Fdtxjr

How do I prevent inappropriate ads from appearing in my game?

Would a primitive species be able to learn English from reading books alone?

What is the tangent at a sharp point on a curve?

What is this high flying aircraft over Pennsylvania?

Do people actually use the word "kaputt" in conversation?

What is it called when someone votes for an option that's not their first choice?

Why is implicit conversion not ambiguous for non-primitive types?

Offset in split text content

Reason why a kingside attack is not justified

How to split IPA spelling into syllables

How do I lift the insulation blower into the attic?

Unfrosted light bulb

Why is "la Gestapo" feminine?

Does capillary rise violate hydrostatic paradox?

What do the positive and negative (+/-) transmit and receive pins mean on Ethernet cables?

What properties make a magic weapon befit a Rogue more than a DEX-based Fighter?

Mortal danger in mid-grade literature

I keep switching characters, how do I stop?

Checking @@ROWCOUNT failing

Should a narrator ever describe things based on a character's view instead of facts?

Why is indicated airspeed rather than ground speed used during the takeoff roll?

Reasons for having MCU pin-states default to pull-up/down out of reset

What should be the ideal length of sentences in a blog post for ease of reading?

Highest stage count that are used one right after the other?

Does $-frac12cos(2x)=sin^2(x)$? [duplicate]

Trig integral $int cosx + sinxcosx dx $Proof of $cos theta+cos 2theta+cos 3theta+cdots+cos ntheta=fracsinfrac12nthetasinfrac12thetacosfrac12(n+1)theta$Solving $sin^2015x+cos^2015x=frac12$How to integrate $int_0^2pi frac12 sin(t) (1- cos(t)) sqrtfrac12 - frac12 cos(t),dt$solve for $x: cos(x) cos(15^circ) - sin (x) sin (15^circ) = frac12$Issues computing $intcos^2x~mathrmdx$ without using $sin^2(x)=1-cos^2(x)$Calculating $lim limits_x to infty fracx+frac12cos xx-frac12sin x$ using the sandwich theoremConverting a function of $cos^2$ to a complete elliptic integral of the first kindEvaluating $int_0^inftyfraccos(x/a)-1xcdotcos(x)ln(x) dx$How to evaluate $intfrac sin^8 x - cos^8x 1 - 2sin^2 x cdot cos^2 x mathrmdx$?Prove that $sin a + sin b + sin(a+b) = 4 sinfrac12(a+b) cos frac12a cosfrac12b$

0

$begingroup$

This question already has an answer here:

• 
  Trig integral $int cosx + sinxcosx dx $
  
  2 answers
  
  

$$int2sin xcos xmathrm dx$$
$1$:Consider $2sin xcos x=sin(2x)$ Then the integration is $-frac12cos(2x)$

$2$: Consider $2sin xcos x=fracmathrm dmathrm dxsin^2(x)$ Then the integration is $sin^2(x)$

If it is true then
$-frac12cos(2x)=sin^2(x)$
then
$-frac12(1+2sin^2(x))=sin^2(x)$,$-frac12=sin^2x$, $-frac12=1$

integration trigonometry trigonometric-integrals

share|cite|improve this question

edited Mar 13 at 18:28

mrtaurho

6,04051641

asked Mar 13 at 18:24

alfatih Abdullahalfatih Abdullah

42

New contributor

alfatih Abdullah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

marked as duplicate by Hans Lundmark, Cesareo, mrtaurho, M. Vinay, Vinyl_cape_jawa Mar 14 at 20:52

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.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Well, $-frac12$ and $1$ are the same up to a constant of integration...
  $endgroup$
  – mrtaurho
  Mar 13 at 18:29
  

  
  
  
  

add a comment | 

0

$begingroup$

This question already has an answer here:

• 
  Trig integral $int cosx + sinxcosx dx $
  
  2 answers
  
  

$$int2sin xcos xmathrm dx$$
$1$:Consider $2sin xcos x=sin(2x)$ Then the integration is $-frac12cos(2x)$

$2$: Consider $2sin xcos x=fracmathrm dmathrm dxsin^2(x)$ Then the integration is $sin^2(x)$

If it is true then
$-frac12cos(2x)=sin^2(x)$
then
$-frac12(1+2sin^2(x))=sin^2(x)$,$-frac12=sin^2x$, $-frac12=1$

integration trigonometry trigonometric-integrals

share|cite|improve this question

edited Mar 13 at 18:28

mrtaurho

6,04051641

asked Mar 13 at 18:24

alfatih Abdullahalfatih Abdullah

42

New contributor

alfatih Abdullah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

marked as duplicate by Hans Lundmark, Cesareo, mrtaurho, M. Vinay, Vinyl_cape_jawa Mar 14 at 20:52

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.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Well, $-frac12$ and $1$ are the same up to a constant of integration...
  $endgroup$
  – mrtaurho
  Mar 13 at 18:29
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Trig integral $int cosx + sinxcosx dx $
  
  2 answers
  
  

$$int2sin xcos xmathrm dx$$
$1$:Consider $2sin xcos x=sin(2x)$ Then the integration is $-frac12cos(2x)$

$2$: Consider $2sin xcos x=fracmathrm dmathrm dxsin^2(x)$ Then the integration is $sin^2(x)$

If it is true then
$-frac12cos(2x)=sin^2(x)$
then
$-frac12(1+2sin^2(x))=sin^2(x)$,$-frac12=sin^2x$, $-frac12=1$

integration trigonometry trigonometric-integrals

share|cite|improve this question

edited Mar 13 at 18:28

mrtaurho

6,04051641

asked Mar 13 at 18:24

alfatih Abdullahalfatih Abdullah

42

New contributor

alfatih Abdullah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 13 at 18:28

mrtaurho

6,04051641

asked Mar 13 at 18:24

alfatih Abdullahalfatih Abdullah

42

New contributor

alfatih Abdullah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 13 at 18:28

mrtaurho

6,04051641

asked Mar 13 at 18:24

alfatih Abdullahalfatih Abdullah

42

New contributor

alfatih Abdullah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 13 at 18:28

mrtaurho

6,04051641

edited Mar 13 at 18:28

mrtaurho

6,04051641

asked Mar 13 at 18:24

alfatih Abdullahalfatih Abdullah

42

New contributor

alfatih Abdullah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 13 at 18:24

alfatih Abdullahalfatih Abdullah

42

1 Answer
1

active

oldest

votes

1

$begingroup$

You are dealing with indefinite integral therefore you have to take into account that while integrating we have to add a constant of integration. To be precise we got that

beginalign*
int 2sin xcos xmathrm dx&=int 2sin xcos xmathrm dx\
int sin(2x)mathrm dx&=int left(fracmathrm dmathrm dxsin^2(x)right)mathrm dx\
-frac12cos(2x)+c_1&=sin^2(x)+c_2\
-frac12cos(2x)&=sin^2(x)+c
endalign*

To determine the constant of integration in this case we can follow your own attempt by expanding the $cos(2x)$, even though you have made a minor mistake there, term to get

beginalign*
-frac12cos(2x)&=sin^2(x)+c\
-frac12(1-2sin^2(x))&=sin^2(x)+c\
c&=-frac12
endalign*

We just proved the Half-Angle Formula since we can now conclude that

$$sin^2(x)=frac12-frac12cos(2x)=frac1-cos(2x)2$$

Which is basically all what was missing.

share|cite|improve this answer

answered Mar 13 at 18:37

mrtaurhomrtaurho

6,04051641

$endgroup$

add a comment | 

1

$begingroup$

You are dealing with indefinite integral therefore you have to take into account that while integrating we have to add a constant of integration. To be precise we got that

beginalign*
int 2sin xcos xmathrm dx&=int 2sin xcos xmathrm dx\
int sin(2x)mathrm dx&=int left(fracmathrm dmathrm dxsin^2(x)right)mathrm dx\
-frac12cos(2x)+c_1&=sin^2(x)+c_2\
-frac12cos(2x)&=sin^2(x)+c
endalign*

To determine the constant of integration in this case we can follow your own attempt by expanding the $cos(2x)$, even though you have made a minor mistake there, term to get

beginalign*
-frac12cos(2x)&=sin^2(x)+c\
-frac12(1-2sin^2(x))&=sin^2(x)+c\
c&=-frac12
endalign*

We just proved the Half-Angle Formula since we can now conclude that

$$sin^2(x)=frac12-frac12cos(2x)=frac1-cos(2x)2$$

Which is basically all what was missing.

share|cite|improve this answer

answered Mar 13 at 18:37

mrtaurhomrtaurho

6,04051641

$endgroup$

add a comment | 

1

$begingroup$

You are dealing with indefinite integral therefore you have to take into account that while integrating we have to add a constant of integration. To be precise we got that

beginalign*
int 2sin xcos xmathrm dx&=int 2sin xcos xmathrm dx\
int sin(2x)mathrm dx&=int left(fracmathrm dmathrm dxsin^2(x)right)mathrm dx\
-frac12cos(2x)+c_1&=sin^2(x)+c_2\
-frac12cos(2x)&=sin^2(x)+c
endalign*

To determine the constant of integration in this case we can follow your own attempt by expanding the $cos(2x)$, even though you have made a minor mistake there, term to get

beginalign*
-frac12cos(2x)&=sin^2(x)+c\
-frac12(1-2sin^2(x))&=sin^2(x)+c\
c&=-frac12
endalign*

We just proved the Half-Angle Formula since we can now conclude that

$$sin^2(x)=frac12-frac12cos(2x)=frac1-cos(2x)2$$

Which is basically all what was missing.

share|cite|improve this answer

answered Mar 13 at 18:37

mrtaurhomrtaurho

6,04051641

$endgroup$

add a comment | 

$begingroup$

You are dealing with indefinite integral therefore you have to take into account that while integrating we have to add a constant of integration. To be precise we got that

beginalign*
int 2sin xcos xmathrm dx&=int 2sin xcos xmathrm dx\
int sin(2x)mathrm dx&=int left(fracmathrm dmathrm dxsin^2(x)right)mathrm dx\
-frac12cos(2x)+c_1&=sin^2(x)+c_2\
-frac12cos(2x)&=sin^2(x)+c
endalign*

To determine the constant of integration in this case we can follow your own attempt by expanding the $cos(2x)$, even though you have made a minor mistake there, term to get

beginalign*
-frac12cos(2x)&=sin^2(x)+c\
-frac12(1-2sin^2(x))&=sin^2(x)+c\
c&=-frac12
endalign*

We just proved the Half-Angle Formula since we can now conclude that

$$sin^2(x)=frac12-frac12cos(2x)=frac1-cos(2x)2$$

Which is basically all what was missing.

share|cite|improve this answer

answered Mar 13 at 18:37

mrtaurhomrtaurho

6,04051641

$endgroup$

share|cite|improve this answer

answered Mar 13 at 18:37

mrtaurhomrtaurho

6,04051641

answered Mar 13 at 18:37

mrtaurhomrtaurho

6,04051641

answered Mar 13 at 18:37

mrtaurhomrtaurho

6,04051641