Integrating $int sec xdx$: Why is $ln|text{sec}x + text{tan}x| + C$ preferred over $tanh^{-1}(sin x) + C$?












2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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










  • $begingroup$
    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
















2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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










  • $begingroup$
    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














2












2








2





$begingroup$


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







share|cite|improve this question











$endgroup$




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






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edited Dec 10 '18 at 0:06









Batominovski

33.1k33293




33.1k33293










asked Dec 9 '18 at 19:32









CyborgOctopusCyborgOctopus

1206




1206












  • $begingroup$
    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










  • $begingroup$
    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










2 Answers
2






active

oldest

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3












$begingroup$

$$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.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    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.






    share|cite|improve this answer









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      Your Answer





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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      3












      $begingroup$

      $$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.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        $$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.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          $$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.






          share|cite|improve this answer











          $endgroup$



          $$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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 9 '18 at 20:01









          Blue

          49.1k870156




          49.1k870156










          answered Dec 9 '18 at 19:42









          Shubham JohriShubham Johri

          5,347718




          5,347718























              0












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  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.






                  share|cite|improve this answer









                  $endgroup$



                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 9 '18 at 22:43









                  user2661923user2661923

                  558112




                  558112






























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