Integral of sec x: Solving Techniques and Tricks

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The integral of sec x can be expressed as ln|tan(x) + sec(x)| + C. A common technique to solve this integral involves multiplying sec(x) by a cleverly chosen form of 1, specifically sec(x)(sec(x) + tan(x))/(tan(x) + sec(x)). This manipulation transforms the integral into a simpler form, allowing it to be expressed as f'(x)/f(x), which is straightforward to integrate. The method proposed by shmoe is highlighted as particularly memorable and effective for this integral. Overall, these techniques streamline the process of integrating sec x.
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I know the integral of sec x is

ln|tan(x)+sec(x)|+C,

but how would you do it? I tried all the techniques and tricks I've learned and nothing came up.
 
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Did you try multiplying by 1?

sec(x)=\displaystyle sec(x)\frac{sec(x)+tan(x)}{tan(x)+sec(x)}=\displaystyle\frac{sec^{2}(x)+sec(x)tan(x)}{tan(x)+sec(x)}. Your integral is now of the form \displaystyle\frac{f'(x)}{f(x)}, easy to handle.
 
Yup. shmoe's method is the easiest one to remember.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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