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Problem with value of integral

Problem with calculation this integral: $int_0^pi frac{dx}{1+3sin^2x}$[Integral][Please identify problem] $displaystyleint cfrac{1}{1+x^4}>mathrm{d} x$Trig substitution integralHelp with maximization problem$int sqrt{frac{x}{x+1}}dx$Problem with definite integral $int _0^{frac{pi }{2}}sin left(arctan left(xright)+xright)dx$Why isn't $int frac{dx}{(1+x^2)sqrt{1+x^2}} = - sin(arctan(x))$?Integral of $frac{sqrt {x}}{x^2+x}$Problem with calculation this integral: $int_0^pi frac{dx}{1+3sin^2x}$Integral involving $phi$Evaluating an integral 2.Integration, get stuck at x=tan($theta$) when calculating arc length

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

edited 5 hours ago

clathratus

4,715337

asked 5 hours ago

LucianLucian

266

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

edited 5 hours ago

clathratus

4,715337

asked 5 hours ago

LucianLucian

266

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

edited 5 hours ago

clathratus

4,715337

asked 5 hours ago

LucianLucian

266

$endgroup$

share|cite|improve this question

edited 5 hours ago

clathratus

4,715337

asked 5 hours ago

LucianLucian

266

share|cite|improve this question

edited 5 hours ago

clathratus

4,715337

asked 5 hours ago

LucianLucian

266

edited 5 hours ago

clathratus

4,715337

edited 5 hours ago

clathratus

4,715337

asked 5 hours ago

LucianLucian

266

asked 5 hours ago

LucianLucian

266

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

answered 5 hours ago

clathratusclathratus

4,715337

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

answered 4 hours ago

LanceLance

64229

$endgroup$

add a comment | 

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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

answered 5 hours ago

clathratusclathratus

4,715337

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

answered 5 hours ago

clathratusclathratus

4,715337

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

answered 5 hours ago

clathratusclathratus

4,715337

$endgroup$

share|cite|improve this answer

answered 5 hours ago

clathratusclathratus

4,715337

answered 5 hours ago

clathratusclathratus

4,715337

answered 5 hours ago

clathratusclathratus

4,715337

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

answered 4 hours ago

LanceLance

64229

$endgroup$

add a comment | 

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

answered 4 hours ago

LanceLance

64229

$endgroup$

add a comment | 

$begingroup$

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

answered 4 hours ago

LanceLance

64229

$endgroup$

share|cite|improve this answer

answered 4 hours ago

LanceLance

64229

answered 4 hours ago

LanceLance

64229

answered 4 hours ago

LanceLance

64229