Integral of $sec(x)$ using $u$ sub












2












$begingroup$


I've just begun learning how to integrate and I wanted to see if I could integrate $sec(x)$ by $u$-substitution. After getting my answer, I was told it couldn't be in complex form, but why, and if so, where did I go wrong? Here is how I did it:
$$int sec(x) dx$$
$$int frac{1}{cos(x)} dx$$
$$int frac{1}{frac{e^{ix}+e^{-ix}}{2i}}dx$$
$$int frac{2i}{e^{2ix}+1}dx$$
$u=2ix,du=2idx$
$$int frac{1}{e^u+1}du$$
$$int frac{e^u}{e^u(e^u+1)}du$$
$v=e^u,dv=e^udu$
$$int frac{1}{v^2+1}dv$$
$[arctan(v)]$



$[arctan(e^u)]$



$arctan(e^{2ix}) + C$






integration indefinite-integrals trigonometric-integrals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Is this right? I was told I couldn't leave my answer in complex form. If so, is that the only answer, or are there other answers?
    $endgroup$
    – Jon due
    Jan 4 at 2:51






  • 4




    $begingroup$
    $$intfrac{1}{cos(x)}dx=intfrac{cos(x)}{cos^2(x)}dx=intfrac{cos(x)}{1-sin^2(x)}dx$$... and we are done.
    $endgroup$
    – Ixion
    Jan 4 at 2:53








  • 2




    $begingroup$
    Your solution is incorrect: the error is in going from $$int frac{1}{frac{e^{ix}+e^{-ix}}{2i}}dx ;;;text{to} ;;;int frac{2i}{e^{2ix}+1}dx$$ you do two things: you moved the $2i$ on top, and multiplied the denominator by $e^{ix}$. However, you need to also multiply the top by $e^{ix}$ as well to maintain equality. Thus the latter integral should be $$int frac{2i cdot e^{ix}}{e^{2ix}+1}dx$$
    $endgroup$
    – Eevee Trainer
    Jan 4 at 2:57






  • 2




    $begingroup$
    Also, to touch on the matter on why you can't use the complex form ... well, as far I know there's no reason you can't, it's just that the original integrand is a function of real values, which outputs values which are also real values. That's generally something I believe we desire in a derivative or antiderivative, i.e. we also want the antiderivative here to be a function of real variables, where $e^{2ix}$ is going to put complex values into the $tan^{-1}$ function. Not that it's invalid, necessarily, as far as I know, just not desirable.
    $endgroup$
    – Eevee Trainer
    Jan 4 at 3:04








  • 2




    $begingroup$
    There are also two more mistakes. First, that's the wrong complex form for $cos$; it should just be $frac{e^{ix}+e^{-ix}}{2}$, no division by $i$. Second, $v(v+1)=v^2+v$, not $v^2+1$. Fixing all of the mistakes, a correct complex-form antiderivative for $sec x$ is $-2iarctan(e^{ix})+C$. Note that the arctangent of a complex number with absolute value $1$ lies on one of the vertical lines with $x$-coordinate $frac{pi}{4}$ or $-frac{pi}{4}$ plus a multiple of $pi$. This is not a very convenient form.
    $endgroup$
    – jmerry
    Jan 4 at 4:18
















2












$begingroup$


I've just begun learning how to integrate and I wanted to see if I could integrate $sec(x)$ by $u$-substitution. After getting my answer, I was told it couldn't be in complex form, but why, and if so, where did I go wrong? Here is how I did it:
$$int sec(x) dx$$
$$int frac{1}{cos(x)} dx$$
$$int frac{1}{frac{e^{ix}+e^{-ix}}{2i}}dx$$
$$int frac{2i}{e^{2ix}+1}dx$$
$u=2ix,du=2idx$
$$int frac{1}{e^u+1}du$$
$$int frac{e^u}{e^u(e^u+1)}du$$
$v=e^u,dv=e^udu$
$$int frac{1}{v^2+1}dv$$
$[arctan(v)]$



$[arctan(e^u)]$



$arctan(e^{2ix}) + C$






integration indefinite-integrals trigonometric-integrals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Is this right? I was told I couldn't leave my answer in complex form. If so, is that the only answer, or are there other answers?
    $endgroup$
    – Jon due
    Jan 4 at 2:51






  • 4




    $begingroup$
    $$intfrac{1}{cos(x)}dx=intfrac{cos(x)}{cos^2(x)}dx=intfrac{cos(x)}{1-sin^2(x)}dx$$... and we are done.
    $endgroup$
    – Ixion
    Jan 4 at 2:53








  • 2




    $begingroup$
    Your solution is incorrect: the error is in going from $$int frac{1}{frac{e^{ix}+e^{-ix}}{2i}}dx ;;;text{to} ;;;int frac{2i}{e^{2ix}+1}dx$$ you do two things: you moved the $2i$ on top, and multiplied the denominator by $e^{ix}$. However, you need to also multiply the top by $e^{ix}$ as well to maintain equality. Thus the latter integral should be $$int frac{2i cdot e^{ix}}{e^{2ix}+1}dx$$
    $endgroup$
    – Eevee Trainer
    Jan 4 at 2:57






  • 2




    $begingroup$
    Also, to touch on the matter on why you can't use the complex form ... well, as far I know there's no reason you can't, it's just that the original integrand is a function of real values, which outputs values which are also real values. That's generally something I believe we desire in a derivative or antiderivative, i.e. we also want the antiderivative here to be a function of real variables, where $e^{2ix}$ is going to put complex values into the $tan^{-1}$ function. Not that it's invalid, necessarily, as far as I know, just not desirable.
    $endgroup$
    – Eevee Trainer
    Jan 4 at 3:04








  • 2




    $begingroup$
    There are also two more mistakes. First, that's the wrong complex form for $cos$; it should just be $frac{e^{ix}+e^{-ix}}{2}$, no division by $i$. Second, $v(v+1)=v^2+v$, not $v^2+1$. Fixing all of the mistakes, a correct complex-form antiderivative for $sec x$ is $-2iarctan(e^{ix})+C$. Note that the arctangent of a complex number with absolute value $1$ lies on one of the vertical lines with $x$-coordinate $frac{pi}{4}$ or $-frac{pi}{4}$ plus a multiple of $pi$. This is not a very convenient form.
    $endgroup$
    – jmerry
    Jan 4 at 4:18














2












2








2





$begingroup$


I've just begun learning how to integrate and I wanted to see if I could integrate $sec(x)$ by $u$-substitution. After getting my answer, I was told it couldn't be in complex form, but why, and if so, where did I go wrong? Here is how I did it:
$$int sec(x) dx$$
$$int frac{1}{cos(x)} dx$$
$$int frac{1}{frac{e^{ix}+e^{-ix}}{2i}}dx$$
$$int frac{2i}{e^{2ix}+1}dx$$
$u=2ix,du=2idx$
$$int frac{1}{e^u+1}du$$
$$int frac{e^u}{e^u(e^u+1)}du$$
$v=e^u,dv=e^udu$
$$int frac{1}{v^2+1}dv$$
$[arctan(v)]$



$[arctan(e^u)]$



$arctan(e^{2ix}) + C$






integration indefinite-integrals trigonometric-integrals







share|cite|improve this question











$endgroup$




I've just begun learning how to integrate and I wanted to see if I could integrate $sec(x)$ by $u$-substitution. After getting my answer, I was told it couldn't be in complex form, but why, and if so, where did I go wrong? Here is how I did it:
$$int sec(x) dx$$
$$int frac{1}{cos(x)} dx$$
$$int frac{1}{frac{e^{ix}+e^{-ix}}{2i}}dx$$
$$int frac{2i}{e^{2ix}+1}dx$$
$u=2ix,du=2idx$
$$int frac{1}{e^u+1}du$$
$$int frac{e^u}{e^u(e^u+1)}du$$
$v=e^u,dv=e^udu$
$$int frac{1}{v^2+1}dv$$
$[arctan(v)]$



$[arctan(e^u)]$



$arctan(e^{2ix}) + C$






integration indefinite-integrals trigonometric-integrals




integration indefinite-integrals trigonometric-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 4:18









max_zorn

3,36361329




3,36361329










asked Jan 4 at 2:49









Jon dueJon due

788




788








  • 1




    $begingroup$
    Is this right? I was told I couldn't leave my answer in complex form. If so, is that the only answer, or are there other answers?
    $endgroup$
    – Jon due
    Jan 4 at 2:51






  • 4




    $begingroup$
    $$intfrac{1}{cos(x)}dx=intfrac{cos(x)}{cos^2(x)}dx=intfrac{cos(x)}{1-sin^2(x)}dx$$... and we are done.
    $endgroup$
    – Ixion
    Jan 4 at 2:53








  • 2




    $begingroup$
    Your solution is incorrect: the error is in going from $$int frac{1}{frac{e^{ix}+e^{-ix}}{2i}}dx ;;;text{to} ;;;int frac{2i}{e^{2ix}+1}dx$$ you do two things: you moved the $2i$ on top, and multiplied the denominator by $e^{ix}$. However, you need to also multiply the top by $e^{ix}$ as well to maintain equality. Thus the latter integral should be $$int frac{2i cdot e^{ix}}{e^{2ix}+1}dx$$
    $endgroup$
    – Eevee Trainer
    Jan 4 at 2:57






  • 2




    $begingroup$
    Also, to touch on the matter on why you can't use the complex form ... well, as far I know there's no reason you can't, it's just that the original integrand is a function of real values, which outputs values which are also real values. That's generally something I believe we desire in a derivative or antiderivative, i.e. we also want the antiderivative here to be a function of real variables, where $e^{2ix}$ is going to put complex values into the $tan^{-1}$ function. Not that it's invalid, necessarily, as far as I know, just not desirable.
    $endgroup$
    – Eevee Trainer
    Jan 4 at 3:04








  • 2




    $begingroup$
    There are also two more mistakes. First, that's the wrong complex form for $cos$; it should just be $frac{e^{ix}+e^{-ix}}{2}$, no division by $i$. Second, $v(v+1)=v^2+v$, not $v^2+1$. Fixing all of the mistakes, a correct complex-form antiderivative for $sec x$ is $-2iarctan(e^{ix})+C$. Note that the arctangent of a complex number with absolute value $1$ lies on one of the vertical lines with $x$-coordinate $frac{pi}{4}$ or $-frac{pi}{4}$ plus a multiple of $pi$. This is not a very convenient form.
    $endgroup$
    – jmerry
    Jan 4 at 4:18














  • 1




    $begingroup$
    Is this right? I was told I couldn't leave my answer in complex form. If so, is that the only answer, or are there other answers?
    $endgroup$
    – Jon due
    Jan 4 at 2:51






  • 4




    $begingroup$
    $$intfrac{1}{cos(x)}dx=intfrac{cos(x)}{cos^2(x)}dx=intfrac{cos(x)}{1-sin^2(x)}dx$$... and we are done.
    $endgroup$
    – Ixion
    Jan 4 at 2:53








  • 2




    $begingroup$
    Your solution is incorrect: the error is in going from $$int frac{1}{frac{e^{ix}+e^{-ix}}{2i}}dx ;;;text{to} ;;;int frac{2i}{e^{2ix}+1}dx$$ you do two things: you moved the $2i$ on top, and multiplied the denominator by $e^{ix}$. However, you need to also multiply the top by $e^{ix}$ as well to maintain equality. Thus the latter integral should be $$int frac{2i cdot e^{ix}}{e^{2ix}+1}dx$$
    $endgroup$
    – Eevee Trainer
    Jan 4 at 2:57






  • 2




    $begingroup$
    Also, to touch on the matter on why you can't use the complex form ... well, as far I know there's no reason you can't, it's just that the original integrand is a function of real values, which outputs values which are also real values. That's generally something I believe we desire in a derivative or antiderivative, i.e. we also want the antiderivative here to be a function of real variables, where $e^{2ix}$ is going to put complex values into the $tan^{-1}$ function. Not that it's invalid, necessarily, as far as I know, just not desirable.
    $endgroup$
    – Eevee Trainer
    Jan 4 at 3:04








  • 2




    $begingroup$
    There are also two more mistakes. First, that's the wrong complex form for $cos$; it should just be $frac{e^{ix}+e^{-ix}}{2}$, no division by $i$. Second, $v(v+1)=v^2+v$, not $v^2+1$. Fixing all of the mistakes, a correct complex-form antiderivative for $sec x$ is $-2iarctan(e^{ix})+C$. Note that the arctangent of a complex number with absolute value $1$ lies on one of the vertical lines with $x$-coordinate $frac{pi}{4}$ or $-frac{pi}{4}$ plus a multiple of $pi$. This is not a very convenient form.
    $endgroup$
    – jmerry
    Jan 4 at 4:18








1




1




$begingroup$
Is this right? I was told I couldn't leave my answer in complex form. If so, is that the only answer, or are there other answers?
$endgroup$
– Jon due
Jan 4 at 2:51




$begingroup$
Is this right? I was told I couldn't leave my answer in complex form. If so, is that the only answer, or are there other answers?
$endgroup$
– Jon due
Jan 4 at 2:51




4




4




$begingroup$
$$intfrac{1}{cos(x)}dx=intfrac{cos(x)}{cos^2(x)}dx=intfrac{cos(x)}{1-sin^2(x)}dx$$... and we are done.
$endgroup$
– Ixion
Jan 4 at 2:53






$begingroup$
$$intfrac{1}{cos(x)}dx=intfrac{cos(x)}{cos^2(x)}dx=intfrac{cos(x)}{1-sin^2(x)}dx$$... and we are done.
$endgroup$
– Ixion
Jan 4 at 2:53






2




2




$begingroup$
Your solution is incorrect: the error is in going from $$int frac{1}{frac{e^{ix}+e^{-ix}}{2i}}dx ;;;text{to} ;;;int frac{2i}{e^{2ix}+1}dx$$ you do two things: you moved the $2i$ on top, and multiplied the denominator by $e^{ix}$. However, you need to also multiply the top by $e^{ix}$ as well to maintain equality. Thus the latter integral should be $$int frac{2i cdot e^{ix}}{e^{2ix}+1}dx$$
$endgroup$
– Eevee Trainer
Jan 4 at 2:57




$begingroup$
Your solution is incorrect: the error is in going from $$int frac{1}{frac{e^{ix}+e^{-ix}}{2i}}dx ;;;text{to} ;;;int frac{2i}{e^{2ix}+1}dx$$ you do two things: you moved the $2i$ on top, and multiplied the denominator by $e^{ix}$. However, you need to also multiply the top by $e^{ix}$ as well to maintain equality. Thus the latter integral should be $$int frac{2i cdot e^{ix}}{e^{2ix}+1}dx$$
$endgroup$
– Eevee Trainer
Jan 4 at 2:57




2




2




$begingroup$
Also, to touch on the matter on why you can't use the complex form ... well, as far I know there's no reason you can't, it's just that the original integrand is a function of real values, which outputs values which are also real values. That's generally something I believe we desire in a derivative or antiderivative, i.e. we also want the antiderivative here to be a function of real variables, where $e^{2ix}$ is going to put complex values into the $tan^{-1}$ function. Not that it's invalid, necessarily, as far as I know, just not desirable.
$endgroup$
– Eevee Trainer
Jan 4 at 3:04






$begingroup$
Also, to touch on the matter on why you can't use the complex form ... well, as far I know there's no reason you can't, it's just that the original integrand is a function of real values, which outputs values which are also real values. That's generally something I believe we desire in a derivative or antiderivative, i.e. we also want the antiderivative here to be a function of real variables, where $e^{2ix}$ is going to put complex values into the $tan^{-1}$ function. Not that it's invalid, necessarily, as far as I know, just not desirable.
$endgroup$
– Eevee Trainer
Jan 4 at 3:04






2




2




$begingroup$
There are also two more mistakes. First, that's the wrong complex form for $cos$; it should just be $frac{e^{ix}+e^{-ix}}{2}$, no division by $i$. Second, $v(v+1)=v^2+v$, not $v^2+1$. Fixing all of the mistakes, a correct complex-form antiderivative for $sec x$ is $-2iarctan(e^{ix})+C$. Note that the arctangent of a complex number with absolute value $1$ lies on one of the vertical lines with $x$-coordinate $frac{pi}{4}$ or $-frac{pi}{4}$ plus a multiple of $pi$. This is not a very convenient form.
$endgroup$
– jmerry
Jan 4 at 4:18




$begingroup$
There are also two more mistakes. First, that's the wrong complex form for $cos$; it should just be $frac{e^{ix}+e^{-ix}}{2}$, no division by $i$. Second, $v(v+1)=v^2+v$, not $v^2+1$. Fixing all of the mistakes, a correct complex-form antiderivative for $sec x$ is $-2iarctan(e^{ix})+C$. Note that the arctangent of a complex number with absolute value $1$ lies on one of the vertical lines with $x$-coordinate $frac{pi}{4}$ or $-frac{pi}{4}$ plus a multiple of $pi$. This is not a very convenient form.
$endgroup$
– jmerry
Jan 4 at 4:18










2 Answers
2






active

oldest

votes


















3












$begingroup$

Going from your $3$rd line to the $4$th, you have multiplied the denominator by $e^{ix}$ but not the numerator. Thus, your work after that will not be correct. Also, in general, as the original question is in reals, your final answer should also be only in reals, with this indicating why your final answer can't be correct.



Note I saw that Eevee Trainer just wrote about the mistake in the comments, along with a more detailed explanation.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I didn't multiply the denominator, I factored out a $e^{ix}$
    $endgroup$
    – Jon due
    Jan 4 at 2:58










  • $begingroup$
    Where/how exactly did you factor? Because I see no way to interpret the matter otherwise. (And even if you did factor it out, where did it go? It doesn't just disappear.)
    $endgroup$
    – Eevee Trainer
    Jan 4 at 3:00












  • $begingroup$
    I see the problem, never mind.
    $endgroup$
    – Jon due
    Jan 4 at 3:01










  • $begingroup$
    @Jondue Your original denominator was $frac{e_{ix} + e{-ix}}{2i}$, so to get to $e^{2ix} + 1$, you multiplied by $2i e^{ix}$. However, you only multiplied the numerator by $2i$.
    $endgroup$
    – John Omielan
    Jan 4 at 3:02





















3












$begingroup$

Just in case you were unaware there is another great trick to solving this integral:



begin{align}
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
end{align}



Here let $u = sec(x) + tan(x) rightarrow frac{du}{dx} = sec(x)tan(x) + sec^2(x)$ and thus:



begin{align}
int sec(x):dx &= int frac{sec^2(x) + sec(x)tan(x)}{u}cdot frac{du}{sec(x)tan(x) + sec^2(x)}\
&= int frac{1}{u}:du = ln|u| + C = lnleft|sec(x) + tan(x) right| + C
end{align}



Where $C$ is the constant of integration.



Personally, for someone new to integrating this is more of a 'special case' method rather than a formal method. But important to be be aware of nonetheless.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Going from your $3$rd line to the $4$th, you have multiplied the denominator by $e^{ix}$ but not the numerator. Thus, your work after that will not be correct. Also, in general, as the original question is in reals, your final answer should also be only in reals, with this indicating why your final answer can't be correct.



    Note I saw that Eevee Trainer just wrote about the mistake in the comments, along with a more detailed explanation.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I didn't multiply the denominator, I factored out a $e^{ix}$
      $endgroup$
      – Jon due
      Jan 4 at 2:58










    • $begingroup$
      Where/how exactly did you factor? Because I see no way to interpret the matter otherwise. (And even if you did factor it out, where did it go? It doesn't just disappear.)
      $endgroup$
      – Eevee Trainer
      Jan 4 at 3:00












    • $begingroup$
      I see the problem, never mind.
      $endgroup$
      – Jon due
      Jan 4 at 3:01










    • $begingroup$
      @Jondue Your original denominator was $frac{e_{ix} + e{-ix}}{2i}$, so to get to $e^{2ix} + 1$, you multiplied by $2i e^{ix}$. However, you only multiplied the numerator by $2i$.
      $endgroup$
      – John Omielan
      Jan 4 at 3:02


















    3












    $begingroup$

    Going from your $3$rd line to the $4$th, you have multiplied the denominator by $e^{ix}$ but not the numerator. Thus, your work after that will not be correct. Also, in general, as the original question is in reals, your final answer should also be only in reals, with this indicating why your final answer can't be correct.



    Note I saw that Eevee Trainer just wrote about the mistake in the comments, along with a more detailed explanation.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I didn't multiply the denominator, I factored out a $e^{ix}$
      $endgroup$
      – Jon due
      Jan 4 at 2:58










    • $begingroup$
      Where/how exactly did you factor? Because I see no way to interpret the matter otherwise. (And even if you did factor it out, where did it go? It doesn't just disappear.)
      $endgroup$
      – Eevee Trainer
      Jan 4 at 3:00












    • $begingroup$
      I see the problem, never mind.
      $endgroup$
      – Jon due
      Jan 4 at 3:01










    • $begingroup$
      @Jondue Your original denominator was $frac{e_{ix} + e{-ix}}{2i}$, so to get to $e^{2ix} + 1$, you multiplied by $2i e^{ix}$. However, you only multiplied the numerator by $2i$.
      $endgroup$
      – John Omielan
      Jan 4 at 3:02
















    3












    3








    3





    $begingroup$

    Going from your $3$rd line to the $4$th, you have multiplied the denominator by $e^{ix}$ but not the numerator. Thus, your work after that will not be correct. Also, in general, as the original question is in reals, your final answer should also be only in reals, with this indicating why your final answer can't be correct.



    Note I saw that Eevee Trainer just wrote about the mistake in the comments, along with a more detailed explanation.






    share|cite|improve this answer











    $endgroup$



    Going from your $3$rd line to the $4$th, you have multiplied the denominator by $e^{ix}$ but not the numerator. Thus, your work after that will not be correct. Also, in general, as the original question is in reals, your final answer should also be only in reals, with this indicating why your final answer can't be correct.



    Note I saw that Eevee Trainer just wrote about the mistake in the comments, along with a more detailed explanation.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 4 at 3:03

























    answered Jan 4 at 2:57









    John OmielanJohn Omielan

    1,951210




    1,951210












    • $begingroup$
      I didn't multiply the denominator, I factored out a $e^{ix}$
      $endgroup$
      – Jon due
      Jan 4 at 2:58










    • $begingroup$
      Where/how exactly did you factor? Because I see no way to interpret the matter otherwise. (And even if you did factor it out, where did it go? It doesn't just disappear.)
      $endgroup$
      – Eevee Trainer
      Jan 4 at 3:00












    • $begingroup$
      I see the problem, never mind.
      $endgroup$
      – Jon due
      Jan 4 at 3:01










    • $begingroup$
      @Jondue Your original denominator was $frac{e_{ix} + e{-ix}}{2i}$, so to get to $e^{2ix} + 1$, you multiplied by $2i e^{ix}$. However, you only multiplied the numerator by $2i$.
      $endgroup$
      – John Omielan
      Jan 4 at 3:02




















    • $begingroup$
      I didn't multiply the denominator, I factored out a $e^{ix}$
      $endgroup$
      – Jon due
      Jan 4 at 2:58










    • $begingroup$
      Where/how exactly did you factor? Because I see no way to interpret the matter otherwise. (And even if you did factor it out, where did it go? It doesn't just disappear.)
      $endgroup$
      – Eevee Trainer
      Jan 4 at 3:00












    • $begingroup$
      I see the problem, never mind.
      $endgroup$
      – Jon due
      Jan 4 at 3:01










    • $begingroup$
      @Jondue Your original denominator was $frac{e_{ix} + e{-ix}}{2i}$, so to get to $e^{2ix} + 1$, you multiplied by $2i e^{ix}$. However, you only multiplied the numerator by $2i$.
      $endgroup$
      – John Omielan
      Jan 4 at 3:02


















    $begingroup$
    I didn't multiply the denominator, I factored out a $e^{ix}$
    $endgroup$
    – Jon due
    Jan 4 at 2:58




    $begingroup$
    I didn't multiply the denominator, I factored out a $e^{ix}$
    $endgroup$
    – Jon due
    Jan 4 at 2:58












    $begingroup$
    Where/how exactly did you factor? Because I see no way to interpret the matter otherwise. (And even if you did factor it out, where did it go? It doesn't just disappear.)
    $endgroup$
    – Eevee Trainer
    Jan 4 at 3:00






    $begingroup$
    Where/how exactly did you factor? Because I see no way to interpret the matter otherwise. (And even if you did factor it out, where did it go? It doesn't just disappear.)
    $endgroup$
    – Eevee Trainer
    Jan 4 at 3:00














    $begingroup$
    I see the problem, never mind.
    $endgroup$
    – Jon due
    Jan 4 at 3:01




    $begingroup$
    I see the problem, never mind.
    $endgroup$
    – Jon due
    Jan 4 at 3:01












    $begingroup$
    @Jondue Your original denominator was $frac{e_{ix} + e{-ix}}{2i}$, so to get to $e^{2ix} + 1$, you multiplied by $2i e^{ix}$. However, you only multiplied the numerator by $2i$.
    $endgroup$
    – John Omielan
    Jan 4 at 3:02






    $begingroup$
    @Jondue Your original denominator was $frac{e_{ix} + e{-ix}}{2i}$, so to get to $e^{2ix} + 1$, you multiplied by $2i e^{ix}$. However, you only multiplied the numerator by $2i$.
    $endgroup$
    – John Omielan
    Jan 4 at 3:02













    3












    $begingroup$

    Just in case you were unaware there is another great trick to solving this integral:



    begin{align}
    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
    end{align}



    Here let $u = sec(x) + tan(x) rightarrow frac{du}{dx} = sec(x)tan(x) + sec^2(x)$ and thus:



    begin{align}
    int sec(x):dx &= int frac{sec^2(x) + sec(x)tan(x)}{u}cdot frac{du}{sec(x)tan(x) + sec^2(x)}\
    &= int frac{1}{u}:du = ln|u| + C = lnleft|sec(x) + tan(x) right| + C
    end{align}



    Where $C$ is the constant of integration.



    Personally, for someone new to integrating this is more of a 'special case' method rather than a formal method. But important to be be aware of nonetheless.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      Just in case you were unaware there is another great trick to solving this integral:



      begin{align}
      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
      end{align}



      Here let $u = sec(x) + tan(x) rightarrow frac{du}{dx} = sec(x)tan(x) + sec^2(x)$ and thus:



      begin{align}
      int sec(x):dx &= int frac{sec^2(x) + sec(x)tan(x)}{u}cdot frac{du}{sec(x)tan(x) + sec^2(x)}\
      &= int frac{1}{u}:du = ln|u| + C = lnleft|sec(x) + tan(x) right| + C
      end{align}



      Where $C$ is the constant of integration.



      Personally, for someone new to integrating this is more of a 'special case' method rather than a formal method. But important to be be aware of nonetheless.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        Just in case you were unaware there is another great trick to solving this integral:



        begin{align}
        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
        end{align}



        Here let $u = sec(x) + tan(x) rightarrow frac{du}{dx} = sec(x)tan(x) + sec^2(x)$ and thus:



        begin{align}
        int sec(x):dx &= int frac{sec^2(x) + sec(x)tan(x)}{u}cdot frac{du}{sec(x)tan(x) + sec^2(x)}\
        &= int frac{1}{u}:du = ln|u| + C = lnleft|sec(x) + tan(x) right| + C
        end{align}



        Where $C$ is the constant of integration.



        Personally, for someone new to integrating this is more of a 'special case' method rather than a formal method. But important to be be aware of nonetheless.






        share|cite|improve this answer









        $endgroup$



        Just in case you were unaware there is another great trick to solving this integral:



        begin{align}
        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
        end{align}



        Here let $u = sec(x) + tan(x) rightarrow frac{du}{dx} = sec(x)tan(x) + sec^2(x)$ and thus:



        begin{align}
        int sec(x):dx &= int frac{sec^2(x) + sec(x)tan(x)}{u}cdot frac{du}{sec(x)tan(x) + sec^2(x)}\
        &= int frac{1}{u}:du = ln|u| + C = lnleft|sec(x) + tan(x) right| + C
        end{align}



        Where $C$ is the constant of integration.



        Personally, for someone new to integrating this is more of a 'special case' method rather than a formal method. But important to be be aware of nonetheless.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 6:52









        DavidGDavidG

        2,2651721




        2,2651721






























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