Integral of $sec(x)$ using $u$ sub
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
|
show 6 more comments
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
1
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?
– Jon due
Jan 4 at 2:51
4
$$intfrac{1}{cos(x)}dx=intfrac{cos(x)}{cos^2(x)}dx=intfrac{cos(x)}{1-sin^2(x)}dx$$... and we are done.
– Ixion
Jan 4 at 2:53
2
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 $$int frac{2i cdot e^{ix}}{e^{2ix}+1}dx$$
– Eevee Trainer
Jan 4 at 2:57
2
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.
– Eevee Trainer
Jan 4 at 3:04
2
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.
– jmerry
Jan 4 at 4:18
|
show 6 more comments
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
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
edited Jan 4 at 4:18
max_zorn
3,29361328
3,29361328
asked Jan 4 at 2:49
Jon dueJon due
738
738
1
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?
– Jon due
Jan 4 at 2:51
4
$$intfrac{1}{cos(x)}dx=intfrac{cos(x)}{cos^2(x)}dx=intfrac{cos(x)}{1-sin^2(x)}dx$$... and we are done.
– Ixion
Jan 4 at 2:53
2
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 $$int frac{2i cdot e^{ix}}{e^{2ix}+1}dx$$
– Eevee Trainer
Jan 4 at 2:57
2
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.
– Eevee Trainer
Jan 4 at 3:04
2
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.
– jmerry
Jan 4 at 4:18
|
show 6 more comments
1
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?
– Jon due
Jan 4 at 2:51
4
$$intfrac{1}{cos(x)}dx=intfrac{cos(x)}{cos^2(x)}dx=intfrac{cos(x)}{1-sin^2(x)}dx$$... and we are done.
– Ixion
Jan 4 at 2:53
2
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 $$int frac{2i cdot e^{ix}}{e^{2ix}+1}dx$$
– Eevee Trainer
Jan 4 at 2:57
2
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.
– Eevee Trainer
Jan 4 at 3:04
2
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.
– jmerry
Jan 4 at 4:18
1
1
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?
– Jon due
Jan 4 at 2:51
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?
– Jon due
Jan 4 at 2:51
4
4
$$intfrac{1}{cos(x)}dx=intfrac{cos(x)}{cos^2(x)}dx=intfrac{cos(x)}{1-sin^2(x)}dx$$... and we are done.
– Ixion
Jan 4 at 2:53
$$intfrac{1}{cos(x)}dx=intfrac{cos(x)}{cos^2(x)}dx=intfrac{cos(x)}{1-sin^2(x)}dx$$... and we are done.
– Ixion
Jan 4 at 2:53
2
2
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 $$int frac{2i cdot e^{ix}}{e^{2ix}+1}dx$$
– Eevee Trainer
Jan 4 at 2:57
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 $$int frac{2i cdot e^{ix}}{e^{2ix}+1}dx$$
– Eevee Trainer
Jan 4 at 2:57
2
2
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.
– Eevee Trainer
Jan 4 at 3:04
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.
– Eevee Trainer
Jan 4 at 3:04
2
2
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.
– jmerry
Jan 4 at 4:18
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.
– jmerry
Jan 4 at 4:18
|
show 6 more comments
2 Answers
2
active
oldest
votes
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.
I didn't multiply the denominator, I factored out a $e^{ix}$
– Jon due
Jan 4 at 2:58
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.)
– Eevee Trainer
Jan 4 at 3:00
I see the problem, never mind.
– Jon due
Jan 4 at 3:01
@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 $2i$.
– John Omielan
Jan 4 at 3:02
add a comment |
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.
add a comment |
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2 Answers
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2 Answers
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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.
I didn't multiply the denominator, I factored out a $e^{ix}$
– Jon due
Jan 4 at 2:58
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.)
– Eevee Trainer
Jan 4 at 3:00
I see the problem, never mind.
– Jon due
Jan 4 at 3:01
@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 $2i$.
– John Omielan
Jan 4 at 3:02
add a comment |
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.
I didn't multiply the denominator, I factored out a $e^{ix}$
– Jon due
Jan 4 at 2:58
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.)
– Eevee Trainer
Jan 4 at 3:00
I see the problem, never mind.
– Jon due
Jan 4 at 3:01
@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 $2i$.
– John Omielan
Jan 4 at 3:02
add a comment |
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.
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.
edited Jan 4 at 3:03
answered Jan 4 at 2:57
John OmielanJohn Omielan
1,13918
1,13918
I didn't multiply the denominator, I factored out a $e^{ix}$
– Jon due
Jan 4 at 2:58
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.)
– Eevee Trainer
Jan 4 at 3:00
I see the problem, never mind.
– Jon due
Jan 4 at 3:01
@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 $2i$.
– John Omielan
Jan 4 at 3:02
add a comment |
I didn't multiply the denominator, I factored out a $e^{ix}$
– Jon due
Jan 4 at 2:58
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.)
– Eevee Trainer
Jan 4 at 3:00
I see the problem, never mind.
– Jon due
Jan 4 at 3:01
@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 $2i$.
– John Omielan
Jan 4 at 3:02
I didn't multiply the denominator, I factored out a $e^{ix}$
– Jon due
Jan 4 at 2:58
I didn't multiply the denominator, I factored out a $e^{ix}$
– Jon due
Jan 4 at 2:58
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.)
– Eevee Trainer
Jan 4 at 3:00
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.)
– Eevee Trainer
Jan 4 at 3:00
I see the problem, never mind.
– Jon due
Jan 4 at 3:01
I see the problem, never mind.
– Jon due
Jan 4 at 3:01
@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 $2i$.
– John Omielan
Jan 4 at 3:02
@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 $2i$.
– John Omielan
Jan 4 at 3:02
add a comment |
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.
add a comment |
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.
add a comment |
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.
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.
answered Jan 4 at 6:52
DavidGDavidG
1,830620
1,830620
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1
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?
– Jon due
Jan 4 at 2:51
4
$$intfrac{1}{cos(x)}dx=intfrac{cos(x)}{cos^2(x)}dx=intfrac{cos(x)}{1-sin^2(x)}dx$$... and we are done.
– Ixion
Jan 4 at 2:53
2
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 $$int frac{2i cdot e^{ix}}{e^{2ix}+1}dx$$
– Eevee Trainer
Jan 4 at 2:57
2
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.
– Eevee Trainer
Jan 4 at 3:04
2
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.
– jmerry
Jan 4 at 4:18