How to calculate $intlimits_{0}^{frac{pi}{2}}frac{mathrm{d}x}{1+tan^{a}(x)}$ [duplicate]

0

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This question already has an answer here:

• 
  Integrate $int_0^{pi/2} frac{1}{1+tan^alpha{x}},mathrm{d}x$
  
  4 answers
  
  


• 
  Evaluate the integral $int^{frac{pi}{2}}_0 frac{sin^3x}{sin^3x+cos^3x},mathrm dx$. [duplicate]
  
  3 answers
  
  

How to calculate
$$
intlimits_{0}^{frac{pi}{2}}frac{mathrm{d}x}{1+tan^{a}(x)} ;,
$$
where $a$ is a constant? Any hint will be appreciated.

definite-integrals

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edited Jan 28 '17 at 12:42

Björn Friedrich

2,70661831

asked Jan 28 '17 at 12:38

Good boyGood boy

562310

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marked as duplicate by lab bhattacharjee, C. Falcon, Claude Leibovici, Lucian, Daniel W. Farlow Jan 28 '17 at 23:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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0

$begingroup$

This question already has an answer here:

• 
  Integrate $int_0^{pi/2} frac{1}{1+tan^alpha{x}},mathrm{d}x$
  
  4 answers
  
  


• 
  Evaluate the integral $int^{frac{pi}{2}}_0 frac{sin^3x}{sin^3x+cos^3x},mathrm dx$. [duplicate]
  
  3 answers
  
  

How to calculate
$$
intlimits_{0}^{frac{pi}{2}}frac{mathrm{d}x}{1+tan^{a}(x)} ;,
$$
where $a$ is a constant? Any hint will be appreciated.

definite-integrals

share|cite|improve this question

edited Jan 28 '17 at 12:42

Björn Friedrich

2,70661831

asked Jan 28 '17 at 12:38

Good boyGood boy

562310

$endgroup$

marked as duplicate by lab bhattacharjee, C. Falcon, Claude Leibovici, Lucian, Daniel W. Farlow Jan 28 '17 at 23:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

0

0

0

$begingroup$

This question already has an answer here:

• 
  Integrate $int_0^{pi/2} frac{1}{1+tan^alpha{x}},mathrm{d}x$
  
  4 answers
  
  


• 
  Evaluate the integral $int^{frac{pi}{2}}_0 frac{sin^3x}{sin^3x+cos^3x},mathrm dx$. [duplicate]
  
  3 answers
  
  

How to calculate
$$
intlimits_{0}^{frac{pi}{2}}frac{mathrm{d}x}{1+tan^{a}(x)} ;,
$$
where $a$ is a constant? Any hint will be appreciated.

definite-integrals

share|cite|improve this question

edited Jan 28 '17 at 12:42

Björn Friedrich

2,70661831

asked Jan 28 '17 at 12:38

Good boyGood boy

562310

$endgroup$

This question already has an answer here:

• 
  Integrate $int_0^{pi/2} frac{1}{1+tan^alpha{x}},mathrm{d}x$
  
  4 answers
  
  


• 
  Evaluate the integral $int^{frac{pi}{2}}_0 frac{sin^3x}{sin^3x+cos^3x},mathrm dx$. [duplicate]
  
  3 answers
  
  

How to calculate
$$
intlimits_{0}^{frac{pi}{2}}frac{mathrm{d}x}{1+tan^{a}(x)} ;,
$$
where $a$ is a constant? Any hint will be appreciated.

This question already has an answer here:

• 
  Integrate $int_0^{pi/2} frac{1}{1+tan^alpha{x}},mathrm{d}x$
  
  4 answers
  
  


• 
  Evaluate the integral $int^{frac{pi}{2}}_0 frac{sin^3x}{sin^3x+cos^3x},mathrm dx$. [duplicate]
  
  3 answers
  
  

definite-integrals

definite-integrals

share|cite|improve this question

edited Jan 28 '17 at 12:42

Björn Friedrich

2,70661831

asked Jan 28 '17 at 12:38

Good boyGood boy

562310

share|cite|improve this question

edited Jan 28 '17 at 12:42

Björn Friedrich

2,70661831

asked Jan 28 '17 at 12:38

Good boyGood boy

562310

share|cite|improve this question

share|cite|improve this question

edited Jan 28 '17 at 12:42

Björn Friedrich

2,70661831

edited Jan 28 '17 at 12:42

Björn Friedrich

2,70661831

edited Jan 28 '17 at 12:42

Björn Friedrich

2,70661831

2,70661831

asked Jan 28 '17 at 12:38

Good boyGood boy

562310

asked Jan 28 '17 at 12:38

Good boyGood boy

562310

asked Jan 28 '17 at 12:38

Good boyGood boy

562310

562310

marked as duplicate by lab bhattacharjee, C. Falcon, Claude Leibovici, Lucian, Daniel W. Farlow Jan 28 '17 at 23:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by lab bhattacharjee, C. Falcon, Claude Leibovici, Lucian, Daniel W. Farlow Jan 28 '17 at 23:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

add a comment | 

1 Answer
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Let $$I = int^{frac{pi}{2}}_{0}frac{1}{1+tan^n (x)}dxcdots cdots (1)$$

Replace $displaystyle xrightarrow frac{pi}{2}-x$

So $$I = int^{frac{pi}{2}}_{0}frac{1}{1+tan^n left(frac{pi}{2}-x right)}dx = int^{frac{pi}{2}}_{0}frac{1}{1+cot^{n}(x)}cdots cdots (2)$$

So $$2I = int^{frac{pi}{2}}_{0}frac{1+tan^{n}(x)}{1+tan^{n}(x)}dx = frac{pi}{2}Rightarrow I = frac{pi}{4}.$$

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answered Jan 28 '17 at 12:56

juantheronjuantheron

34.4k1151143

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

$begingroup$

Let $$I = int^{frac{pi}{2}}_{0}frac{1}{1+tan^n (x)}dxcdots cdots (1)$$

Replace $displaystyle xrightarrow frac{pi}{2}-x$

So $$I = int^{frac{pi}{2}}_{0}frac{1}{1+tan^n left(frac{pi}{2}-x right)}dx = int^{frac{pi}{2}}_{0}frac{1}{1+cot^{n}(x)}cdots cdots (2)$$

So $$2I = int^{frac{pi}{2}}_{0}frac{1+tan^{n}(x)}{1+tan^{n}(x)}dx = frac{pi}{2}Rightarrow I = frac{pi}{4}.$$

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answered Jan 28 '17 at 12:56

juantheronjuantheron

34.4k1151143

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2

$begingroup$

Let $$I = int^{frac{pi}{2}}_{0}frac{1}{1+tan^n (x)}dxcdots cdots (1)$$

Replace $displaystyle xrightarrow frac{pi}{2}-x$

So $$I = int^{frac{pi}{2}}_{0}frac{1}{1+tan^n left(frac{pi}{2}-x right)}dx = int^{frac{pi}{2}}_{0}frac{1}{1+cot^{n}(x)}cdots cdots (2)$$

So $$2I = int^{frac{pi}{2}}_{0}frac{1+tan^{n}(x)}{1+tan^{n}(x)}dx = frac{pi}{2}Rightarrow I = frac{pi}{4}.$$

share|cite|improve this answer

answered Jan 28 '17 at 12:56

juantheronjuantheron

34.4k1151143

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2

2

2

$begingroup$

Let $$I = int^{frac{pi}{2}}_{0}frac{1}{1+tan^n (x)}dxcdots cdots (1)$$

Replace $displaystyle xrightarrow frac{pi}{2}-x$

So $$I = int^{frac{pi}{2}}_{0}frac{1}{1+tan^n left(frac{pi}{2}-x right)}dx = int^{frac{pi}{2}}_{0}frac{1}{1+cot^{n}(x)}cdots cdots (2)$$

So $$2I = int^{frac{pi}{2}}_{0}frac{1+tan^{n}(x)}{1+tan^{n}(x)}dx = frac{pi}{2}Rightarrow I = frac{pi}{4}.$$

share|cite|improve this answer

answered Jan 28 '17 at 12:56

juantheronjuantheron

34.4k1151143

$endgroup$

Let $$I = int^{frac{pi}{2}}_{0}frac{1}{1+tan^n (x)}dxcdots cdots (1)$$

Replace $displaystyle xrightarrow frac{pi}{2}-x$

So $$I = int^{frac{pi}{2}}_{0}frac{1}{1+tan^n left(frac{pi}{2}-x right)}dx = int^{frac{pi}{2}}_{0}frac{1}{1+cot^{n}(x)}cdots cdots (2)$$

So $$2I = int^{frac{pi}{2}}_{0}frac{1+tan^{n}(x)}{1+tan^{n}(x)}dx = frac{pi}{2}Rightarrow I = frac{pi}{4}.$$

share|cite|improve this answer

answered Jan 28 '17 at 12:56

juantheronjuantheron

34.4k1151143

share|cite|improve this answer

share|cite|improve this answer

answered Jan 28 '17 at 12:56

juantheronjuantheron

34.4k1151143

answered Jan 28 '17 at 12:56

juantheronjuantheron

34.4k1151143

answered Jan 28 '17 at 12:56

juantheronjuantheron

34.4k1151143

34.4k1151143

add a comment | 

add a comment | 

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