Problem with value of integral












2












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I calculate $int frac{dx}{sin^2x+1}=frac{1}{sqrt{2}}arctan(sqrt{2}tan x)+c.$ And then I want to calculate $$int_{0}^{pi} frac{dx}{sin^2x+1}$$. But $tanpi=tan0=0$. So it seems that $int_{0}^{pi} frac{dx}{sin^2x+1}=0$, but it's not true. Where is mistake in my justification ?






real-analysis calculus integration proof-verification







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  • 1




    $begingroup$
    math.stackexchange.com/questions/2229955/…
    $endgroup$
    – Olivier Oloa
    7 hours ago






  • 2




    $begingroup$
    The function you found as the integral is discontinuous at $pi/2$.
    $endgroup$
    – Bernard Massé
    7 hours ago
















2












$begingroup$


I calculate $int frac{dx}{sin^2x+1}=frac{1}{sqrt{2}}arctan(sqrt{2}tan x)+c.$ And then I want to calculate $$int_{0}^{pi} frac{dx}{sin^2x+1}$$. But $tanpi=tan0=0$. So it seems that $int_{0}^{pi} frac{dx}{sin^2x+1}=0$, but it's not true. Where is mistake in my justification ?






real-analysis calculus integration proof-verification







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    math.stackexchange.com/questions/2229955/…
    $endgroup$
    – Olivier Oloa
    7 hours ago






  • 2




    $begingroup$
    The function you found as the integral is discontinuous at $pi/2$.
    $endgroup$
    – Bernard Massé
    7 hours ago














2












2








2


1



$begingroup$


I calculate $int frac{dx}{sin^2x+1}=frac{1}{sqrt{2}}arctan(sqrt{2}tan x)+c.$ And then I want to calculate $$int_{0}^{pi} frac{dx}{sin^2x+1}$$. But $tanpi=tan0=0$. So it seems that $int_{0}^{pi} frac{dx}{sin^2x+1}=0$, but it's not true. Where is mistake in my justification ?






real-analysis calculus integration proof-verification







share|cite|improve this question











$endgroup$




I calculate $int frac{dx}{sin^2x+1}=frac{1}{sqrt{2}}arctan(sqrt{2}tan x)+c.$ And then I want to calculate $$int_{0}^{pi} frac{dx}{sin^2x+1}$$. But $tanpi=tan0=0$. So it seems that $int_{0}^{pi} frac{dx}{sin^2x+1}=0$, but it's not true. Where is mistake in my justification ?






real-analysis calculus integration proof-verification




real-analysis calculus integration proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 7 hours ago









clathratus

4,715337




4,715337










asked 7 hours ago









LucianLucian

266




266








  • 1




    $begingroup$
    math.stackexchange.com/questions/2229955/…
    $endgroup$
    – Olivier Oloa
    7 hours ago






  • 2




    $begingroup$
    The function you found as the integral is discontinuous at $pi/2$.
    $endgroup$
    – Bernard Massé
    7 hours ago














  • 1




    $begingroup$
    math.stackexchange.com/questions/2229955/…
    $endgroup$
    – Olivier Oloa
    7 hours ago






  • 2




    $begingroup$
    The function you found as the integral is discontinuous at $pi/2$.
    $endgroup$
    – Bernard Massé
    7 hours ago








1




1




$begingroup$
math.stackexchange.com/questions/2229955/…
$endgroup$
– Olivier Oloa
7 hours ago




$begingroup$
math.stackexchange.com/questions/2229955/…
$endgroup$
– Olivier Oloa
7 hours ago




2




2




$begingroup$
The function you found as the integral is discontinuous at $pi/2$.
$endgroup$
– Bernard Massé
7 hours ago




$begingroup$
The function you found as the integral is discontinuous at $pi/2$.
$endgroup$
– Bernard Massé
7 hours ago










2 Answers
2






active

oldest

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5












$begingroup$

Hint:
$$int_0^pi frac{dx}{1+sin(x)^2}=2int_0^{pi/2}frac{dx}{1+sin(x)^2}$$
And $arctan infty=pi/2$. Can you take it from here?






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Sure ! I now see my mistake. Thank you very much for answer !
    $endgroup$
    – Lucian
    6 hours ago










  • $begingroup$
    You are very welcome :)
    $endgroup$
    – clathratus
    5 hours ago



















2












$begingroup$

I made the same mistake before. Refer to [Integral][Please identify problem] $displaystyleint cfrac{1}{1+x^4}>mathrm{d} x$



The reason of the problem is that $arctan(sqrt{2}tan x)$ has a jump but the integral should be continuous, so in the two branches of the function (splitted by the jump point), you need to pick 2 different constants to make it continuous. It also apply to similar situations.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    Hint:
    $$int_0^pi frac{dx}{1+sin(x)^2}=2int_0^{pi/2}frac{dx}{1+sin(x)^2}$$
    And $arctan infty=pi/2$. Can you take it from here?






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Sure ! I now see my mistake. Thank you very much for answer !
      $endgroup$
      – Lucian
      6 hours ago










    • $begingroup$
      You are very welcome :)
      $endgroup$
      – clathratus
      5 hours ago
















    5












    $begingroup$

    Hint:
    $$int_0^pi frac{dx}{1+sin(x)^2}=2int_0^{pi/2}frac{dx}{1+sin(x)^2}$$
    And $arctan infty=pi/2$. Can you take it from here?






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Sure ! I now see my mistake. Thank you very much for answer !
      $endgroup$
      – Lucian
      6 hours ago










    • $begingroup$
      You are very welcome :)
      $endgroup$
      – clathratus
      5 hours ago














    5












    5








    5





    $begingroup$

    Hint:
    $$int_0^pi frac{dx}{1+sin(x)^2}=2int_0^{pi/2}frac{dx}{1+sin(x)^2}$$
    And $arctan infty=pi/2$. Can you take it from here?






    share|cite|improve this answer









    $endgroup$



    Hint:
    $$int_0^pi frac{dx}{1+sin(x)^2}=2int_0^{pi/2}frac{dx}{1+sin(x)^2}$$
    And $arctan infty=pi/2$. Can you take it from here?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 7 hours ago









    clathratusclathratus

    4,715337




    4,715337








    • 1




      $begingroup$
      Sure ! I now see my mistake. Thank you very much for answer !
      $endgroup$
      – Lucian
      6 hours ago










    • $begingroup$
      You are very welcome :)
      $endgroup$
      – clathratus
      5 hours ago














    • 1




      $begingroup$
      Sure ! I now see my mistake. Thank you very much for answer !
      $endgroup$
      – Lucian
      6 hours ago










    • $begingroup$
      You are very welcome :)
      $endgroup$
      – clathratus
      5 hours ago








    1




    1




    $begingroup$
    Sure ! I now see my mistake. Thank you very much for answer !
    $endgroup$
    – Lucian
    6 hours ago




    $begingroup$
    Sure ! I now see my mistake. Thank you very much for answer !
    $endgroup$
    – Lucian
    6 hours ago












    $begingroup$
    You are very welcome :)
    $endgroup$
    – clathratus
    5 hours ago




    $begingroup$
    You are very welcome :)
    $endgroup$
    – clathratus
    5 hours ago











    2












    $begingroup$

    I made the same mistake before. Refer to [Integral][Please identify problem] $displaystyleint cfrac{1}{1+x^4}>mathrm{d} x$



    The reason of the problem is that $arctan(sqrt{2}tan x)$ has a jump but the integral should be continuous, so in the two branches of the function (splitted by the jump point), you need to pick 2 different constants to make it continuous. It also apply to similar situations.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      I made the same mistake before. Refer to [Integral][Please identify problem] $displaystyleint cfrac{1}{1+x^4}>mathrm{d} x$



      The reason of the problem is that $arctan(sqrt{2}tan x)$ has a jump but the integral should be continuous, so in the two branches of the function (splitted by the jump point), you need to pick 2 different constants to make it continuous. It also apply to similar situations.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        I made the same mistake before. Refer to [Integral][Please identify problem] $displaystyleint cfrac{1}{1+x^4}>mathrm{d} x$



        The reason of the problem is that $arctan(sqrt{2}tan x)$ has a jump but the integral should be continuous, so in the two branches of the function (splitted by the jump point), you need to pick 2 different constants to make it continuous. It also apply to similar situations.






        share|cite|improve this answer









        $endgroup$



        I made the same mistake before. Refer to [Integral][Please identify problem] $displaystyleint cfrac{1}{1+x^4}>mathrm{d} x$



        The reason of the problem is that $arctan(sqrt{2}tan x)$ has a jump but the integral should be continuous, so in the two branches of the function (splitted by the jump point), you need to pick 2 different constants to make it continuous. It also apply to similar situations.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 6 hours ago









        LanceLance

        64229




        64229






























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