Mixing
Why is the domain for sec trig subs the regions where the signs are the same? Work check
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I am reading my textbook and I see this:
So why are we allowed to just restrict the domain like this? I can see that we're trying to make it so that $a|tan(theta)| = atan(theta)$ but why are we allowed to do this?
Here's a problem I am trying to solve:
$$int frac{sqrt{x^2 - 4}}{x}$$ and $x = 2sec{theta}$
I have the answer, but why can't this domain include $pi/2$ to $pi$?
calculus
edited Feb 2 at 18:28
Ted
22.2k13361
asked Feb 2 at 18:11
Jwan622Jwan622
2,39011632
$endgroup$
add a comment |
$begingroup$
I am reading my textbook and I see this:
So why are we allowed to just restrict the domain like this? I can see that we're trying to make it so that $a|tan(theta)| = atan(theta)$ but why are we allowed to do this?
Here's a problem I am trying to solve:
$$int frac{sqrt{x^2 - 4}}{x}$$ and $x = 2sec{theta}$
I have the answer, but why can't this domain include $pi/2$ to $pi$?
calculus
edited Feb 2 at 18:28
Ted
22.2k13361
asked Feb 2 at 18:11
Jwan622Jwan622
2,39011632
$endgroup$
add a comment |
$begingroup$
I am reading my textbook and I see this:
So why are we allowed to just restrict the domain like this? I can see that we're trying to make it so that $a|tan(theta)| = atan(theta)$ but why are we allowed to do this?
Here's a problem I am trying to solve:
$$int frac{sqrt{x^2 - 4}}{x}$$ and $x = 2sec{theta}$
I have the answer, but why can't this domain include $pi/2$ to $pi$?
calculus
edited Feb 2 at 18:28
Ted
22.2k13361
asked Feb 2 at 18:11
Jwan622Jwan622
2,39011632
$endgroup$
I am reading my textbook and I see this:
So why are we allowed to just restrict the domain like this? I can see that we're trying to make it so that $a|tan(theta)| = atan(theta)$ but why are we allowed to do this?
Here's a problem I am trying to solve:
$$int frac{sqrt{x^2 - 4}}{x}$$ and $x = 2sec{theta}$
I have the answer, but why can't this domain include $pi/2$ to $pi$?
calculus
calculus
edited Feb 2 at 18:28
Ted
22.2k13361
asked Feb 2 at 18:11
Jwan622Jwan622
2,39011632
edited Feb 2 at 18:28
Ted
22.2k13361
asked Feb 2 at 18:11
Jwan622Jwan622
2,39011632
edited Feb 2 at 18:28
Ted
22.2k13361
edited Feb 2 at 18:28
Ted
22.2k13361
edited Feb 2 at 18:28
Ted
22.2k13361
22.2k13361
asked Feb 2 at 18:11
Jwan622Jwan622
2,39011632
asked Feb 2 at 18:11
Jwan622Jwan622
2,39011632
asked Feb 2 at 18:11
Jwan622Jwan622
2,39011632
2,39011632
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The reason to make this restriction is so that there is a unique $theta$ such that $x = a sec theta$. Effectively, we want an inverse function of $sec$, but (like all trig functions) $sec$ is not one-to-one, so we must restrict the domain of $sec$ to make it one-to-one. Notice that the range of $sec$ consists of two disconnected pieces: the positives $[1, infty)$ and the negatives $(-infty, -1]$. It is natural to choose $0 le theta < pi/2$ to get the positive values of $sec$, but for the negative values there are two reasonable choices: $pi/2 < theta le pi$ or $pi < theta le 3pi/2$. (Notice there is a discontinuity of $sec$ at $pi/2$ so either way we do not get a connected interval.) For this problem, we choose the second one because $|tan theta| = tantheta$ is a convenient property for this problem.
You can also do the problem with the other choice, but then you end up with a sign factor $|tan theta|/tantheta$ in the integral which you have to resolve by splitting into two cases.
answered Feb 2 at 18:46
TedTed
22.2k13361
$endgroup$
add a comment |
$begingroup$
The book solution is glossing over an important point--the function $1/sqrt{x^2-a^2}$, and thus its antiderivative, is only defined on the set $(-infty,-a)cup(a,infty)$, which is not a connected set. So to find an antiderivative properly, each connected component (in this case $(-infty,-a)$ and $(a,infty)$) should be handled separately.
It just so happens that the change of variables $xrightarrow asectheta$ is helpful for both connected components. For $(a,infty)$, this either maps to $(0,pi/2)$ or $(pi/2,pi)$, while for $(-a,-infty)$, this maps to either $(pi,3pi/2)$ or $(3pi/2,2pi)$. For each component, either choice is valid, but you have to make one or the other, because otherwise the mapping isn't unique. The book merely chooses $(0,pi/2)$ for $(a,infty)$ and $(pi,3pi/2)$ for $(-a,-infty)$ because it makes the algebra simpler.
In fact, because it's sloppy in handling the connected components, the book actually has an error. The general form of the function that satisfies $f'(x) = 1/sqrt{x^2-a^2}$ is
$$
f(x) = begin{cases}lnleft|x+sqrt{x^2-a^2}right| + C_1 & x < -a\lnleft|x+sqrt{x^2-a^2}right|+ C_2 & x > aend{cases}.
$$
Because the domain isn't connected, there's no continuity requirement across the gap, so the two regions can have different constants of integration.
answered Feb 2 at 19:04
eyeballfrogeyeballfrog
7,212633
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The reason to make this restriction is so that there is a unique $theta$ such that $x = a sec theta$. Effectively, we want an inverse function of $sec$, but (like all trig functions) $sec$ is not one-to-one, so we must restrict the domain of $sec$ to make it one-to-one. Notice that the range of $sec$ consists of two disconnected pieces: the positives $[1, infty)$ and the negatives $(-infty, -1]$. It is natural to choose $0 le theta < pi/2$ to get the positive values of $sec$, but for the negative values there are two reasonable choices: $pi/2 < theta le pi$ or $pi < theta le 3pi/2$. (Notice there is a discontinuity of $sec$ at $pi/2$ so either way we do not get a connected interval.) For this problem, we choose the second one because $|tan theta| = tantheta$ is a convenient property for this problem.
You can also do the problem with the other choice, but then you end up with a sign factor $|tan theta|/tantheta$ in the integral which you have to resolve by splitting into two cases.
answered Feb 2 at 18:46
TedTed
22.2k13361
$endgroup$
add a comment |
$begingroup$
The reason to make this restriction is so that there is a unique $theta$ such that $x = a sec theta$. Effectively, we want an inverse function of $sec$, but (like all trig functions) $sec$ is not one-to-one, so we must restrict the domain of $sec$ to make it one-to-one. Notice that the range of $sec$ consists of two disconnected pieces: the positives $[1, infty)$ and the negatives $(-infty, -1]$. It is natural to choose $0 le theta < pi/2$ to get the positive values of $sec$, but for the negative values there are two reasonable choices: $pi/2 < theta le pi$ or $pi < theta le 3pi/2$. (Notice there is a discontinuity of $sec$ at $pi/2$ so either way we do not get a connected interval.) For this problem, we choose the second one because $|tan theta| = tantheta$ is a convenient property for this problem.
You can also do the problem with the other choice, but then you end up with a sign factor $|tan theta|/tantheta$ in the integral which you have to resolve by splitting into two cases.
answered Feb 2 at 18:46
TedTed
22.2k13361
$endgroup$
add a comment |
$begingroup$
The reason to make this restriction is so that there is a unique $theta$ such that $x = a sec theta$. Effectively, we want an inverse function of $sec$, but (like all trig functions) $sec$ is not one-to-one, so we must restrict the domain of $sec$ to make it one-to-one. Notice that the range of $sec$ consists of two disconnected pieces: the positives $[1, infty)$ and the negatives $(-infty, -1]$. It is natural to choose $0 le theta < pi/2$ to get the positive values of $sec$, but for the negative values there are two reasonable choices: $pi/2 < theta le pi$ or $pi < theta le 3pi/2$. (Notice there is a discontinuity of $sec$ at $pi/2$ so either way we do not get a connected interval.) For this problem, we choose the second one because $|tan theta| = tantheta$ is a convenient property for this problem.
You can also do the problem with the other choice, but then you end up with a sign factor $|tan theta|/tantheta$ in the integral which you have to resolve by splitting into two cases.
answered Feb 2 at 18:46
TedTed
22.2k13361
$endgroup$
The reason to make this restriction is so that there is a unique $theta$ such that $x = a sec theta$. Effectively, we want an inverse function of $sec$, but (like all trig functions) $sec$ is not one-to-one, so we must restrict the domain of $sec$ to make it one-to-one. Notice that the range of $sec$ consists of two disconnected pieces: the positives $[1, infty)$ and the negatives $(-infty, -1]$. It is natural to choose $0 le theta < pi/2$ to get the positive values of $sec$, but for the negative values there are two reasonable choices: $pi/2 < theta le pi$ or $pi < theta le 3pi/2$. (Notice there is a discontinuity of $sec$ at $pi/2$ so either way we do not get a connected interval.) For this problem, we choose the second one because $|tan theta| = tantheta$ is a convenient property for this problem.
You can also do the problem with the other choice, but then you end up with a sign factor $|tan theta|/tantheta$ in the integral which you have to resolve by splitting into two cases.
answered Feb 2 at 18:46
TedTed
22.2k13361
answered Feb 2 at 18:46
TedTed
22.2k13361
answered Feb 2 at 18:46
TedTed
22.2k13361
answered Feb 2 at 18:46
TedTed
22.2k13361
22.2k13361
add a comment |
add a comment |
$begingroup$
The book solution is glossing over an important point--the function $1/sqrt{x^2-a^2}$, and thus its antiderivative, is only defined on the set $(-infty,-a)cup(a,infty)$, which is not a connected set. So to find an antiderivative properly, each connected component (in this case $(-infty,-a)$ and $(a,infty)$) should be handled separately.
It just so happens that the change of variables $xrightarrow asectheta$ is helpful for both connected components. For $(a,infty)$, this either maps to $(0,pi/2)$ or $(pi/2,pi)$, while for $(-a,-infty)$, this maps to either $(pi,3pi/2)$ or $(3pi/2,2pi)$. For each component, either choice is valid, but you have to make one or the other, because otherwise the mapping isn't unique. The book merely chooses $(0,pi/2)$ for $(a,infty)$ and $(pi,3pi/2)$ for $(-a,-infty)$ because it makes the algebra simpler.
In fact, because it's sloppy in handling the connected components, the book actually has an error. The general form of the function that satisfies $f'(x) = 1/sqrt{x^2-a^2}$ is
$$
f(x) = begin{cases}lnleft|x+sqrt{x^2-a^2}right| + C_1 & x < -a\lnleft|x+sqrt{x^2-a^2}right|+ C_2 & x > aend{cases}.
$$
Because the domain isn't connected, there's no continuity requirement across the gap, so the two regions can have different constants of integration.
answered Feb 2 at 19:04
eyeballfrogeyeballfrog
7,212633
$endgroup$
add a comment |
$begingroup$
The book solution is glossing over an important point--the function $1/sqrt{x^2-a^2}$, and thus its antiderivative, is only defined on the set $(-infty,-a)cup(a,infty)$, which is not a connected set. So to find an antiderivative properly, each connected component (in this case $(-infty,-a)$ and $(a,infty)$) should be handled separately.
It just so happens that the change of variables $xrightarrow asectheta$ is helpful for both connected components. For $(a,infty)$, this either maps to $(0,pi/2)$ or $(pi/2,pi)$, while for $(-a,-infty)$, this maps to either $(pi,3pi/2)$ or $(3pi/2,2pi)$. For each component, either choice is valid, but you have to make one or the other, because otherwise the mapping isn't unique. The book merely chooses $(0,pi/2)$ for $(a,infty)$ and $(pi,3pi/2)$ for $(-a,-infty)$ because it makes the algebra simpler.
In fact, because it's sloppy in handling the connected components, the book actually has an error. The general form of the function that satisfies $f'(x) = 1/sqrt{x^2-a^2}$ is
$$
f(x) = begin{cases}lnleft|x+sqrt{x^2-a^2}right| + C_1 & x < -a\lnleft|x+sqrt{x^2-a^2}right|+ C_2 & x > aend{cases}.
$$
Because the domain isn't connected, there's no continuity requirement across the gap, so the two regions can have different constants of integration.
answered Feb 2 at 19:04
eyeballfrogeyeballfrog
7,212633
$endgroup$
add a comment |
$begingroup$
The book solution is glossing over an important point--the function $1/sqrt{x^2-a^2}$, and thus its antiderivative, is only defined on the set $(-infty,-a)cup(a,infty)$, which is not a connected set. So to find an antiderivative properly, each connected component (in this case $(-infty,-a)$ and $(a,infty)$) should be handled separately.
It just so happens that the change of variables $xrightarrow asectheta$ is helpful for both connected components. For $(a,infty)$, this either maps to $(0,pi/2)$ or $(pi/2,pi)$, while for $(-a,-infty)$, this maps to either $(pi,3pi/2)$ or $(3pi/2,2pi)$. For each component, either choice is valid, but you have to make one or the other, because otherwise the mapping isn't unique. The book merely chooses $(0,pi/2)$ for $(a,infty)$ and $(pi,3pi/2)$ for $(-a,-infty)$ because it makes the algebra simpler.
In fact, because it's sloppy in handling the connected components, the book actually has an error. The general form of the function that satisfies $f'(x) = 1/sqrt{x^2-a^2}$ is
$$
f(x) = begin{cases}lnleft|x+sqrt{x^2-a^2}right| + C_1 & x < -a\lnleft|x+sqrt{x^2-a^2}right|+ C_2 & x > aend{cases}.
$$
Because the domain isn't connected, there's no continuity requirement across the gap, so the two regions can have different constants of integration.
answered Feb 2 at 19:04
eyeballfrogeyeballfrog
7,212633
$endgroup$
The book solution is glossing over an important point--the function $1/sqrt{x^2-a^2}$, and thus its antiderivative, is only defined on the set $(-infty,-a)cup(a,infty)$, which is not a connected set. So to find an antiderivative properly, each connected component (in this case $(-infty,-a)$ and $(a,infty)$) should be handled separately.
It just so happens that the change of variables $xrightarrow asectheta$ is helpful for both connected components. For $(a,infty)$, this either maps to $(0,pi/2)$ or $(pi/2,pi)$, while for $(-a,-infty)$, this maps to either $(pi,3pi/2)$ or $(3pi/2,2pi)$. For each component, either choice is valid, but you have to make one or the other, because otherwise the mapping isn't unique. The book merely chooses $(0,pi/2)$ for $(a,infty)$ and $(pi,3pi/2)$ for $(-a,-infty)$ because it makes the algebra simpler.
In fact, because it's sloppy in handling the connected components, the book actually has an error. The general form of the function that satisfies $f'(x) = 1/sqrt{x^2-a^2}$ is
$$
f(x) = begin{cases}lnleft|x+sqrt{x^2-a^2}right| + C_1 & x < -a\lnleft|x+sqrt{x^2-a^2}right|+ C_2 & x > aend{cases}.
$$
Because the domain isn't connected, there's no continuity requirement across the gap, so the two regions can have different constants of integration.
answered Feb 2 at 19:04
eyeballfrogeyeballfrog
7,212633
edited Feb 2 at 19:14
answered Feb 2 at 19:04
eyeballfrogeyeballfrog
7,212633
answered Feb 2 at 19:04
eyeballfrogeyeballfrog
7,212633
answered Feb 2 at 19:04
eyeballfrogeyeballfrog
7,212633
7,212633
add a comment |
add a comment |
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