec^2 (x)$$

So adding results and factoring gives,

$$\frac{d}{dx}(\sec x+ \tan x)=\sec x(\sec x+\tan x)$$

Which means,

$$\frac{\frac{d}{dx}(\sec x + \tan x)}{\sec x + \tan x}=\sec x$$

But $\frac{1}{u}\frac{du}{dx}$ is the logarithmic derivative, it is equal to $\frac{d}{dx}\ln u$ by the chain rule. So we have,

$$\frac{d}{dx}(\ln (\sec x+\tan x))=\sec x$$

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The derivative of $\ln|\cos x|$ is $\frac{-\sin x}{\cos x}=-\tan x$. More generally, it's $\frac{f'(x)}{f(x)}$ that integrates to $\ln|f(x)|$, not $\frac1{f(x)}$. This is due to the chain rule.

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1) Multiply top and bottom by $$\sec(x) + \tan(x)$$

2) Let $u ~=~ \sec(x) + \tan(x)$

3) Notice that $\sec(x) \cdot (\sec(x) + \tan(x)) ~\mathrm d~x ~=~ \mathrm d~u$

4) You end up having to integrate $\frac{\mathrm d~u}u = \ln\lvert u\rvert$

Overall, you obtain $\ln \lvert\sec(x) + \tan(x)\rvert + C$

The reason why your process is incorrect is because you are using a wrong identity.

$$\int \frac{1}{x} dx = ln(x)$$

only holds in that form and no other form. You can't do

$$\int \frac{1}{f(x)} dx = ln\left\lvert f(x)\right\rvert$$

That is simply wrong.

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here is my totally different approach to your integral:

Verify $\int\sec x\ dx=\frac12 \ln \left\lvert\frac{1+\sin x}{1-\sin x}\right\rvert + C$

Just another way of looking at it. There are actually many ways to do this integral. Another nice way of looking at it , is to multiply top and bottom by $\cos{x}$ and use $\cos^2{x}=1-\sin^2{x}$ in the denominator. Then perform a u-sub. Can you see which one?

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