# Math Help - Integral secant and csc

1. ## Integral secant and csc

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

2. Originally Posted by RockHard
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
The trick is to multiply secant by $\frac{\sec x+\tan x}{\sec x+\tan x}$ and cosecant by $\frac{\csc x+\cot x}{\csc x+\cot x}$. The substitution for these integrals should then become obvious.

3. Originally Posted by RockHard
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 ...

$\int \sec{x} \, dx$

$\int \sec{x} \cdot \frac{\sec{x}+\tan{x}}{\sec{x}+\tan{x}} \, dx$

$\int \frac{\sec^2{x}+\sec{x}\tan{x}}{\sec{x}+\tan{x}} \, dx$

let $u = \sec{x}+\tan{x}$

$du = (\sec{x}\tan{x} + \sec^2{x}) \, dx$

substitute ...

$\int \frac{du}{u}$

$\ln|u| + C$

back substitute ...

$\ln|\sec{x}+\tan{x}| + C$

4. 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.

5. Originally Posted by skeeter
can't tell you how someone came up with it, but is sure is elegant ...

$\int \sec{x} \, dx$

$\int \sec{x} \cdot \frac{\sec{x}+\tan{x}}{\sec{x}+\tan{x}} \, dx$
Indeed!

Had I not been aware of this, I would have done it differently. The idea is similar, I would have multiplied by $\frac {\cos x}{\cos x}$. More intuitive, and a bit more work and certainly not as elegant. But you get to the same thing eventually.

6. or

$\sec(x) = \sec(x) \cdot \frac{\sqrt{ \sec^2(x)-1}}{\sqrt{ \sec^2(x)-1}}$

$= \frac{ \sec(x) \tan(x) }{\sqrt{ \sec^2(x)-1}}$

$= \frac{d(\sec(x))}{dx}\cdot \frac{1}{\sqrt{ \sec^2(x)-1}}$

$\int \sec(x) ~dx = \int \frac{d(\sec(x))}{\sqrt{ \sec^2(x)-1}}$

$=\cosh^{-1}(\sec(x)) + C$

7. Originally Posted by RockHard
.... I just wanted to know how do you manage or who decided this will work, as in the reasoning.
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.