Integrate $int frac11+ tan xdx$ [duplicate]

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This question already has an answer here:

• 
  Simplest way to integrate $int frac11+tan xdx,$
  
  4 answers
  
  

Does this integral have a closed form?

$$int frac11+ tan x,dx$$

My attempt:

$$int frac11+ tan x,dx=ln (sin x + cos x) +int fractan x1+ tan x,dx$$

What is next?

calculus

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edited Jul 26 at 21:11

Michael Hardy

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asked Jul 26 at 20:28

sirous dehmami

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Jul 27 at 0:19

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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up vote
1
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favorite

1

This question already has an answer here:

• 
  Simplest way to integrate $int frac11+tan xdx,$
  
  4 answers
  
  

Does this integral have a closed form?

$$int frac11+ tan x,dx$$

My attempt:

$$int frac11+ tan x,dx=ln (sin x + cos x) +int fractan x1+ tan x,dx$$

What is next?

calculus

share|cite|improve this question

edited Jul 26 at 21:11

Michael Hardy

204k23186461

asked Jul 26 at 20:28

sirous dehmami

294

marked as duplicate by gt6989b, Clayton, Batominovski, amWhy calculus
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Jul 27 at 0:19

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 | 

up vote
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down vote

favorite

1

up vote
1
down vote

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1

This question already has an answer here:

• 
  Simplest way to integrate $int frac11+tan xdx,$
  
  4 answers
  
  

Does this integral have a closed form?

$$int frac11+ tan x,dx$$

My attempt:

$$int frac11+ tan x,dx=ln (sin x + cos x) +int fractan x1+ tan x,dx$$

What is next?

calculus

share|cite|improve this question

edited Jul 26 at 21:11

Michael Hardy

204k23186461

asked Jul 26 at 20:28

sirous dehmami

294

This question already has an answer here:

• 
  Simplest way to integrate $int frac11+tan xdx,$
  
  4 answers
  
  

Does this integral have a closed form?

$$int frac11+ tan x,dx$$

My attempt:

$$int frac11+ tan x,dx=ln (sin x + cos x) +int fractan x1+ tan x,dx$$

What is next?

This question already has an answer here:

• 
  Simplest way to integrate $int frac11+tan xdx,$
  
  4 answers
  
  

calculus

share|cite|improve this question

edited Jul 26 at 21:11

Michael Hardy

204k23186461

asked Jul 26 at 20:28

sirous dehmami

294

share|cite|improve this question

share|cite|improve this question

edited Jul 26 at 21:11

Michael Hardy

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edited Jul 26 at 21:11

Michael Hardy

204k23186461

edited Jul 26 at 21:11

Michael Hardy

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asked Jul 26 at 20:28

sirous dehmami

294

asked Jul 26 at 20:28

sirous dehmami

294

asked Jul 26 at 20:28

sirous dehmami

294

294

marked as duplicate by gt6989b, Clayton, Batominovski, amWhy calculus
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Jul 27 at 0:19

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7 Answers
7

active

oldest

votes

up vote
7
down vote

$$
I=int frac 1 1+tan x , dx=int frac cos xcos x +sin x , dx\
J=int frac sin xcos x +sin x , dx\
I+J=int frac cos x+sin xcos x +sin x , dx = x+C_1\ \
I-J=int frac cos x-sin xcos x +sin x ,dx=int frac dleft( sin x +cos xright) cos x +sin x =ln left| sin x +cos x right| + C_2 \
2I = x+ln left| sin x +cos xright| + C_3 \
I=frac 1 2 left[ x+ln left| sin x +cos x right| + C_3 right]
$$

share|cite|improve this answer

edited Jul 26 at 22:22

Michael Hardy

204k23186461

answered Jul 26 at 20:33

haqnatural

20.5k72457

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Writing int frac cos x + sin x cos x + sin x dx where int fraccos x + sin xcos x + sin x , dx would suffice doesn't serve a purpose and makes editing harder.
  – Michael Hardy
  Jul 26 at 22:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MichaelHardy, thank you for editing, but I am still can't find better-writing program rather than this one s1.daumcdn.net/editor/fp/service_nc/pencil/…
  – haqnatural
  Jul 26 at 22:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Notice this code: int fraccos x + sin xcos x + sin x , dx. That's not really complicated. Do you need a program to do that?
  – Michael Hardy
  Jul 26 at 22:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes, it is not, but many such formulas take a track of time to do so
  – haqnatural
  Jul 26 at 22:47
  

  
  
  
  

add a comment | 

up vote
1
down vote

Notice that $$int fractanx1+tanx= int 1-frac11+tanx$$

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answered Jul 26 at 21:21

idk

9431414

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up vote
1
down vote

Continuation from the Work in the Question
$$
beginalign
intfrac11+tan(x),mathrmdx
&=log(sin(x)+cos(x))+intfractan(x)1+tan(x),mathrmdxtag1\
&=frac12log(sin(x)+cos(x))+frac12intfrac1+tan(x)1+tan(x),mathrmdxtag2\
&=frac12(x+log(sin(x)+cos(x)))+Ctag3
endalign
$$
Explanation:

$(1)$: copied from the question

$(2)$: average the left and right sides of $(1)$

$(3)$: $int1,mathrmdx=x+C$

---

Another Approach

One can also try $u=tan(x)$ and partial fractions:
$$
beginalign
intfracmathrmdx1+tan(x)
&=intfracmathrmdtan(x)(1+tan(x))left(1+tan^2(x)right)tag4\
&=intfrac12left(frac11+u+frac1-u1+u^2right)mathrmdutag5\
&=frac12left(log(1+u)+tan^-1(u)-frac12logleft(1+u^2right)right)+Ctag6\
&=frac12(x+log(sin(x)+cos(x)))+Ctag7
endalign
$$
Explanation:

$(4)$: $mathrmdtan(x)=left(1+tan^2(x)right),mathrmdx$

$(5)$: $u=tan(x)$ and partial fractions

$(6)$: integrate

$(7)$: backsubstitute

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edited Jul 27 at 6:46

answered Jul 26 at 23:04

robjohn♦

258k25297612

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up vote
0
down vote

$$ tan x=frac 2tan x/21+tan^2 x/2$$
Solves your integral with $$u=tan x/2$$

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answered Jul 26 at 21:03

Mohammad Riazi-Kermani

27.3k41851

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up vote
0
down vote

Notice that

$$(log(sin xpmcos x))'=fraccos xmpsin xsin x+cos x$$

Then

$$(log(sin x+cos x)+log(sin x-cos x))'=(log(sin^2x-cos^2x))'=frac2cos xsin x+cos x.$$

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answered Jul 26 at 21:15

Yves Daoust

110k665203

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up vote
0
down vote

How about:
$$intfrac11+tan(x)dx= ln(sin(x)+cos(x)) +intfractan(x)1+tan(x)dx= ln(sin(x)+cos(x)) +intleft(1-frac11+tan(x)right)dx$$
$$therefore I= ln(sin(x)+cos(x))+x-I$$
$$therefore I=fracln(sin(x)+cos(x))+x2+C $$

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answered Jul 26 at 21:21

Henry Lee

49210

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0
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If the numerator and denominator are both polynomials of trigonometric functions, then the substitution $y= tan(x/2)$ should always work. It is not necessarily the shortest solution, but it works.

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answered Jul 26 at 22:15

A. Pongrácz

1,574115

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7 Answers
7

active

oldest

votes

7 Answers
7

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
7
down vote

$$
I=int frac 1 1+tan x , dx=int frac cos xcos x +sin x , dx\
J=int frac sin xcos x +sin x , dx\
I+J=int frac cos x+sin xcos x +sin x , dx = x+C_1\ \
I-J=int frac cos x-sin xcos x +sin x ,dx=int frac dleft( sin x +cos xright) cos x +sin x =ln left| sin x +cos x right| + C_2 \
2I = x+ln left| sin x +cos xright| + C_3 \
I=frac 1 2 left[ x+ln left| sin x +cos x right| + C_3 right]
$$

share|cite|improve this answer

edited Jul 26 at 22:22

Michael Hardy

204k23186461

answered Jul 26 at 20:33

haqnatural

20.5k72457

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Writing int frac cos x + sin x cos x + sin x dx where int fraccos x + sin xcos x + sin x , dx would suffice doesn't serve a purpose and makes editing harder.
  – Michael Hardy
  Jul 26 at 22:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MichaelHardy, thank you for editing, but I am still can't find better-writing program rather than this one s1.daumcdn.net/editor/fp/service_nc/pencil/…
  – haqnatural
  Jul 26 at 22:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Notice this code: int fraccos x + sin xcos x + sin x , dx. That's not really complicated. Do you need a program to do that?
  – Michael Hardy
  Jul 26 at 22:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes, it is not, but many such formulas take a track of time to do so
  – haqnatural
  Jul 26 at 22:47
  

  
  
  
  

add a comment | 

up vote
7
down vote

$$
I=int frac 1 1+tan x , dx=int frac cos xcos x +sin x , dx\
J=int frac sin xcos x +sin x , dx\
I+J=int frac cos x+sin xcos x +sin x , dx = x+C_1\ \
I-J=int frac cos x-sin xcos x +sin x ,dx=int frac dleft( sin x +cos xright) cos x +sin x =ln left| sin x +cos x right| + C_2 \
2I = x+ln left| sin x +cos xright| + C_3 \
I=frac 1 2 left[ x+ln left| sin x +cos x right| + C_3 right]
$$

share|cite|improve this answer

edited Jul 26 at 22:22

Michael Hardy

204k23186461

answered Jul 26 at 20:33

haqnatural

20.5k72457

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Writing int frac cos x + sin x cos x + sin x dx where int fraccos x + sin xcos x + sin x , dx would suffice doesn't serve a purpose and makes editing harder.
  – Michael Hardy
  Jul 26 at 22:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MichaelHardy, thank you for editing, but I am still can't find better-writing program rather than this one s1.daumcdn.net/editor/fp/service_nc/pencil/…
  – haqnatural
  Jul 26 at 22:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Notice this code: int fraccos x + sin xcos x + sin x , dx. That's not really complicated. Do you need a program to do that?
  – Michael Hardy
  Jul 26 at 22:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes, it is not, but many such formulas take a track of time to do so
  – haqnatural
  Jul 26 at 22:47
  

  
  
  
  

add a comment | 

up vote
7
down vote

up vote
7
down vote

$$
I=int frac 1 1+tan x , dx=int frac cos xcos x +sin x , dx\
J=int frac sin xcos x +sin x , dx\
I+J=int frac cos x+sin xcos x +sin x , dx = x+C_1\ \
I-J=int frac cos x-sin xcos x +sin x ,dx=int frac dleft( sin x +cos xright) cos x +sin x =ln left| sin x +cos x right| + C_2 \
2I = x+ln left| sin x +cos xright| + C_3 \
I=frac 1 2 left[ x+ln left| sin x +cos x right| + C_3 right]
$$

share|cite|improve this answer

edited Jul 26 at 22:22

Michael Hardy

204k23186461

answered Jul 26 at 20:33

haqnatural

20.5k72457

$$
I=int frac 1 1+tan x , dx=int frac cos xcos x +sin x , dx\
J=int frac sin xcos x +sin x , dx\
I+J=int frac cos x+sin xcos x +sin x , dx = x+C_1\ \
I-J=int frac cos x-sin xcos x +sin x ,dx=int frac dleft( sin x +cos xright) cos x +sin x =ln left| sin x +cos x right| + C_2 \
2I = x+ln left| sin x +cos xright| + C_3 \
I=frac 1 2 left[ x+ln left| sin x +cos x right| + C_3 right]
$$

share|cite|improve this answer

edited Jul 26 at 22:22

Michael Hardy

204k23186461

answered Jul 26 at 20:33

haqnatural

20.5k72457

share|cite|improve this answer

share|cite|improve this answer

edited Jul 26 at 22:22

Michael Hardy

204k23186461

edited Jul 26 at 22:22

Michael Hardy

204k23186461

edited Jul 26 at 22:22

Michael Hardy

204k23186461

204k23186461

answered Jul 26 at 20:33

haqnatural

20.5k72457

answered Jul 26 at 20:33

haqnatural

20.5k72457

answered Jul 26 at 20:33

haqnatural

20.5k72457

20.5k72457

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Writing int frac cos x + sin x cos x + sin x dx where int fraccos x + sin xcos x + sin x , dx would suffice doesn't serve a purpose and makes editing harder.
  – Michael Hardy
  Jul 26 at 22:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MichaelHardy, thank you for editing, but I am still can't find better-writing program rather than this one s1.daumcdn.net/editor/fp/service_nc/pencil/…
  – haqnatural
  Jul 26 at 22:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Notice this code: int fraccos x + sin xcos x + sin x , dx. That's not really complicated. Do you need a program to do that?
  – Michael Hardy
  Jul 26 at 22:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes, it is not, but many such formulas take a track of time to do so
  – haqnatural
  Jul 26 at 22:47
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Writing int frac cos x + sin x cos x + sin x dx where int fraccos x + sin xcos x + sin x , dx would suffice doesn't serve a purpose and makes editing harder.
  – Michael Hardy
  Jul 26 at 22:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MichaelHardy, thank you for editing, but I am still can't find better-writing program rather than this one s1.daumcdn.net/editor/fp/service_nc/pencil/…
  – haqnatural
  Jul 26 at 22:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Notice this code: int fraccos x + sin xcos x + sin x , dx. That's not really complicated. Do you need a program to do that?
  – Michael Hardy
  Jul 26 at 22:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes, it is not, but many such formulas take a track of time to do so
  – haqnatural
  Jul 26 at 22:47
  

  
  
  
  

Writing int frac cos x + sin x cos x + sin x dx where int fraccos x + sin xcos x + sin x , dx would suffice doesn't serve a purpose and makes editing harder.
– Michael Hardy
Jul 26 at 22:26

Writing int frac cos x + sin x cos x + sin x dx where int fraccos x + sin xcos x + sin x , dx would suffice doesn't serve a purpose and makes editing harder.
– Michael Hardy
Jul 26 at 22:26

@MichaelHardy, thank you for editing, but I am still can't find better-writing program rather than this one s1.daumcdn.net/editor/fp/service_nc/pencil/…
– haqnatural
Jul 26 at 22:32

@MichaelHardy, thank you for editing, but I am still can't find better-writing program rather than this one s1.daumcdn.net/editor/fp/service_nc/pencil/…
– haqnatural
Jul 26 at 22:32

Notice this code: int fraccos x + sin xcos x + sin x , dx. That's not really complicated. Do you need a program to do that?
– Michael Hardy
Jul 26 at 22:37

Notice this code: int fraccos x + sin xcos x + sin x , dx. That's not really complicated. Do you need a program to do that?
– Michael Hardy
Jul 26 at 22:37

yes, it is not, but many such formulas take a track of time to do so
– haqnatural
Jul 26 at 22:47

yes, it is not, but many such formulas take a track of time to do so
– haqnatural
Jul 26 at 22:47

add a comment | 

up vote
1
down vote

Notice that $$int fractanx1+tanx= int 1-frac11+tanx$$

share|cite|improve this answer

answered Jul 26 at 21:21

idk

9431414

add a comment | 

up vote
1
down vote

Notice that $$int fractanx1+tanx= int 1-frac11+tanx$$

share|cite|improve this answer

answered Jul 26 at 21:21

idk

9431414

add a comment | 

up vote
1
down vote

up vote
1
down vote

Notice that $$int fractanx1+tanx= int 1-frac11+tanx$$

share|cite|improve this answer

answered Jul 26 at 21:21

idk

9431414

Notice that $$int fractanx1+tanx= int 1-frac11+tanx$$

share|cite|improve this answer

answered Jul 26 at 21:21

idk

9431414

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 21:21

idk

9431414

answered Jul 26 at 21:21

idk

9431414

answered Jul 26 at 21:21

idk

9431414

9431414

add a comment | 

add a comment | 

up vote
1
down vote

Continuation from the Work in the Question
$$
beginalign
intfrac11+tan(x),mathrmdx
&=log(sin(x)+cos(x))+intfractan(x)1+tan(x),mathrmdxtag1\
&=frac12log(sin(x)+cos(x))+frac12intfrac1+tan(x)1+tan(x),mathrmdxtag2\
&=frac12(x+log(sin(x)+cos(x)))+Ctag3
endalign
$$
Explanation:

$(1)$: copied from the question

$(2)$: average the left and right sides of $(1)$

$(3)$: $int1,mathrmdx=x+C$

---

Another Approach

One can also try $u=tan(x)$ and partial fractions:
$$
beginalign
intfracmathrmdx1+tan(x)
&=intfracmathrmdtan(x)(1+tan(x))left(1+tan^2(x)right)tag4\
&=intfrac12left(frac11+u+frac1-u1+u^2right)mathrmdutag5\
&=frac12left(log(1+u)+tan^-1(u)-frac12logleft(1+u^2right)right)+Ctag6\
&=frac12(x+log(sin(x)+cos(x)))+Ctag7
endalign
$$
Explanation:

$(4)$: $mathrmdtan(x)=left(1+tan^2(x)right),mathrmdx$

$(5)$: $u=tan(x)$ and partial fractions

$(6)$: integrate

$(7)$: backsubstitute

share|cite|improve this answer

edited Jul 27 at 6:46

answered Jul 26 at 23:04

robjohn♦

258k25297612

add a comment | 

up vote
1
down vote

Continuation from the Work in the Question
$$
beginalign
intfrac11+tan(x),mathrmdx
&=log(sin(x)+cos(x))+intfractan(x)1+tan(x),mathrmdxtag1\
&=frac12log(sin(x)+cos(x))+frac12intfrac1+tan(x)1+tan(x),mathrmdxtag2\
&=frac12(x+log(sin(x)+cos(x)))+Ctag3
endalign
$$
Explanation:

$(1)$: copied from the question

$(2)$: average the left and right sides of $(1)$

$(3)$: $int1,mathrmdx=x+C$

---

Another Approach

One can also try $u=tan(x)$ and partial fractions:
$$
beginalign
intfracmathrmdx1+tan(x)
&=intfracmathrmdtan(x)(1+tan(x))left(1+tan^2(x)right)tag4\
&=intfrac12left(frac11+u+frac1-u1+u^2right)mathrmdutag5\
&=frac12left(log(1+u)+tan^-1(u)-frac12logleft(1+u^2right)right)+Ctag6\
&=frac12(x+log(sin(x)+cos(x)))+Ctag7
endalign
$$
Explanation:

$(4)$: $mathrmdtan(x)=left(1+tan^2(x)right),mathrmdx$

$(5)$: $u=tan(x)$ and partial fractions

$(6)$: integrate

$(7)$: backsubstitute

share|cite|improve this answer

edited Jul 27 at 6:46

answered Jul 26 at 23:04

robjohn♦

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Continuation from the Work in the Question
$$
beginalign
intfrac11+tan(x),mathrmdx
&=log(sin(x)+cos(x))+intfractan(x)1+tan(x),mathrmdxtag1\
&=frac12log(sin(x)+cos(x))+frac12intfrac1+tan(x)1+tan(x),mathrmdxtag2\
&=frac12(x+log(sin(x)+cos(x)))+Ctag3
endalign
$$
Explanation:

$(1)$: copied from the question

$(2)$: average the left and right sides of $(1)$

$(3)$: $int1,mathrmdx=x+C$

---

Another Approach

One can also try $u=tan(x)$ and partial fractions:
$$
beginalign
intfracmathrmdx1+tan(x)
&=intfracmathrmdtan(x)(1+tan(x))left(1+tan^2(x)right)tag4\
&=intfrac12left(frac11+u+frac1-u1+u^2right)mathrmdutag5\
&=frac12left(log(1+u)+tan^-1(u)-frac12logleft(1+u^2right)right)+Ctag6\
&=frac12(x+log(sin(x)+cos(x)))+Ctag7
endalign
$$
Explanation:

$(4)$: $mathrmdtan(x)=left(1+tan^2(x)right),mathrmdx$

$(5)$: $u=tan(x)$ and partial fractions

$(6)$: integrate

$(7)$: backsubstitute

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edited Jul 27 at 6:46

answered Jul 26 at 23:04

robjohn♦

258k25297612

Continuation from the Work in the Question
$$
beginalign
intfrac11+tan(x),mathrmdx
&=log(sin(x)+cos(x))+intfractan(x)1+tan(x),mathrmdxtag1\
&=frac12log(sin(x)+cos(x))+frac12intfrac1+tan(x)1+tan(x),mathrmdxtag2\
&=frac12(x+log(sin(x)+cos(x)))+Ctag3
endalign
$$
Explanation:

$(1)$: copied from the question

$(2)$: average the left and right sides of $(1)$

$(3)$: $int1,mathrmdx=x+C$

---

Another Approach

One can also try $u=tan(x)$ and partial fractions:
$$
beginalign
intfracmathrmdx1+tan(x)
&=intfracmathrmdtan(x)(1+tan(x))left(1+tan^2(x)right)tag4\
&=intfrac12left(frac11+u+frac1-u1+u^2right)mathrmdutag5\
&=frac12left(log(1+u)+tan^-1(u)-frac12logleft(1+u^2right)right)+Ctag6\
&=frac12(x+log(sin(x)+cos(x)))+Ctag7
endalign
$$
Explanation:

$(4)$: $mathrmdtan(x)=left(1+tan^2(x)right),mathrmdx$

$(5)$: $u=tan(x)$ and partial fractions

$(6)$: integrate

$(7)$: backsubstitute

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edited Jul 27 at 6:46

answered Jul 26 at 23:04

robjohn♦

258k25297612

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share|cite|improve this answer

edited Jul 27 at 6:46

edited Jul 27 at 6:46

edited Jul 27 at 6:46

answered Jul 26 at 23:04

robjohn♦

258k25297612

answered Jul 26 at 23:04

robjohn♦

258k25297612

answered Jul 26 at 23:04

robjohn♦

258k25297612

258k25297612

add a comment | 

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$$ tan x=frac 2tan x/21+tan^2 x/2$$
Solves your integral with $$u=tan x/2$$

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answered Jul 26 at 21:03

Mohammad Riazi-Kermani

27.3k41851

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$$ tan x=frac 2tan x/21+tan^2 x/2$$
Solves your integral with $$u=tan x/2$$

share|cite|improve this answer

answered Jul 26 at 21:03

Mohammad Riazi-Kermani

27.3k41851

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up vote
0
down vote

up vote
0
down vote

$$ tan x=frac 2tan x/21+tan^2 x/2$$
Solves your integral with $$u=tan x/2$$

share|cite|improve this answer

answered Jul 26 at 21:03

Mohammad Riazi-Kermani

27.3k41851

$$ tan x=frac 2tan x/21+tan^2 x/2$$
Solves your integral with $$u=tan x/2$$

share|cite|improve this answer

answered Jul 26 at 21:03

Mohammad Riazi-Kermani

27.3k41851

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share|cite|improve this answer

answered Jul 26 at 21:03

Mohammad Riazi-Kermani

27.3k41851

answered Jul 26 at 21:03

Mohammad Riazi-Kermani

27.3k41851

answered Jul 26 at 21:03

Mohammad Riazi-Kermani

27.3k41851

27.3k41851

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Notice that

$$(log(sin xpmcos x))'=fraccos xmpsin xsin x+cos x$$

Then

$$(log(sin x+cos x)+log(sin x-cos x))'=(log(sin^2x-cos^2x))'=frac2cos xsin x+cos x.$$

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answered Jul 26 at 21:15

Yves Daoust

110k665203

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Notice that

$$(log(sin xpmcos x))'=fraccos xmpsin xsin x+cos x$$

Then

$$(log(sin x+cos x)+log(sin x-cos x))'=(log(sin^2x-cos^2x))'=frac2cos xsin x+cos x.$$

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answered Jul 26 at 21:15

Yves Daoust

110k665203

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up vote
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Notice that

$$(log(sin xpmcos x))'=fraccos xmpsin xsin x+cos x$$

Then

$$(log(sin x+cos x)+log(sin x-cos x))'=(log(sin^2x-cos^2x))'=frac2cos xsin x+cos x.$$

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answered Jul 26 at 21:15

Yves Daoust

110k665203

Notice that

$$(log(sin xpmcos x))'=fraccos xmpsin xsin x+cos x$$

Then

$$(log(sin x+cos x)+log(sin x-cos x))'=(log(sin^2x-cos^2x))'=frac2cos xsin x+cos x.$$

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answered Jul 26 at 21:15

Yves Daoust

110k665203

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share|cite|improve this answer

answered Jul 26 at 21:15

Yves Daoust

110k665203

answered Jul 26 at 21:15

Yves Daoust

110k665203

answered Jul 26 at 21:15

Yves Daoust

110k665203

110k665203

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How about:
$$intfrac11+tan(x)dx= ln(sin(x)+cos(x)) +intfractan(x)1+tan(x)dx= ln(sin(x)+cos(x)) +intleft(1-frac11+tan(x)right)dx$$
$$therefore I= ln(sin(x)+cos(x))+x-I$$
$$therefore I=fracln(sin(x)+cos(x))+x2+C $$

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answered Jul 26 at 21:21

Henry Lee

49210

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How about:
$$intfrac11+tan(x)dx= ln(sin(x)+cos(x)) +intfractan(x)1+tan(x)dx= ln(sin(x)+cos(x)) +intleft(1-frac11+tan(x)right)dx$$
$$therefore I= ln(sin(x)+cos(x))+x-I$$
$$therefore I=fracln(sin(x)+cos(x))+x2+C $$

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answered Jul 26 at 21:21

Henry Lee

49210

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How about:
$$intfrac11+tan(x)dx= ln(sin(x)+cos(x)) +intfractan(x)1+tan(x)dx= ln(sin(x)+cos(x)) +intleft(1-frac11+tan(x)right)dx$$
$$therefore I= ln(sin(x)+cos(x))+x-I$$
$$therefore I=fracln(sin(x)+cos(x))+x2+C $$

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answered Jul 26 at 21:21

Henry Lee

49210

How about:
$$intfrac11+tan(x)dx= ln(sin(x)+cos(x)) +intfractan(x)1+tan(x)dx= ln(sin(x)+cos(x)) +intleft(1-frac11+tan(x)right)dx$$
$$therefore I= ln(sin(x)+cos(x))+x-I$$
$$therefore I=fracln(sin(x)+cos(x))+x2+C $$

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answered Jul 26 at 21:21

Henry Lee

49210

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share|cite|improve this answer

answered Jul 26 at 21:21

Henry Lee

49210

answered Jul 26 at 21:21

Henry Lee

49210

answered Jul 26 at 21:21

Henry Lee

49210

49210

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0
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If the numerator and denominator are both polynomials of trigonometric functions, then the substitution $y= tan(x/2)$ should always work. It is not necessarily the shortest solution, but it works.

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answered Jul 26 at 22:15

A. Pongrácz

1,574115

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up vote
0
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If the numerator and denominator are both polynomials of trigonometric functions, then the substitution $y= tan(x/2)$ should always work. It is not necessarily the shortest solution, but it works.

share|cite|improve this answer

answered Jul 26 at 22:15

A. Pongrácz

1,574115

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up vote
0
down vote

up vote
0
down vote

If the numerator and denominator are both polynomials of trigonometric functions, then the substitution $y= tan(x/2)$ should always work. It is not necessarily the shortest solution, but it works.

share|cite|improve this answer

answered Jul 26 at 22:15

A. Pongrácz

1,574115

If the numerator and denominator are both polynomials of trigonometric functions, then the substitution $y= tan(x/2)$ should always work. It is not necessarily the shortest solution, but it works.

share|cite|improve this answer

answered Jul 26 at 22:15

A. Pongrácz

1,574115

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share|cite|improve this answer

answered Jul 26 at 22:15

A. Pongrácz

1,574115

answered Jul 26 at 22:15

A. Pongrácz

1,574115

answered Jul 26 at 22:15

A. Pongrácz

1,574115

1,574115

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