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You have to multiply with $\frac{\sec(x) + \tan(x)}{\sec(x) + \tan(x)}$ (http://math2.org/math/integrals/more/sec.htm), but how do you come up with this idea? Is there a specific reason for that step, or is it just mathematical intuition?

J. M. ain't a mathematician
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user3144065
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    MathJax hint: if you put a backslash before common functions, they come out in the right font and spacing, so \sec(x) gives $\sec (x)$ – Ross Millikan Oct 06 '16 at 14:06
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    It took a long time for that particular integral to be solved. So I'd say you really just have to guess it. But once you've seen it, you understand why it is good guess given you know the antiderivatives of $\sec^2(x)$ and $\sec(x)\tan(x)$. –  Oct 06 '16 at 14:07

2 Answers2

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Before I learned about $\sec(x)$, the way I used to integrate $\frac{1}{\cos x}$ is to multiply with $\frac{\cos x}{\cos x}$ and substitute $u = \sin x$. You may find that a bit more "natural".

trang1618
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Well, we want to multiply it by some $\frac{f(x)}{f(x)}$ so that $f'(x)=\sec(x)f(x)$ for the $u$-sub.

Let $f(x)$ be of the form $g(x)+h(x)$. We'd want to find some $g(x)$ and $h(x)$ such that $\sec(x)g(x)$ and $\sec(x)h(x)$ have known antiderivatives.

Hm... so what derivatives do we know of that involve $\sec(x)$ multiplied by something...?

Well, that's not particularly hard...

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

$$\frac{d}{dx}\tan(x)=\sec^2(x)$$

So we would have $g(x)=\tan(x)$ and $h(x)=\sec(x)$, giving us our $f(x)$.

Then the rest is easy.

Simply Beautiful Art
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