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As far as I understood, the Fundamental Theorem of Calculus states that the integral of a function is its anti-derivative. And yet, although the integral of $\tan x$ is $\sec x$, the derivative of $\sec x$ is $\tan x\sec x$. I understand the calculation and you get $\tan x\sec x$ as the derivative, but how does it make sense in light of the fundamental theorem? What am I missing here?

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    $\begingroup$ The integral of $\tan{x}$ is NOT $\sec{x}$. $\endgroup$ – David H Jun 22 '14 at 8:11
  • $\begingroup$ Well, embarrassingly enough, you are right... Thanks. $\endgroup$ – Uziya Jun 22 '14 at 8:29
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The actual anti-derivative of $\tan{x}$ is:

$$\int\tan{x}\,\mathrm{d}x=\int\frac{\sin{x}}{\cos{x}}\,\mathrm{d}x=\int\frac{-\mathrm{d}(\cos{x})}{\cos{x}}=-\ln{(\cos{x})}+\text{constant}.$$

This gives us the definite integral,

$$\int_{0}^{x}\tan{u}\,\mathrm{d}u=-\ln{(\cos{x})}=\ln{\left(\frac{1}{\cos{x}}\right)}=\ln{(\sec{x})}.$$

Applying the fundamental theorem of calculus to this integral gives us:

$$\begin{align} \frac{d}{dx}\int_{0}^{x}\tan{u}\,\mathrm{d}u&=\frac{d}{dx}\ln{(\sec{x})}\\ \implies \tan{x}&=\frac{\frac{d}{dx}(\sec{x})}{\sec{x}}. \end{align}$$

Multiplying both sides by $\sec{x}$ yields:

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

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