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In trig substitutions, why favor $sin$, $tan$, $sec$, $sinh$, $cosh$, $tanh$ over $cos$, $cot$, $csc$, etc?
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Exemplified by this question and its comment, there seems to be a near-universal preference for the substitutions $sin(theta)$, $tan(theta)$, $sec(theta)$, $sinh(theta)$, $cosh(theta)$, and $tanh(theta)$ as opposed to $cos(theta)$, $csc(theta)$, $cot(theta)$, $mathrmsech(theta)$, $mathrmcsch(theta)$, or $tanh(theta)$.
Why the preference of one half dozen over the other, and why is there an asymmetry between the specific trig functions and the specific hyperbolic functions that are preferred (I am not asking about the general difference between the two categories, as asked in the linked question, but why, i.e., $sec(theta)$ is privileged over $mathrmsech(theta)$)? Finally, is it viable to use any of the more exotic trig functions or their hyperbolic counterparts, or even an inverse trig/hyberbolic function for such a substitution?
integration trigonometry substitution hyperbolic-functions
edited Jul 25 at 20:00
Blue
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asked Jul 16 at 16:29
user10478
1749
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up vote
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Exemplified by this question and its comment, there seems to be a near-universal preference for the substitutions $sin(theta)$, $tan(theta)$, $sec(theta)$, $sinh(theta)$, $cosh(theta)$, and $tanh(theta)$ as opposed to $cos(theta)$, $csc(theta)$, $cot(theta)$, $mathrmsech(theta)$, $mathrmcsch(theta)$, or $tanh(theta)$.
Why the preference of one half dozen over the other, and why is there an asymmetry between the specific trig functions and the specific hyperbolic functions that are preferred (I am not asking about the general difference between the two categories, as asked in the linked question, but why, i.e., $sec(theta)$ is privileged over $mathrmsech(theta)$)? Finally, is it viable to use any of the more exotic trig functions or their hyperbolic counterparts, or even an inverse trig/hyberbolic function for such a substitution?
integration trigonometry substitution hyperbolic-functions
edited Jul 25 at 20:00
Blue
43.7k868141
asked Jul 16 at 16:29
user10478
1749
I don't know if I understand your question right, but substituting with $cos$ or with $sin$ is basically the same, because the functions only differ by a phase.
– klirk
Jul 16 at 16:47
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Exemplified by this question and its comment, there seems to be a near-universal preference for the substitutions $sin(theta)$, $tan(theta)$, $sec(theta)$, $sinh(theta)$, $cosh(theta)$, and $tanh(theta)$ as opposed to $cos(theta)$, $csc(theta)$, $cot(theta)$, $mathrmsech(theta)$, $mathrmcsch(theta)$, or $tanh(theta)$.
Why the preference of one half dozen over the other, and why is there an asymmetry between the specific trig functions and the specific hyperbolic functions that are preferred (I am not asking about the general difference between the two categories, as asked in the linked question, but why, i.e., $sec(theta)$ is privileged over $mathrmsech(theta)$)? Finally, is it viable to use any of the more exotic trig functions or their hyperbolic counterparts, or even an inverse trig/hyberbolic function for such a substitution?
integration trigonometry substitution hyperbolic-functions
edited Jul 25 at 20:00
Blue
43.7k868141
asked Jul 16 at 16:29
user10478
1749
Exemplified by this question and its comment, there seems to be a near-universal preference for the substitutions $sin(theta)$, $tan(theta)$, $sec(theta)$, $sinh(theta)$, $cosh(theta)$, and $tanh(theta)$ as opposed to $cos(theta)$, $csc(theta)$, $cot(theta)$, $mathrmsech(theta)$, $mathrmcsch(theta)$, or $tanh(theta)$.
Why the preference of one half dozen over the other, and why is there an asymmetry between the specific trig functions and the specific hyperbolic functions that are preferred (I am not asking about the general difference between the two categories, as asked in the linked question, but why, i.e., $sec(theta)$ is privileged over $mathrmsech(theta)$)? Finally, is it viable to use any of the more exotic trig functions or their hyperbolic counterparts, or even an inverse trig/hyberbolic function for such a substitution?
integration trigonometry substitution hyperbolic-functions
edited Jul 25 at 20:00
Blue
43.7k868141
asked Jul 16 at 16:29
user10478
1749
edited Jul 25 at 20:00
Blue
43.7k868141
edited Jul 25 at 20:00
Blue
43.7k868141
edited Jul 25 at 20:00
Blue
43.7k868141
43.7k868141
asked Jul 16 at 16:29
user10478
1749
asked Jul 16 at 16:29
user10478
1749
asked Jul 16 at 16:29
user10478
1749
1749
I don't know if I understand your question right, but substituting with $cos$ or with $sin$ is basically the same, because the functions only differ by a phase.
– klirk
Jul 16 at 16:47
add a comment |Â
I don't know if I understand your question right, but substituting with $cos$ or with $sin$ is basically the same, because the functions only differ by a phase.
– klirk
Jul 16 at 16:47
I don't know if I understand your question right, but substituting with $cos$ or with $sin$ is basically the same, because the functions only differ by a phase.
– klirk
Jul 16 at 16:47
I don't know if I understand your question right, but substituting with $cos$ or with $sin$ is basically the same, because the functions only differ by a phase.
– klirk
Jul 16 at 16:47
add a comment |Â
1 Answer
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Say I evaluate $int_0^1fracdxsqrt1-x^2$; the advantage of $x=sintheta$ over $x=costheta$ is it's order-preserving. That doesn't sound like much, but order-reversing substitutions can often trip people up with more complicated problems. The same logic explains using $tantheta$ over $csctheta$ etc.
The Gudermannian function $operatornamegd(x)$ defined by $tanfrac12operatornamegdx=tanhfracx2$ implies a number of connections between circular and hyperbolic functions, e.g. $sinoperatornamegdx=tanh x$. This explains some of the observed flexibility in substitutions; for example, with the above integral I could have instead used $x=tanh t$. As you'll see if you try that one, you get a much harder problem than with $x=sintheta$ - the kind of problem, in fact, that furthers understanding of the Gudermannian.
There are times a hyperbolic substitution is worthwhile, e.g. $x=sinh t$ to compute $intfracdxsqrt1+x^2=int dt=operatornamearsinhx+C$. Or to take a less obvious example, $intsqrt1+x^2dx$ is probably easier with $x=sinh t$ than $x=tantheta$, since everyone knows how to integrate $cosh^2 t$, whereas $sec^3theta$ is so tricky it gets its own Wikipedia article.
answered Jul 16 at 17:09
J.G.
13.2k11424
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Say I evaluate $int_0^1fracdxsqrt1-x^2$; the advantage of $x=sintheta$ over $x=costheta$ is it's order-preserving. That doesn't sound like much, but order-reversing substitutions can often trip people up with more complicated problems. The same logic explains using $tantheta$ over $csctheta$ etc.
The Gudermannian function $operatornamegd(x)$ defined by $tanfrac12operatornamegdx=tanhfracx2$ implies a number of connections between circular and hyperbolic functions, e.g. $sinoperatornamegdx=tanh x$. This explains some of the observed flexibility in substitutions; for example, with the above integral I could have instead used $x=tanh t$. As you'll see if you try that one, you get a much harder problem than with $x=sintheta$ - the kind of problem, in fact, that furthers understanding of the Gudermannian.
There are times a hyperbolic substitution is worthwhile, e.g. $x=sinh t$ to compute $intfracdxsqrt1+x^2=int dt=operatornamearsinhx+C$. Or to take a less obvious example, $intsqrt1+x^2dx$ is probably easier with $x=sinh t$ than $x=tantheta$, since everyone knows how to integrate $cosh^2 t$, whereas $sec^3theta$ is so tricky it gets its own Wikipedia article.
answered Jul 16 at 17:09
J.G.
13.2k11424
add a comment |Â
up vote
1
down vote
accepted
Say I evaluate $int_0^1fracdxsqrt1-x^2$; the advantage of $x=sintheta$ over $x=costheta$ is it's order-preserving. That doesn't sound like much, but order-reversing substitutions can often trip people up with more complicated problems. The same logic explains using $tantheta$ over $csctheta$ etc.
The Gudermannian function $operatornamegd(x)$ defined by $tanfrac12operatornamegdx=tanhfracx2$ implies a number of connections between circular and hyperbolic functions, e.g. $sinoperatornamegdx=tanh x$. This explains some of the observed flexibility in substitutions; for example, with the above integral I could have instead used $x=tanh t$. As you'll see if you try that one, you get a much harder problem than with $x=sintheta$ - the kind of problem, in fact, that furthers understanding of the Gudermannian.
There are times a hyperbolic substitution is worthwhile, e.g. $x=sinh t$ to compute $intfracdxsqrt1+x^2=int dt=operatornamearsinhx+C$. Or to take a less obvious example, $intsqrt1+x^2dx$ is probably easier with $x=sinh t$ than $x=tantheta$, since everyone knows how to integrate $cosh^2 t$, whereas $sec^3theta$ is so tricky it gets its own Wikipedia article.
answered Jul 16 at 17:09
J.G.
13.2k11424
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Say I evaluate $int_0^1fracdxsqrt1-x^2$; the advantage of $x=sintheta$ over $x=costheta$ is it's order-preserving. That doesn't sound like much, but order-reversing substitutions can often trip people up with more complicated problems. The same logic explains using $tantheta$ over $csctheta$ etc.
The Gudermannian function $operatornamegd(x)$ defined by $tanfrac12operatornamegdx=tanhfracx2$ implies a number of connections between circular and hyperbolic functions, e.g. $sinoperatornamegdx=tanh x$. This explains some of the observed flexibility in substitutions; for example, with the above integral I could have instead used $x=tanh t$. As you'll see if you try that one, you get a much harder problem than with $x=sintheta$ - the kind of problem, in fact, that furthers understanding of the Gudermannian.
There are times a hyperbolic substitution is worthwhile, e.g. $x=sinh t$ to compute $intfracdxsqrt1+x^2=int dt=operatornamearsinhx+C$. Or to take a less obvious example, $intsqrt1+x^2dx$ is probably easier with $x=sinh t$ than $x=tantheta$, since everyone knows how to integrate $cosh^2 t$, whereas $sec^3theta$ is so tricky it gets its own Wikipedia article.
answered Jul 16 at 17:09
J.G.
13.2k11424
Say I evaluate $int_0^1fracdxsqrt1-x^2$; the advantage of $x=sintheta$ over $x=costheta$ is it's order-preserving. That doesn't sound like much, but order-reversing substitutions can often trip people up with more complicated problems. The same logic explains using $tantheta$ over $csctheta$ etc.
The Gudermannian function $operatornamegd(x)$ defined by $tanfrac12operatornamegdx=tanhfracx2$ implies a number of connections between circular and hyperbolic functions, e.g. $sinoperatornamegdx=tanh x$. This explains some of the observed flexibility in substitutions; for example, with the above integral I could have instead used $x=tanh t$. As you'll see if you try that one, you get a much harder problem than with $x=sintheta$ - the kind of problem, in fact, that furthers understanding of the Gudermannian.
There are times a hyperbolic substitution is worthwhile, e.g. $x=sinh t$ to compute $intfracdxsqrt1+x^2=int dt=operatornamearsinhx+C$. Or to take a less obvious example, $intsqrt1+x^2dx$ is probably easier with $x=sinh t$ than $x=tantheta$, since everyone knows how to integrate $cosh^2 t$, whereas $sec^3theta$ is so tricky it gets its own Wikipedia article.
answered Jul 16 at 17:09
J.G.
13.2k11424
edited Jul 25 at 19:31
answered Jul 16 at 17:09
J.G.
13.2k11424
answered Jul 16 at 17:09
J.G.
13.2k11424
answered Jul 16 at 17:09
J.G.
13.2k11424
13.2k11424
add a comment |Â
add a comment |Â
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I don't know if I understand your question right, but substituting with $cos$ or with $sin$ is basically the same, because the functions only differ by a phase.
– klirk
Jul 16 at 16:47