Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicates:
Evaluating $\int P(\sin x, \cos x) \text{d}x$
Ways to evaluate $\int \sec \theta d \theta$

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.

share|improve this question
2  
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
2  
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
show 2 more comments

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2 Answers

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

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

share|improve this answer
add comment

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

share|improve this answer
add comment

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