Integral secant and csc

**RockHard** · December 5th 2009, 07:34 PM

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

**Chris L T521** · December 5th 2009, 07:40 PM

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.

**skeeter** · December 5th 2009, 07:45 PM

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 ...

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

**RockHard** · December 5th 2009, 07:53 PM

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.

**Jhevon** · December 5th 2009, 08:20 PM

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.

**simplependulum** · December 5th 2009, 08:25 PM

or

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

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

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

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

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

**Jhevon** · December 5th 2009, 08:26 PM

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.

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