Using tangent substitution on $Bbb R$
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How can you calculate de integral of $f:Bbb RtoBbb R$ , $f(x)=dfrac1{3cos x+sin x+1}$? I have done it with tangent substitution but that works if $x$ is in $(-pi,pi)$ and I don't know how to extend that to $Bbb R$.
calculus integration
edited Dec 2 at 23:09
Jean-Claude Arbaut
14.7k63363
asked Dec 2 at 20:09
Gaboru
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up vote
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How can you calculate de integral of $f:Bbb RtoBbb R$ , $f(x)=dfrac1{3cos x+sin x+1}$? I have done it with tangent substitution but that works if $x$ is in $(-pi,pi)$ and I don't know how to extend that to $Bbb R$.
calculus integration
edited Dec 2 at 23:09
Jean-Claude Arbaut
14.7k63363
asked Dec 2 at 20:09
Gaboru
84
New contributor
Gaboru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Note that $f$ is periodic with any period of length $2pi$. Thus knowing an antiderivative on $[-pi,pi]$ is enough...
– Jean Marie
Dec 2 at 21:20
On -pi and pi would you need to put c1 and c2 and calculate them from the continuity?
– Gaboru
Dec 2 at 21:41
The issue is that if your function possesses asymptotes, for example near $x=2.1$. Any antiderivative makes sense only in intervals of continuity... Keep your answer as it is, adding it an arbitrary constant.
– Jean Marie
Dec 2 at 22:15
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How can you calculate de integral of $f:Bbb RtoBbb R$ , $f(x)=dfrac1{3cos x+sin x+1}$? I have done it with tangent substitution but that works if $x$ is in $(-pi,pi)$ and I don't know how to extend that to $Bbb R$.
calculus integration
edited Dec 2 at 23:09
Jean-Claude Arbaut
14.7k63363
asked Dec 2 at 20:09
Gaboru
84
New contributor
Gaboru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
How can you calculate de integral of $f:Bbb RtoBbb R$ , $f(x)=dfrac1{3cos x+sin x+1}$? I have done it with tangent substitution but that works if $x$ is in $(-pi,pi)$ and I don't know how to extend that to $Bbb R$.
calculus integration
calculus integration
edited Dec 2 at 23:09
Jean-Claude Arbaut
14.7k63363
asked Dec 2 at 20:09
Gaboru
84
New contributor
Gaboru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 2 at 23:09
Jean-Claude Arbaut
14.7k63363
asked Dec 2 at 20:09
Gaboru
84
New contributor
Gaboru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 2 at 23:09
Jean-Claude Arbaut
14.7k63363
edited Dec 2 at 23:09
Jean-Claude Arbaut
14.7k63363
edited Dec 2 at 23:09
Jean-Claude Arbaut
14.7k63363
14.7k63363
asked Dec 2 at 20:09
Gaboru
84
New contributor
Gaboru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 2 at 20:09
Gaboru
84
asked Dec 2 at 20:09
Gaboru
84
84
New contributor
Gaboru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Gaboru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Gaboru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Note that $f$ is periodic with any period of length $2pi$. Thus knowing an antiderivative on $[-pi,pi]$ is enough...
– Jean Marie
Dec 2 at 21:20
On -pi and pi would you need to put c1 and c2 and calculate them from the continuity?
– Gaboru
Dec 2 at 21:41
The issue is that if your function possesses asymptotes, for example near $x=2.1$. Any antiderivative makes sense only in intervals of continuity... Keep your answer as it is, adding it an arbitrary constant.
– Jean Marie
Dec 2 at 22:15
add a comment |
Note that $f$ is periodic with any period of length $2pi$. Thus knowing an antiderivative on $[-pi,pi]$ is enough...
– Jean Marie
Dec 2 at 21:20
On -pi and pi would you need to put c1 and c2 and calculate them from the continuity?
– Gaboru
Dec 2 at 21:41
The issue is that if your function possesses asymptotes, for example near $x=2.1$. Any antiderivative makes sense only in intervals of continuity... Keep your answer as it is, adding it an arbitrary constant.
– Jean Marie
Dec 2 at 22:15
Note that $f$ is periodic with any period of length $2pi$. Thus knowing an antiderivative on $[-pi,pi]$ is enough...
– Jean Marie
Dec 2 at 21:20
Note that $f$ is periodic with any period of length $2pi$. Thus knowing an antiderivative on $[-pi,pi]$ is enough...
– Jean Marie
Dec 2 at 21:20
On -pi and pi would you need to put c1 and c2 and calculate them from the continuity?
– Gaboru
Dec 2 at 21:41
On -pi and pi would you need to put c1 and c2 and calculate them from the continuity?
– Gaboru
Dec 2 at 21:41
The issue is that if your function possesses asymptotes, for example near $x=2.1$. Any antiderivative makes sense only in intervals of continuity... Keep your answer as it is, adding it an arbitrary constant.
– Jean Marie
Dec 2 at 22:15
The issue is that if your function possesses asymptotes, for example near $x=2.1$. Any antiderivative makes sense only in intervals of continuity... Keep your answer as it is, adding it an arbitrary constant.
– Jean Marie
Dec 2 at 22:15
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Have a look at the discontinuous curve associated with your function :
It possesses separated domains $D_k$, due to asymptotes : on each one, if $F(x)$ is your antiderivative, you can give the answer $F(x)+C_k$, each time with a specific $C_k$.
answered Dec 2 at 23:07
Jean Marie
28.2k41848
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Have a look at the discontinuous curve associated with your function :
It possesses separated domains $D_k$, due to asymptotes : on each one, if $F(x)$ is your antiderivative, you can give the answer $F(x)+C_k$, each time with a specific $C_k$.
answered Dec 2 at 23:07
Jean Marie
28.2k41848
add a comment |
up vote
0
down vote
Have a look at the discontinuous curve associated with your function :
It possesses separated domains $D_k$, due to asymptotes : on each one, if $F(x)$ is your antiderivative, you can give the answer $F(x)+C_k$, each time with a specific $C_k$.
answered Dec 2 at 23:07
Jean Marie
28.2k41848
add a comment |
up vote
0
down vote
up vote
0
down vote
Have a look at the discontinuous curve associated with your function :
It possesses separated domains $D_k$, due to asymptotes : on each one, if $F(x)$ is your antiderivative, you can give the answer $F(x)+C_k$, each time with a specific $C_k$.
answered Dec 2 at 23:07
Jean Marie
28.2k41848
Have a look at the discontinuous curve associated with your function :
It possesses separated domains $D_k$, due to asymptotes : on each one, if $F(x)$ is your antiderivative, you can give the answer $F(x)+C_k$, each time with a specific $C_k$.
answered Dec 2 at 23:07
Jean Marie
28.2k41848
answered Dec 2 at 23:07
Jean Marie
28.2k41848
answered Dec 2 at 23:07
Jean Marie
28.2k41848
answered Dec 2 at 23:07
Jean Marie
28.2k41848
28.2k41848
add a comment |
add a comment |
Gaboru is a new contributor. Be nice, and check out our Code of Conduct.
Gaboru is a new contributor. Be nice, and check out our Code of Conduct.
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Note that $f$ is periodic with any period of length $2pi$. Thus knowing an antiderivative on $[-pi,pi]$ is enough...
– Jean Marie
Dec 2 at 21:20
On -pi and pi would you need to put c1 and c2 and calculate them from the continuity?
– Gaboru
Dec 2 at 21:41
The issue is that if your function possesses asymptotes, for example near $x=2.1$. Any antiderivative makes sense only in intervals of continuity... Keep your answer as it is, adding it an arbitrary constant.
– Jean Marie
Dec 2 at 22:15