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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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    Ah thanks. I need to do a better job of searching next time. ><2011-03-31

2 Answers 2

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$\begin{align} \int\sec x\;dx &=\int\sec x\cdot\frac{\sec x+\tan x}{\sec x+\tan x}\;dx \\ &=\int\frac{\sec^2x+\sec x\tan x}{\sec x+\tan x}\;dx \\ (\text{Letting }u=\sec x+\tan x&\text{ and }du=\sec x\tan x+\sec^2 x\;dx) \\ &=\int\frac{du}{u} \\ &=\log|u|+C \\ &=\log|\sec x+\tan x|+C \end{align}$

Now, the output I get from Mathematica is: $\begin{align} -\log(\cos\frac{x}{2}-\sin\frac{x}{2})+\log(\cos\frac{x}{2}+\sin\frac{x}{2}) &=\log\left(\frac{\cos\frac{x}{2}+\sin\frac{x}{2}}{\cos\frac{x}{2}-\sin\frac{x}{2}}\right) \\ &=\log\left(\frac{(\cos\frac{x}{2}+\sin\frac{x}{2})^2}{(\cos\frac{x}{2}-\sin\frac{x}{2})(\cos\frac{x}{2}+\sin\frac{x}{2})}\right) \\ &=\log\left(\frac{\cos^2\frac{x}{2}+\sin^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}}{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}\right) \\ &=\log\left(\frac{1+\sin(2\cdot\frac{x}{2})}{\cos(2\cdot\frac{x}{2})}\right) \\ &=\log\left(\frac{1+\sin x}{\cos x}\right) \\ &=\log(\sec x+\tan x) \end{align}$

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$\sec x\tan x=\frac{\sin x}{\cos^2 x}=-\frac{du}{dx}\frac{1}{u^2}$ where $u=\cos x$