# Antiderivative of sec(x) [duplicate]

Using Mathematica to get the antiderivative for sec(x), I get $$-\log(\cos\frac{x}{2}-\sin\frac{x}{2})+\log(\cos\frac{x}{2}+\sin\frac{x}{2}).$$

This doesn't look familiar, so, I'm thinking there's probably some identity or other way to transform this...

Any insight would be appreciated.

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## marked as duplicate by Arturo Magidin, t.b., Américo Tavares, Aryabhata, Qiaochu YuanMar 31 '11 at 19:07

This falls under math.stackexchange.com/questions/29980/… –  Arturo Magidin Mar 31 '11 at 18:34
For $\sec(x)\tan(x)$, this is the derivative of $\sec(x)$. For $\sec(x)$ it's more complicated, but Weierstrass substitution works (in the worse case scenario). –  Arturo Magidin Mar 31 '11 at 18:38
@Arturo: I updated 29980 to include rational functions. I believe your current answer addresses that, but notifying you, just in case you think it might need editing. –  Aryabhata Mar 31 '11 at 18:48
@Moron: Thanks. –  Arturo Magidin Mar 31 '11 at 18:53
The Antiderivative of $\sec(x)$ was already asked in this question math.stackexchange.com/questions/6695/… "ways to evaluate integral sec" –  Américo Tavares Mar 31 '11 at 18:59

\begin{align} \int\sec x\;dx &=\int\sec x\cdot\frac{\sec x+\tan x}{\sec x+\tan x}\;dx \\ &=\int\frac{\sec^2x+\sec x\tan x}{\sec x+\tan x}\;dx \\ (\text{Letting }u=\sec x+\tan x&\text{ and }du=\sec x\tan x+\sec^2 x\;dx) \\ &=\int\frac{du}{u} \\ &=\log|u|+C \\ &=\log|\sec x+\tan x|+C \end{align}
Now, the output I get from Mathematica is: \begin{align} -\log(\cos\frac{x}{2}-\sin\frac{x}{2})+\log(\cos\frac{x}{2}+\sin\frac{x}{2}) &=\log\left(\frac{\cos\frac{x}{2}+\sin\frac{x}{2}}{\cos\frac{x}{2}-\sin\frac{x}{2}}\right) \\ &=\log\left(\frac{(\cos\frac{x}{2}+\sin\frac{x}{2})^2}{(\cos\frac{x}{2}-\sin\frac{x}{2})(\cos\frac{x}{2}+\sin\frac{x}{2})}\right) \\ &=\log\left(\frac{\cos^2\frac{x}{2}+\sin^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}}{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}\right) \\ &=\log\left(\frac{1+\sin(2\cdot\frac{x}{2})}{\cos(2\cdot\frac{x}{2})}\right) \\ &=\log\left(\frac{1+\sin x}{\cos x}\right) \\ &=\log(\sec x+\tan x) \end{align}
$\sec x\tan x=\frac{\sin x}{\cos^2 x}=-\frac{du}{dx}\frac{1}{u^2}$ where $u=\cos x$