How does $pi$ become $pi^2$?

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I am currently doing integration and cannot seem to understand this question

$xcosleft(pixright)$

I have tried following the following

Formula I have used:$∫f g′= f g − ∫ f′g$

f = $x$
f'= $1$
g =
$dfracsinleft(pixright)pi$

g'= $cos(pix)$

The end result should be the following:

$dfracpixsinleft(pixright)+cosleft(pixright)pi^2$

I cannot seem to understand how that ends up being the end result.

indefinite-integrals

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edited Jul 20 at 13:50

Michael Lugo

17.4k33475

asked Jul 20 at 13:36

KyleMcCann

427

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Which part exactly do you not understand?
  – Sobi
  Jul 20 at 13:37
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Did you integrate $f'g$? If you did, you get $pi^2$ in the denominator. The rest is just finding a common denominator.
  – Dan Sp.
  Jul 20 at 13:43
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

I am currently doing integration and cannot seem to understand this question

$xcosleft(pixright)$

I have tried following the following

Formula I have used:$∫f g′= f g − ∫ f′g$

f = $x$
f'= $1$
g =
$dfracsinleft(pixright)pi$

g'= $cos(pix)$

The end result should be the following:

$dfracpixsinleft(pixright)+cosleft(pixright)pi^2$

I cannot seem to understand how that ends up being the end result.

indefinite-integrals

share|cite|improve this question

edited Jul 20 at 13:50

Michael Lugo

17.4k33475

asked Jul 20 at 13:36

KyleMcCann

427

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Which part exactly do you not understand?
  – Sobi
  Jul 20 at 13:37
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Did you integrate $f'g$? If you did, you get $pi^2$ in the denominator. The rest is just finding a common denominator.
  – Dan Sp.
  Jul 20 at 13:43
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I am currently doing integration and cannot seem to understand this question

$xcosleft(pixright)$

I have tried following the following

Formula I have used:$∫f g′= f g − ∫ f′g$

f = $x$
f'= $1$
g =
$dfracsinleft(pixright)pi$

g'= $cos(pix)$

The end result should be the following:

$dfracpixsinleft(pixright)+cosleft(pixright)pi^2$

I cannot seem to understand how that ends up being the end result.

indefinite-integrals

share|cite|improve this question

edited Jul 20 at 13:50

Michael Lugo

17.4k33475

asked Jul 20 at 13:36

KyleMcCann

427

I am currently doing integration and cannot seem to understand this question

$xcosleft(pixright)$

I have tried following the following

Formula I have used:$∫f g′= f g − ∫ f′g$

f = $x$
f'= $1$
g =
$dfracsinleft(pixright)pi$

g'= $cos(pix)$

The end result should be the following:

$dfracpixsinleft(pixright)+cosleft(pixright)pi^2$

I cannot seem to understand how that ends up being the end result.

indefinite-integrals

share|cite|improve this question

edited Jul 20 at 13:50

Michael Lugo

17.4k33475

asked Jul 20 at 13:36

KyleMcCann

427

share|cite|improve this question

share|cite|improve this question

edited Jul 20 at 13:50

Michael Lugo

17.4k33475

edited Jul 20 at 13:50

Michael Lugo

17.4k33475

edited Jul 20 at 13:50

Michael Lugo

17.4k33475

17.4k33475

asked Jul 20 at 13:36

KyleMcCann

427

asked Jul 20 at 13:36

KyleMcCann

427

asked Jul 20 at 13:36

KyleMcCann

427

427

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Which part exactly do you not understand?
  – Sobi
  Jul 20 at 13:37
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Did you integrate $f'g$? If you did, you get $pi^2$ in the denominator. The rest is just finding a common denominator.
  – Dan Sp.
  Jul 20 at 13:43
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Which part exactly do you not understand?
  – Sobi
  Jul 20 at 13:37
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Did you integrate $f'g$? If you did, you get $pi^2$ in the denominator. The rest is just finding a common denominator.
  – Dan Sp.
  Jul 20 at 13:43
  

  
  
  
  

Which part exactly do you not understand?
– Sobi
Jul 20 at 13:37

Which part exactly do you not understand?
– Sobi
Jul 20 at 13:37

3

3

Did you integrate $f'g$? If you did, you get $pi^2$ in the denominator. The rest is just finding a common denominator.
– Dan Sp.
Jul 20 at 13:43

Did you integrate $f'g$? If you did, you get $pi^2$ in the denominator. The rest is just finding a common denominator.
– Dan Sp.
Jul 20 at 13:43

add a comment | 

2 Answers
2

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oldest

votes

up vote
3
down vote

$$int udv = uv - int vdu$$

$$u = x$$
$$du = dx$$
$$dv = cos(pi x)$$
$$v = fracsin(pi x)pi$$

$$ =fracxsin(pi x)pi - int fracsin(pi x)pidx $$
$$ =fracxsin(pi x)pi + fraccos(pi x)pi^2 + C$$
$$ = fracpi x sin(pi x) + cos(pi x)pi^2 +C $$

share|cite|improve this answer

edited Jul 20 at 15:23

jiaminglimjm

347314

answered Jul 20 at 13:43

Safder

758

add a comment | 

up vote
1
down vote

As you said:
$$int x cos(x) dx = xfracsin(pi x)pi - int fracsin(pi x)pi$$

So $pi^2$ origin from the last integral.
Infact:
$$int fracsin(pi x)pi = - fraccos^2(pi x)pi^2$$

share|cite|improve this answer

answered Jul 20 at 13:46

Luca Savant

763

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote

$$int udv = uv - int vdu$$

$$u = x$$
$$du = dx$$
$$dv = cos(pi x)$$
$$v = fracsin(pi x)pi$$

$$ =fracxsin(pi x)pi - int fracsin(pi x)pidx $$
$$ =fracxsin(pi x)pi + fraccos(pi x)pi^2 + C$$
$$ = fracpi x sin(pi x) + cos(pi x)pi^2 +C $$

share|cite|improve this answer

edited Jul 20 at 15:23

jiaminglimjm

347314

answered Jul 20 at 13:43

Safder

758

add a comment | 

up vote
3
down vote

$$int udv = uv - int vdu$$

$$u = x$$
$$du = dx$$
$$dv = cos(pi x)$$
$$v = fracsin(pi x)pi$$

$$ =fracxsin(pi x)pi - int fracsin(pi x)pidx $$
$$ =fracxsin(pi x)pi + fraccos(pi x)pi^2 + C$$
$$ = fracpi x sin(pi x) + cos(pi x)pi^2 +C $$

share|cite|improve this answer

edited Jul 20 at 15:23

jiaminglimjm

347314

answered Jul 20 at 13:43

Safder

758

add a comment | 

up vote
3
down vote

up vote
3
down vote

$$int udv = uv - int vdu$$

$$u = x$$
$$du = dx$$
$$dv = cos(pi x)$$
$$v = fracsin(pi x)pi$$

$$ =fracxsin(pi x)pi - int fracsin(pi x)pidx $$
$$ =fracxsin(pi x)pi + fraccos(pi x)pi^2 + C$$
$$ = fracpi x sin(pi x) + cos(pi x)pi^2 +C $$

share|cite|improve this answer

edited Jul 20 at 15:23

jiaminglimjm

347314

answered Jul 20 at 13:43

Safder

758

$$int udv = uv - int vdu$$

$$u = x$$
$$du = dx$$
$$dv = cos(pi x)$$
$$v = fracsin(pi x)pi$$

$$ =fracxsin(pi x)pi - int fracsin(pi x)pidx $$
$$ =fracxsin(pi x)pi + fraccos(pi x)pi^2 + C$$
$$ = fracpi x sin(pi x) + cos(pi x)pi^2 +C $$

share|cite|improve this answer

edited Jul 20 at 15:23

jiaminglimjm

347314

answered Jul 20 at 13:43

Safder

758

share|cite|improve this answer

share|cite|improve this answer

edited Jul 20 at 15:23

jiaminglimjm

347314

edited Jul 20 at 15:23

jiaminglimjm

347314

edited Jul 20 at 15:23

jiaminglimjm

347314

347314

answered Jul 20 at 13:43

Safder

758

answered Jul 20 at 13:43

Safder

758

answered Jul 20 at 13:43

Safder

758

758

add a comment | 

add a comment | 

up vote
1
down vote

As you said:
$$int x cos(x) dx = xfracsin(pi x)pi - int fracsin(pi x)pi$$

So $pi^2$ origin from the last integral.
Infact:
$$int fracsin(pi x)pi = - fraccos^2(pi x)pi^2$$

share|cite|improve this answer

answered Jul 20 at 13:46

Luca Savant

763

add a comment | 

up vote
1
down vote

As you said:
$$int x cos(x) dx = xfracsin(pi x)pi - int fracsin(pi x)pi$$

So $pi^2$ origin from the last integral.
Infact:
$$int fracsin(pi x)pi = - fraccos^2(pi x)pi^2$$

share|cite|improve this answer

answered Jul 20 at 13:46

Luca Savant

763

add a comment | 

up vote
1
down vote

up vote
1
down vote

As you said:
$$int x cos(x) dx = xfracsin(pi x)pi - int fracsin(pi x)pi$$

So $pi^2$ origin from the last integral.
Infact:
$$int fracsin(pi x)pi = - fraccos^2(pi x)pi^2$$

share|cite|improve this answer

answered Jul 20 at 13:46

Luca Savant

763

As you said:
$$int x cos(x) dx = xfracsin(pi x)pi - int fracsin(pi x)pi$$

So $pi^2$ origin from the last integral.
Infact:
$$int fracsin(pi x)pi = - fraccos^2(pi x)pi^2$$

share|cite|improve this answer

answered Jul 20 at 13:46

Luca Savant

763

share|cite|improve this answer

share|cite|improve this answer

answered Jul 20 at 13:46

Luca Savant

763

answered Jul 20 at 13:46

Luca Savant

763

answered Jul 20 at 13:46

Luca Savant

763

763

add a comment | 

add a comment | 

 

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