Can someone explain to me the substiution which is made for these integrals, I never find out why, it is they just just do it, and have no reasoning, perhaps a little insight could help
can't tell you how someone came up with it, but is sure is elegant ...
$\displaystyle \int \sec{x} \, dx$
$\displaystyle \int \sec{x} \cdot \frac{\sec{x}+\tan{x}}{\sec{x}+\tan{x}} \, dx$
$\displaystyle \int \frac{\sec^2{x}+\sec{x}\tan{x}}{\sec{x}+\tan{x}} \, dx$
let $\displaystyle u = \sec{x}+\tan{x}$
$\displaystyle du = (\sec{x}\tan{x} + \sec^2{x}) \, dx$
substitute ...
$\displaystyle \int \frac{du}{u}$
$\displaystyle \ln|u| + C$
back substitute ...
$\displaystyle \ln|\sec{x}+\tan{x}| + C$
Yes, I know you use that to arrive by substituting the denominator to get the natural log, however, I am one of those that want to know why rather than this is how you do it and the hows its goes with no reason. sorry to be a pain thanks for the input though. I just wanted to know how do you manage or who decided this will work, as in the reasoning.
or
$\displaystyle \sec(x) = \sec(x) \cdot \frac{\sqrt{ \sec^2(x)-1}}{\sqrt{ \sec^2(x)-1}}$
$\displaystyle = \frac{ \sec(x) \tan(x) }{\sqrt{ \sec^2(x)-1}}$
$\displaystyle = \frac{d(\sec(x))}{dx}\cdot \frac{1}{\sqrt{ \sec^2(x)-1}} $
$\displaystyle \int \sec(x) ~dx = \int \frac{d(\sec(x))}{\sqrt{ \sec^2(x)-1}} $
$\displaystyle =\cosh^{-1}(\sec(x)) + C $
who knows who came up with it? I don't.
math is like that sometimes. sometimes someone recognizes a pattern and applies it backwards to get somewhere. being very familiar with trig derivatives no doubt was a motivating factor here. seeing how that would work backwards might have led to such a method.
sometimes, you don't know where it came from or why someone would even think of that. yet you can't deny that it works, and that it is stunningly elegant.