What's this definite integral?











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I'm confused about solving this definite integral: $int_{pi/6}^{7pi/6} sec{x}tan{x}{dx}$. When I solve it via fundamental theorem of calculus, it's pretty easy to see that the $sec{x}tan{x}$ integrates to $sec{x}$ then one simply solves: $sec{7pi/6}-sec{pi/6}$ to get: $-2.309$. However when I check this answer via a calculator, it says that the integral diverges--the graph of $sec{x}tan{x}$ confirms this. Another is that the result of $-2.309$ doesn't make sense because most of the graph is above the x-axis between the limits, thus the area should be positive. Does anybody know what's going on here, and how can I solve this?










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    $$pi/6<pi/2<7pi/6$$
    – hamam_Abdallah
    Nov 18 at 18:42















up vote
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I'm confused about solving this definite integral: $int_{pi/6}^{7pi/6} sec{x}tan{x}{dx}$. When I solve it via fundamental theorem of calculus, it's pretty easy to see that the $sec{x}tan{x}$ integrates to $sec{x}$ then one simply solves: $sec{7pi/6}-sec{pi/6}$ to get: $-2.309$. However when I check this answer via a calculator, it says that the integral diverges--the graph of $sec{x}tan{x}$ confirms this. Another is that the result of $-2.309$ doesn't make sense because most of the graph is above the x-axis between the limits, thus the area should be positive. Does anybody know what's going on here, and how can I solve this?










share|cite|improve this question


















  • 2




    $$pi/6<pi/2<7pi/6$$
    – hamam_Abdallah
    Nov 18 at 18:42













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I'm confused about solving this definite integral: $int_{pi/6}^{7pi/6} sec{x}tan{x}{dx}$. When I solve it via fundamental theorem of calculus, it's pretty easy to see that the $sec{x}tan{x}$ integrates to $sec{x}$ then one simply solves: $sec{7pi/6}-sec{pi/6}$ to get: $-2.309$. However when I check this answer via a calculator, it says that the integral diverges--the graph of $sec{x}tan{x}$ confirms this. Another is that the result of $-2.309$ doesn't make sense because most of the graph is above the x-axis between the limits, thus the area should be positive. Does anybody know what's going on here, and how can I solve this?










share|cite|improve this question













I'm confused about solving this definite integral: $int_{pi/6}^{7pi/6} sec{x}tan{x}{dx}$. When I solve it via fundamental theorem of calculus, it's pretty easy to see that the $sec{x}tan{x}$ integrates to $sec{x}$ then one simply solves: $sec{7pi/6}-sec{pi/6}$ to get: $-2.309$. However when I check this answer via a calculator, it says that the integral diverges--the graph of $sec{x}tan{x}$ confirms this. Another is that the result of $-2.309$ doesn't make sense because most of the graph is above the x-axis between the limits, thus the area should be positive. Does anybody know what's going on here, and how can I solve this?







calculus integration definite-integrals






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asked Nov 18 at 18:39









Jeamz

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83








  • 2




    $$pi/6<pi/2<7pi/6$$
    – hamam_Abdallah
    Nov 18 at 18:42














  • 2




    $$pi/6<pi/2<7pi/6$$
    – hamam_Abdallah
    Nov 18 at 18:42








2




2




$$pi/6<pi/2<7pi/6$$
– hamam_Abdallah
Nov 18 at 18:42




$$pi/6<pi/2<7pi/6$$
– hamam_Abdallah
Nov 18 at 18:42










2 Answers
2






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2
down vote



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You can't solve this using the fundamental theorem because one of the conditions is continuity of function on the interval of integration, which is violated at point $frac{pi}{2}$, which is inside the domain of integration.






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  • Thanks for clearing that up, this is the first time I had to test for continuity before solving a definite integral.
    – Jeamz
    Nov 18 at 18:56






  • 1




    @Jeamz This is a standard not-quite-a-trick question designed to encourage students to think before "plugging in to a formula".
    – Ethan Bolker
    Nov 18 at 19:31










  • The second FTC does not require the continuity of the function. For instance it works very well with $text{sgn}(x)$. What we have here is more serious: a vertical asymptote, which makes the integral improper.
    – Yves Daoust
    Nov 18 at 19:45


















up vote
2
down vote













Note that $cos (pi/2)=0$ and near $pi/2$ the cosine function behaves asymptotically as $cos(x)sim pi/2-x$.



Therefore $sec(x) tan(x) sim frac{1}{(pi/2 -x)^2}$ for $xto pi/2$.



Inasmuch as the integral $int_{pi/6}^{7pi/6}frac{1}{(pi/2-x)^2},dx$ fails to converge, the integral of interest diverges.





Note that a similarly naive application of the FOC would give the erroneous result



$$int_{pi/6}^{7pi/6}frac1{(pi/2-x)^2},dx=left.left(frac1{pi/2-x}right)right|_{pi/6}^{7pi/6}$$






There is one more point of interest. There are improper integrals that fail to converge, but can be interpreted, and exist, in the sense of their Cauchy Principal Values.




For example, the integral $int_{-1}^{1}frac 1x ,dx$ diverges. However, its Cauchy Principal Value is



$$lim_{epsilon to 0^+} left( int_{-1}^{-epsilon}frac1x ,dx+int_{epsilon}^{1}frac1x ,dxright)=0$$





It is, therefore, important to note that the integral of interest $int_{pi/6}^{7pi/6}sec(x)tan(x),dx$ fails to exist even as a Catchy Principal Value.






share|cite|improve this answer























  • Our calc curriculum doesn't cover this, but I have seen Cauchy Principal values before.
    – Jeamz
    Nov 18 at 19:56










  • Pleased to hear. I hope this answer was useful.
    – Mark Viola
    Nov 18 at 20:06











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










You can't solve this using the fundamental theorem because one of the conditions is continuity of function on the interval of integration, which is violated at point $frac{pi}{2}$, which is inside the domain of integration.






share|cite|improve this answer





















  • Thanks for clearing that up, this is the first time I had to test for continuity before solving a definite integral.
    – Jeamz
    Nov 18 at 18:56






  • 1




    @Jeamz This is a standard not-quite-a-trick question designed to encourage students to think before "plugging in to a formula".
    – Ethan Bolker
    Nov 18 at 19:31










  • The second FTC does not require the continuity of the function. For instance it works very well with $text{sgn}(x)$. What we have here is more serious: a vertical asymptote, which makes the integral improper.
    – Yves Daoust
    Nov 18 at 19:45















up vote
2
down vote



accepted










You can't solve this using the fundamental theorem because one of the conditions is continuity of function on the interval of integration, which is violated at point $frac{pi}{2}$, which is inside the domain of integration.






share|cite|improve this answer





















  • Thanks for clearing that up, this is the first time I had to test for continuity before solving a definite integral.
    – Jeamz
    Nov 18 at 18:56






  • 1




    @Jeamz This is a standard not-quite-a-trick question designed to encourage students to think before "plugging in to a formula".
    – Ethan Bolker
    Nov 18 at 19:31










  • The second FTC does not require the continuity of the function. For instance it works very well with $text{sgn}(x)$. What we have here is more serious: a vertical asymptote, which makes the integral improper.
    – Yves Daoust
    Nov 18 at 19:45













up vote
2
down vote



accepted







up vote
2
down vote



accepted






You can't solve this using the fundamental theorem because one of the conditions is continuity of function on the interval of integration, which is violated at point $frac{pi}{2}$, which is inside the domain of integration.






share|cite|improve this answer












You can't solve this using the fundamental theorem because one of the conditions is continuity of function on the interval of integration, which is violated at point $frac{pi}{2}$, which is inside the domain of integration.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 18 at 18:46









Makina

1,036114




1,036114












  • Thanks for clearing that up, this is the first time I had to test for continuity before solving a definite integral.
    – Jeamz
    Nov 18 at 18:56






  • 1




    @Jeamz This is a standard not-quite-a-trick question designed to encourage students to think before "plugging in to a formula".
    – Ethan Bolker
    Nov 18 at 19:31










  • The second FTC does not require the continuity of the function. For instance it works very well with $text{sgn}(x)$. What we have here is more serious: a vertical asymptote, which makes the integral improper.
    – Yves Daoust
    Nov 18 at 19:45


















  • Thanks for clearing that up, this is the first time I had to test for continuity before solving a definite integral.
    – Jeamz
    Nov 18 at 18:56






  • 1




    @Jeamz This is a standard not-quite-a-trick question designed to encourage students to think before "plugging in to a formula".
    – Ethan Bolker
    Nov 18 at 19:31










  • The second FTC does not require the continuity of the function. For instance it works very well with $text{sgn}(x)$. What we have here is more serious: a vertical asymptote, which makes the integral improper.
    – Yves Daoust
    Nov 18 at 19:45
















Thanks for clearing that up, this is the first time I had to test for continuity before solving a definite integral.
– Jeamz
Nov 18 at 18:56




Thanks for clearing that up, this is the first time I had to test for continuity before solving a definite integral.
– Jeamz
Nov 18 at 18:56




1




1




@Jeamz This is a standard not-quite-a-trick question designed to encourage students to think before "plugging in to a formula".
– Ethan Bolker
Nov 18 at 19:31




@Jeamz This is a standard not-quite-a-trick question designed to encourage students to think before "plugging in to a formula".
– Ethan Bolker
Nov 18 at 19:31












The second FTC does not require the continuity of the function. For instance it works very well with $text{sgn}(x)$. What we have here is more serious: a vertical asymptote, which makes the integral improper.
– Yves Daoust
Nov 18 at 19:45




The second FTC does not require the continuity of the function. For instance it works very well with $text{sgn}(x)$. What we have here is more serious: a vertical asymptote, which makes the integral improper.
– Yves Daoust
Nov 18 at 19:45










up vote
2
down vote













Note that $cos (pi/2)=0$ and near $pi/2$ the cosine function behaves asymptotically as $cos(x)sim pi/2-x$.



Therefore $sec(x) tan(x) sim frac{1}{(pi/2 -x)^2}$ for $xto pi/2$.



Inasmuch as the integral $int_{pi/6}^{7pi/6}frac{1}{(pi/2-x)^2},dx$ fails to converge, the integral of interest diverges.





Note that a similarly naive application of the FOC would give the erroneous result



$$int_{pi/6}^{7pi/6}frac1{(pi/2-x)^2},dx=left.left(frac1{pi/2-x}right)right|_{pi/6}^{7pi/6}$$






There is one more point of interest. There are improper integrals that fail to converge, but can be interpreted, and exist, in the sense of their Cauchy Principal Values.




For example, the integral $int_{-1}^{1}frac 1x ,dx$ diverges. However, its Cauchy Principal Value is



$$lim_{epsilon to 0^+} left( int_{-1}^{-epsilon}frac1x ,dx+int_{epsilon}^{1}frac1x ,dxright)=0$$





It is, therefore, important to note that the integral of interest $int_{pi/6}^{7pi/6}sec(x)tan(x),dx$ fails to exist even as a Catchy Principal Value.






share|cite|improve this answer























  • Our calc curriculum doesn't cover this, but I have seen Cauchy Principal values before.
    – Jeamz
    Nov 18 at 19:56










  • Pleased to hear. I hope this answer was useful.
    – Mark Viola
    Nov 18 at 20:06















up vote
2
down vote













Note that $cos (pi/2)=0$ and near $pi/2$ the cosine function behaves asymptotically as $cos(x)sim pi/2-x$.



Therefore $sec(x) tan(x) sim frac{1}{(pi/2 -x)^2}$ for $xto pi/2$.



Inasmuch as the integral $int_{pi/6}^{7pi/6}frac{1}{(pi/2-x)^2},dx$ fails to converge, the integral of interest diverges.





Note that a similarly naive application of the FOC would give the erroneous result



$$int_{pi/6}^{7pi/6}frac1{(pi/2-x)^2},dx=left.left(frac1{pi/2-x}right)right|_{pi/6}^{7pi/6}$$






There is one more point of interest. There are improper integrals that fail to converge, but can be interpreted, and exist, in the sense of their Cauchy Principal Values.




For example, the integral $int_{-1}^{1}frac 1x ,dx$ diverges. However, its Cauchy Principal Value is



$$lim_{epsilon to 0^+} left( int_{-1}^{-epsilon}frac1x ,dx+int_{epsilon}^{1}frac1x ,dxright)=0$$





It is, therefore, important to note that the integral of interest $int_{pi/6}^{7pi/6}sec(x)tan(x),dx$ fails to exist even as a Catchy Principal Value.






share|cite|improve this answer























  • Our calc curriculum doesn't cover this, but I have seen Cauchy Principal values before.
    – Jeamz
    Nov 18 at 19:56










  • Pleased to hear. I hope this answer was useful.
    – Mark Viola
    Nov 18 at 20:06













up vote
2
down vote










up vote
2
down vote









Note that $cos (pi/2)=0$ and near $pi/2$ the cosine function behaves asymptotically as $cos(x)sim pi/2-x$.



Therefore $sec(x) tan(x) sim frac{1}{(pi/2 -x)^2}$ for $xto pi/2$.



Inasmuch as the integral $int_{pi/6}^{7pi/6}frac{1}{(pi/2-x)^2},dx$ fails to converge, the integral of interest diverges.





Note that a similarly naive application of the FOC would give the erroneous result



$$int_{pi/6}^{7pi/6}frac1{(pi/2-x)^2},dx=left.left(frac1{pi/2-x}right)right|_{pi/6}^{7pi/6}$$






There is one more point of interest. There are improper integrals that fail to converge, but can be interpreted, and exist, in the sense of their Cauchy Principal Values.




For example, the integral $int_{-1}^{1}frac 1x ,dx$ diverges. However, its Cauchy Principal Value is



$$lim_{epsilon to 0^+} left( int_{-1}^{-epsilon}frac1x ,dx+int_{epsilon}^{1}frac1x ,dxright)=0$$





It is, therefore, important to note that the integral of interest $int_{pi/6}^{7pi/6}sec(x)tan(x),dx$ fails to exist even as a Catchy Principal Value.






share|cite|improve this answer














Note that $cos (pi/2)=0$ and near $pi/2$ the cosine function behaves asymptotically as $cos(x)sim pi/2-x$.



Therefore $sec(x) tan(x) sim frac{1}{(pi/2 -x)^2}$ for $xto pi/2$.



Inasmuch as the integral $int_{pi/6}^{7pi/6}frac{1}{(pi/2-x)^2},dx$ fails to converge, the integral of interest diverges.





Note that a similarly naive application of the FOC would give the erroneous result



$$int_{pi/6}^{7pi/6}frac1{(pi/2-x)^2},dx=left.left(frac1{pi/2-x}right)right|_{pi/6}^{7pi/6}$$






There is one more point of interest. There are improper integrals that fail to converge, but can be interpreted, and exist, in the sense of their Cauchy Principal Values.




For example, the integral $int_{-1}^{1}frac 1x ,dx$ diverges. However, its Cauchy Principal Value is



$$lim_{epsilon to 0^+} left( int_{-1}^{-epsilon}frac1x ,dx+int_{epsilon}^{1}frac1x ,dxright)=0$$





It is, therefore, important to note that the integral of interest $int_{pi/6}^{7pi/6}sec(x)tan(x),dx$ fails to exist even as a Catchy Principal Value.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 18 at 19:37

























answered Nov 18 at 18:48









Mark Viola

129k1273170




129k1273170












  • Our calc curriculum doesn't cover this, but I have seen Cauchy Principal values before.
    – Jeamz
    Nov 18 at 19:56










  • Pleased to hear. I hope this answer was useful.
    – Mark Viola
    Nov 18 at 20:06


















  • Our calc curriculum doesn't cover this, but I have seen Cauchy Principal values before.
    – Jeamz
    Nov 18 at 19:56










  • Pleased to hear. I hope this answer was useful.
    – Mark Viola
    Nov 18 at 20:06
















Our calc curriculum doesn't cover this, but I have seen Cauchy Principal values before.
– Jeamz
Nov 18 at 19:56




Our calc curriculum doesn't cover this, but I have seen Cauchy Principal values before.
– Jeamz
Nov 18 at 19:56












Pleased to hear. I hope this answer was useful.
– Mark Viola
Nov 18 at 20:06




Pleased to hear. I hope this answer was useful.
– Mark Viola
Nov 18 at 20:06


















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