Integrating $int sec xdx$: Why is $ln|text{sec}x + text{tan}x| + C$ preferred over $tanh^{-1}(sin x) + C$?
$begingroup$
I was trying to integrate $sec^3x$ and discovered that I would have to integrate $sec x$ in the process. I had not seen the "standard" approach and came up with my own solution, which is apparently quite different:
$$int sec xdx = int frac{dx}{cos x}$$
I substituted $u = sin x$ so that $dx = frac{du}{cos x}$. Then
$$int frac{dx}{cos x} = int frac{du}{cos^2x} = int frac{du}{1 - sin^2x} = int frac{du}{1 - u^2}$$
The solution to this is $tanh^{-1}u + C$. Since $u = sin x$, this means that
$$int sec xdx = tanh^{-1}(sin x) + C tag{1}$$
After looking it up, I found out that the standard form of the integral is $$intsec x dx = ln|text{sec}x + text{tan}x| + C tag{2}$$
I couldn't find anything about the alternate form $(1)$ which is, as far as I can tell, equivalent to $(2)$. So, did I make a mistake here? If not, is there a reason to prefer the usual form $(2)$?
calculus trigonometry soft-question indefinite-integrals trigonometric-integrals
edited Dec 10 '18 at 0:06
Batominovski
1
asked Dec 9 '18 at 19:32
CyborgOctopusCyborgOctopus
756
$endgroup$
add a comment |
$begingroup$
I was trying to integrate $sec^3x$ and discovered that I would have to integrate $sec x$ in the process. I had not seen the "standard" approach and came up with my own solution, which is apparently quite different:
$$int sec xdx = int frac{dx}{cos x}$$
I substituted $u = sin x$ so that $dx = frac{du}{cos x}$. Then
$$int frac{dx}{cos x} = int frac{du}{cos^2x} = int frac{du}{1 - sin^2x} = int frac{du}{1 - u^2}$$
The solution to this is $tanh^{-1}u + C$. Since $u = sin x$, this means that
$$int sec xdx = tanh^{-1}(sin x) + C tag{1}$$
After looking it up, I found out that the standard form of the integral is $$intsec x dx = ln|text{sec}x + text{tan}x| + C tag{2}$$
I couldn't find anything about the alternate form $(1)$ which is, as far as I can tell, equivalent to $(2)$. So, did I make a mistake here? If not, is there a reason to prefer the usual form $(2)$?
calculus trigonometry soft-question indefinite-integrals trigonometric-integrals
edited Dec 10 '18 at 0:06
Batominovski
1
asked Dec 9 '18 at 19:32
CyborgOctopusCyborgOctopus
756
$endgroup$
$begingroup$
You might be interested in reading about the Gudermannian function (see here also). See this 8 October 2009 sci.math post and these notes for connections with the integral of secant.
$endgroup$
– Dave L. Renfro
Dec 9 '18 at 20:38
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See also matheducators.stackexchange.com/questions/14631/…
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– J.G.
Dec 9 '18 at 22:47
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No mistake, you've just come across on of the many amazing relationships that exist between our elementary (and non-elementary) functions. No one form is better or worse than the other. Depending upon the situation they each have their advantages.
$endgroup$
– DavidG
Dec 10 '18 at 1:20
add a comment |
$begingroup$
I was trying to integrate $sec^3x$ and discovered that I would have to integrate $sec x$ in the process. I had not seen the "standard" approach and came up with my own solution, which is apparently quite different:
$$int sec xdx = int frac{dx}{cos x}$$
I substituted $u = sin x$ so that $dx = frac{du}{cos x}$. Then
$$int frac{dx}{cos x} = int frac{du}{cos^2x} = int frac{du}{1 - sin^2x} = int frac{du}{1 - u^2}$$
The solution to this is $tanh^{-1}u + C$. Since $u = sin x$, this means that
$$int sec xdx = tanh^{-1}(sin x) + C tag{1}$$
After looking it up, I found out that the standard form of the integral is $$intsec x dx = ln|text{sec}x + text{tan}x| + C tag{2}$$
I couldn't find anything about the alternate form $(1)$ which is, as far as I can tell, equivalent to $(2)$. So, did I make a mistake here? If not, is there a reason to prefer the usual form $(2)$?
calculus trigonometry soft-question indefinite-integrals trigonometric-integrals
edited Dec 10 '18 at 0:06
Batominovski
1
asked Dec 9 '18 at 19:32
CyborgOctopusCyborgOctopus
756
$endgroup$
I was trying to integrate $sec^3x$ and discovered that I would have to integrate $sec x$ in the process. I had not seen the "standard" approach and came up with my own solution, which is apparently quite different:
$$int sec xdx = int frac{dx}{cos x}$$
I substituted $u = sin x$ so that $dx = frac{du}{cos x}$. Then
$$int frac{dx}{cos x} = int frac{du}{cos^2x} = int frac{du}{1 - sin^2x} = int frac{du}{1 - u^2}$$
The solution to this is $tanh^{-1}u + C$. Since $u = sin x$, this means that
$$int sec xdx = tanh^{-1}(sin x) + C tag{1}$$
After looking it up, I found out that the standard form of the integral is $$intsec x dx = ln|text{sec}x + text{tan}x| + C tag{2}$$
I couldn't find anything about the alternate form $(1)$ which is, as far as I can tell, equivalent to $(2)$. So, did I make a mistake here? If not, is there a reason to prefer the usual form $(2)$?
calculus trigonometry soft-question indefinite-integrals trigonometric-integrals
calculus trigonometry soft-question indefinite-integrals trigonometric-integrals
edited Dec 10 '18 at 0:06
Batominovski
1
asked Dec 9 '18 at 19:32
CyborgOctopusCyborgOctopus
756
edited Dec 10 '18 at 0:06
Batominovski
1
asked Dec 9 '18 at 19:32
CyborgOctopusCyborgOctopus
756
edited Dec 10 '18 at 0:06
Batominovski
1
edited Dec 10 '18 at 0:06
Batominovski
1
edited Dec 10 '18 at 0:06
Batominovski
1
1
asked Dec 9 '18 at 19:32
CyborgOctopusCyborgOctopus
756
asked Dec 9 '18 at 19:32
CyborgOctopusCyborgOctopus
756
asked Dec 9 '18 at 19:32
CyborgOctopusCyborgOctopus
756
756
$begingroup$
You might be interested in reading about the Gudermannian function (see here also). See this 8 October 2009 sci.math post and these notes for connections with the integral of secant.
$endgroup$
– Dave L. Renfro
Dec 9 '18 at 20:38
$begingroup$
See also matheducators.stackexchange.com/questions/14631/…
$endgroup$
– J.G.
Dec 9 '18 at 22:47
$begingroup$
No mistake, you've just come across on of the many amazing relationships that exist between our elementary (and non-elementary) functions. No one form is better or worse than the other. Depending upon the situation they each have their advantages.
$endgroup$
– DavidG
Dec 10 '18 at 1:20
add a comment |
$begingroup$
You might be interested in reading about the Gudermannian function (see here also). See this 8 October 2009 sci.math post and these notes for connections with the integral of secant.
$endgroup$
– Dave L. Renfro
Dec 9 '18 at 20:38
$begingroup$
See also matheducators.stackexchange.com/questions/14631/…
$endgroup$
– J.G.
Dec 9 '18 at 22:47
$begingroup$
No mistake, you've just come across on of the many amazing relationships that exist between our elementary (and non-elementary) functions. No one form is better or worse than the other. Depending upon the situation they each have their advantages.
$endgroup$
– DavidG
Dec 10 '18 at 1:20
$begingroup$
You might be interested in reading about the Gudermannian function (see here also). See this 8 October 2009 sci.math post and these notes for connections with the integral of secant.
$endgroup$
– Dave L. Renfro
Dec 9 '18 at 20:38
$begingroup$
You might be interested in reading about the Gudermannian function (see here also). See this 8 October 2009 sci.math post and these notes for connections with the integral of secant.
$endgroup$
– Dave L. Renfro
Dec 9 '18 at 20:38
$begingroup$
See also matheducators.stackexchange.com/questions/14631/…
$endgroup$
– J.G.
Dec 9 '18 at 22:47
$begingroup$
See also matheducators.stackexchange.com/questions/14631/…
$endgroup$
– J.G.
Dec 9 '18 at 22:47
$begingroup$
No mistake, you've just come across on of the many amazing relationships that exist between our elementary (and non-elementary) functions. No one form is better or worse than the other. Depending upon the situation they each have their advantages.
$endgroup$
– DavidG
Dec 10 '18 at 1:20
$begingroup$
No mistake, you've just come across on of the many amazing relationships that exist between our elementary (and non-elementary) functions. No one form is better or worse than the other. Depending upon the situation they each have their advantages.
$endgroup$
– DavidG
Dec 10 '18 at 1:20
add a comment |
2 Answers
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$begingroup$
$$begin{align}
tanh^{-1}(sin x) &=frac12lnleft|frac{1+sin x}{1-sin x}cdot frac{1/cos x}{1/cos x}right|\[4pt]
&=frac12lnleft|frac{sec x+tan x}{sec x-tan x}right| \[4pt]
&=frac12lnleft|frac{sec x+tan x}{sec x-tan x}cdotfrac{sec x+tan x}{sec x+tan x}right|\[4pt]
&=phantom{frac12}lnleft|sec x+tan xright|
end{align}$$
I believe most introductory calculus books use the equivalent form because their readers are not aware of hyperbolic trigonometric functions or their inverses.
edited Dec 9 '18 at 20:01
Blue
48k870153
answered Dec 9 '18 at 19:42
Shubham JohriShubham Johri
5,097717
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add a comment |
$begingroup$
I was taught to calculate $;int (sec x) dx;$ via the substitution $;u = tan(x/2).;$ As the other answer suggested, in math, you don't want to use a steamroller when a flyswatter will do. Problems involving $;int (sec x) dx;$ (for example $;int (sec x)^3 dx);$ can be solved in a straightforward (if somewhat arduous) manner without using hyperbolic functions.
The convention is to avoid advanced topics (e.g. hyperbolic functions), unless the problem requires it.
answered Dec 9 '18 at 22:43
user2661923user2661923
538112
$endgroup$
add a comment |
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2 Answers
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$begingroup$
$$begin{align}
tanh^{-1}(sin x) &=frac12lnleft|frac{1+sin x}{1-sin x}cdot frac{1/cos x}{1/cos x}right|\[4pt]
&=frac12lnleft|frac{sec x+tan x}{sec x-tan x}right| \[4pt]
&=frac12lnleft|frac{sec x+tan x}{sec x-tan x}cdotfrac{sec x+tan x}{sec x+tan x}right|\[4pt]
&=phantom{frac12}lnleft|sec x+tan xright|
end{align}$$
I believe most introductory calculus books use the equivalent form because their readers are not aware of hyperbolic trigonometric functions or their inverses.
edited Dec 9 '18 at 20:01
Blue
48k870153
answered Dec 9 '18 at 19:42
Shubham JohriShubham Johri
5,097717
$endgroup$
add a comment |
$begingroup$
$$begin{align}
tanh^{-1}(sin x) &=frac12lnleft|frac{1+sin x}{1-sin x}cdot frac{1/cos x}{1/cos x}right|\[4pt]
&=frac12lnleft|frac{sec x+tan x}{sec x-tan x}right| \[4pt]
&=frac12lnleft|frac{sec x+tan x}{sec x-tan x}cdotfrac{sec x+tan x}{sec x+tan x}right|\[4pt]
&=phantom{frac12}lnleft|sec x+tan xright|
end{align}$$
I believe most introductory calculus books use the equivalent form because their readers are not aware of hyperbolic trigonometric functions or their inverses.
edited Dec 9 '18 at 20:01
Blue
48k870153
answered Dec 9 '18 at 19:42
Shubham JohriShubham Johri
5,097717
$endgroup$
add a comment |
$begingroup$
$$begin{align}
tanh^{-1}(sin x) &=frac12lnleft|frac{1+sin x}{1-sin x}cdot frac{1/cos x}{1/cos x}right|\[4pt]
&=frac12lnleft|frac{sec x+tan x}{sec x-tan x}right| \[4pt]
&=frac12lnleft|frac{sec x+tan x}{sec x-tan x}cdotfrac{sec x+tan x}{sec x+tan x}right|\[4pt]
&=phantom{frac12}lnleft|sec x+tan xright|
end{align}$$
I believe most introductory calculus books use the equivalent form because their readers are not aware of hyperbolic trigonometric functions or their inverses.
edited Dec 9 '18 at 20:01
Blue
48k870153
answered Dec 9 '18 at 19:42
Shubham JohriShubham Johri
5,097717
$endgroup$
$$begin{align}
tanh^{-1}(sin x) &=frac12lnleft|frac{1+sin x}{1-sin x}cdot frac{1/cos x}{1/cos x}right|\[4pt]
&=frac12lnleft|frac{sec x+tan x}{sec x-tan x}right| \[4pt]
&=frac12lnleft|frac{sec x+tan x}{sec x-tan x}cdotfrac{sec x+tan x}{sec x+tan x}right|\[4pt]
&=phantom{frac12}lnleft|sec x+tan xright|
end{align}$$
I believe most introductory calculus books use the equivalent form because their readers are not aware of hyperbolic trigonometric functions or their inverses.
edited Dec 9 '18 at 20:01
Blue
48k870153
answered Dec 9 '18 at 19:42
Shubham JohriShubham Johri
5,097717
edited Dec 9 '18 at 20:01
Blue
48k870153
edited Dec 9 '18 at 20:01
Blue
48k870153
edited Dec 9 '18 at 20:01
Blue
48k870153
48k870153
answered Dec 9 '18 at 19:42
Shubham JohriShubham Johri
5,097717
answered Dec 9 '18 at 19:42
Shubham JohriShubham Johri
5,097717
answered Dec 9 '18 at 19:42
Shubham JohriShubham Johri
5,097717
5,097717
add a comment |
add a comment |
$begingroup$
I was taught to calculate $;int (sec x) dx;$ via the substitution $;u = tan(x/2).;$ As the other answer suggested, in math, you don't want to use a steamroller when a flyswatter will do. Problems involving $;int (sec x) dx;$ (for example $;int (sec x)^3 dx);$ can be solved in a straightforward (if somewhat arduous) manner without using hyperbolic functions.
The convention is to avoid advanced topics (e.g. hyperbolic functions), unless the problem requires it.
answered Dec 9 '18 at 22:43
user2661923user2661923
538112
$endgroup$
add a comment |
$begingroup$
I was taught to calculate $;int (sec x) dx;$ via the substitution $;u = tan(x/2).;$ As the other answer suggested, in math, you don't want to use a steamroller when a flyswatter will do. Problems involving $;int (sec x) dx;$ (for example $;int (sec x)^3 dx);$ can be solved in a straightforward (if somewhat arduous) manner without using hyperbolic functions.
The convention is to avoid advanced topics (e.g. hyperbolic functions), unless the problem requires it.
answered Dec 9 '18 at 22:43
user2661923user2661923
538112
$endgroup$
add a comment |
$begingroup$
I was taught to calculate $;int (sec x) dx;$ via the substitution $;u = tan(x/2).;$ As the other answer suggested, in math, you don't want to use a steamroller when a flyswatter will do. Problems involving $;int (sec x) dx;$ (for example $;int (sec x)^3 dx);$ can be solved in a straightforward (if somewhat arduous) manner without using hyperbolic functions.
The convention is to avoid advanced topics (e.g. hyperbolic functions), unless the problem requires it.
answered Dec 9 '18 at 22:43
user2661923user2661923
538112
$endgroup$
I was taught to calculate $;int (sec x) dx;$ via the substitution $;u = tan(x/2).;$ As the other answer suggested, in math, you don't want to use a steamroller when a flyswatter will do. Problems involving $;int (sec x) dx;$ (for example $;int (sec x)^3 dx);$ can be solved in a straightforward (if somewhat arduous) manner without using hyperbolic functions.
The convention is to avoid advanced topics (e.g. hyperbolic functions), unless the problem requires it.
answered Dec 9 '18 at 22:43
user2661923user2661923
538112
answered Dec 9 '18 at 22:43
user2661923user2661923
538112
answered Dec 9 '18 at 22:43
user2661923user2661923
538112
answered Dec 9 '18 at 22:43
user2661923user2661923
538112
538112
add a comment |
add a comment |
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$begingroup$
You might be interested in reading about the Gudermannian function (see here also). See this 8 October 2009 sci.math post and these notes for connections with the integral of secant.
$endgroup$
– Dave L. Renfro
Dec 9 '18 at 20:38
$begingroup$
See also matheducators.stackexchange.com/questions/14631/…
$endgroup$
– J.G.
Dec 9 '18 at 22:47
$begingroup$
No mistake, you've just come across on of the many amazing relationships that exist between our elementary (and non-elementary) functions. No one form is better or worse than the other. Depending upon the situation they each have their advantages.
$endgroup$
– DavidG
Dec 10 '18 at 1:20