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Solving integrals/ (sin(x+2y)/3(x+y^2))dy

Integral of (sin(x+2y)/3(x+y^2))dy dy

The solution

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  0                         
  /                         
 |                          
 |  sin(x + 2*y) /     2\   
 |  ------------*\x + y / dy
 |       3                  
 |                          
/                           
-1

$$\int\limits_{-1}^{0} \frac{\sin{\left(x + 2 y \right)}}{3} \left(x + y^{2}\right)\, dy$$

Integral((sin(x + 2*y)/3)*(x + y^2), (y, -1, 0))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{\sin{\left(x + 2 y \right)}}{3} \left(x + y^{2}\right) = \frac{x \sin{\left(x + 2 y \right)}}{3} + \frac{y^{2} \sin{\left(x + 2 y \right)}}{3}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{x \sin{\left(x + 2 y \right)}}{3}\, dy = \frac{x \int \sin{\left(x + 2 y \right)}\, dy}{3}$
    1. Let $u = x + 2 y$ .
      
      Then let $du = 2 dy$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\sin{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(x + 2 y \right)}}{2}$
    So, the result is: $- \frac{x \cos{\left(x + 2 y \right)}}{6}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{y^{2} \sin{\left(x + 2 y \right)}}{3}\, dy = \frac{\int y^{2} \sin{\left(x + 2 y \right)}\, dy}{3}$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(y \right)} = y^{2}$ and let $\operatorname{dv}{\left(y \right)} = \sin{\left(x + 2 y \right)}$ .
      
      Then $\operatorname{du}{\left(y \right)} = 2 y$ .
      
      To find $v{\left(y \right)}$ :
      
      Let $u = x + 2 y$ .
      
      Then let $du = 2 dy$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\sin{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(x + 2 y \right)}}{2}$
      
      Now evaluate the sub-integral.
    2. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(y \right)} = - y$ and let $\operatorname{dv}{\left(y \right)} = \cos{\left(x + 2 y \right)}$ .
      
      Then $\operatorname{du}{\left(y \right)} = -1$ .
      
      To find $v{\left(y \right)}$ :
      
      Let $u = x + 2 y$ .
      
      Then let $du = 2 dy$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(x + 2 y \right)}}{2}$
      
      Now evaluate the sub-integral.
    3. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\sin{\left(x + 2 y \right)}}{2}\right)\, dy = - \frac{\int \sin{\left(x + 2 y \right)}\, dy}{2}$
      
      Let $u = x + 2 y$ .
      
      Then let $du = 2 dy$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\sin{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(x + 2 y \right)}}{2}$
      
      So, the result is: $\frac{\cos{\left(x + 2 y \right)}}{4}$
    So, the result is: $- \frac{y^{2} \cos{\left(x + 2 y \right)}}{6} + \frac{y \sin{\left(x + 2 y \right)}}{6} + \frac{\cos{\left(x + 2 y \right)}}{12}$
  The result is: $- \frac{x \cos{\left(x + 2 y \right)}}{6} - \frac{y^{2} \cos{\left(x + 2 y \right)}}{6} + \frac{y \sin{\left(x + 2 y \right)}}{6} + \frac{\cos{\left(x + 2 y \right)}}{12}$
Method #2
1. Rewrite the integrand:
  
  $\frac{\sin{\left(x + 2 y \right)}}{3} \left(x + y^{2}\right) = \frac{x \sin{\left(x + 2 y \right)}}{3} + \frac{y^{2} \sin{\left(x + 2 y \right)}}{3}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{x \sin{\left(x + 2 y \right)}}{3}\, dy = \frac{x \int \sin{\left(x + 2 y \right)}\, dy}{3}$
    1. Let $u = x + 2 y$ .
      
      Then let $du = 2 dy$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\sin{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(x + 2 y \right)}}{2}$
    So, the result is: $- \frac{x \cos{\left(x + 2 y \right)}}{6}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{y^{2} \sin{\left(x + 2 y \right)}}{3}\, dy = \frac{\int y^{2} \sin{\left(x + 2 y \right)}\, dy}{3}$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(y \right)} = y^{2}$ and let $\operatorname{dv}{\left(y \right)} = \sin{\left(x + 2 y \right)}$ .
      
      Then $\operatorname{du}{\left(y \right)} = 2 y$ .
      
      To find $v{\left(y \right)}$ :
      
      Let $u = x + 2 y$ .
      
      Then let $du = 2 dy$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\sin{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(x + 2 y \right)}}{2}$
      
      Now evaluate the sub-integral.
    2. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(y \right)} = - y$ and let $\operatorname{dv}{\left(y \right)} = \cos{\left(x + 2 y \right)}$ .
      
      Then $\operatorname{du}{\left(y \right)} = -1$ .
      
      To find $v{\left(y \right)}$ :
      
      Let $u = x + 2 y$ .
      
      Then let $du = 2 dy$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(x + 2 y \right)}}{2}$
      
      Now evaluate the sub-integral.
    3. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\sin{\left(x + 2 y \right)}}{2}\right)\, dy = - \frac{\int \sin{\left(x + 2 y \right)}\, dy}{2}$
      
      Let $u = x + 2 y$ .
      
      Then let $du = 2 dy$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\sin{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(x + 2 y \right)}}{2}$
      
      So, the result is: $\frac{\cos{\left(x + 2 y \right)}}{4}$
    So, the result is: $- \frac{y^{2} \cos{\left(x + 2 y \right)}}{6} + \frac{y \sin{\left(x + 2 y \right)}}{6} + \frac{\cos{\left(x + 2 y \right)}}{12}$
  The result is: $- \frac{x \cos{\left(x + 2 y \right)}}{6} - \frac{y^{2} \cos{\left(x + 2 y \right)}}{6} + \frac{y \sin{\left(x + 2 y \right)}}{6} + \frac{\cos{\left(x + 2 y \right)}}{12}$
Add the constant of integration:

$- \frac{x \cos{\left(x + 2 y \right)}}{6} - \frac{y^{2} \cos{\left(x + 2 y \right)}}{6} + \frac{y \sin{\left(x + 2 y \right)}}{6} + \frac{\cos{\left(x + 2 y \right)}}{12}+ \mathrm{constant}$

The answer is:

$- \frac{x \cos{\left(x + 2 y \right)}}{6} - \frac{y^{2} \cos{\left(x + 2 y \right)}}{6} + \frac{y \sin{\left(x + 2 y \right)}}{6} + \frac{\cos{\left(x + 2 y \right)}}{12}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                               
 |                                                                 2                              
 | sin(x + 2*y) /     2\          cos(x + 2*y)   x*cos(x + 2*y)   y *cos(x + 2*y)   y*sin(x + 2*y)
 | ------------*\x + y / dy = C + ------------ - -------------- - --------------- + --------------
 |      3                              12              6                 6                6       
 |                                                                                                
/

$$\int \frac{\sin{\left(x + 2 y \right)}}{3} \left(x + y^{2}\right)\, dy = C - \frac{x \cos{\left(x + 2 y \right)}}{6} - \frac{y^{2} \cos{\left(x + 2 y \right)}}{6} + \frac{y \sin{\left(x + 2 y \right)}}{6} + \frac{\cos{\left(x + 2 y \right)}}{12}$$

Simplify

The answer [src]

sin(-2 + x)   cos(x)   cos(-2 + x)   x*cos(x)   x*cos(-2 + x)
----------- + ------ + ----------- - -------- + -------------
     6          12          12          6             6

$$- \frac{x \cos{\left(x \right)}}{6} + \frac{x \cos{\left(x - 2 \right)}}{6} + \frac{\sin{\left(x - 2 \right)}}{6} + \frac{\cos{\left(x \right)}}{12} + \frac{\cos{\left(x - 2 \right)}}{12}$$

sin(-2 + x)   cos(x)   cos(-2 + x)   x*cos(x)   x*cos(-2 + x)
----------- + ------ + ----------- - -------- + -------------
     6          12          12          6             6

$$- \frac{x \cos{\left(x \right)}}{6} + \frac{x \cos{\left(x - 2 \right)}}{6} + \frac{\sin{\left(x - 2 \right)}}{6} + \frac{\cos{\left(x \right)}}{12} + \frac{\cos{\left(x - 2 \right)}}{12}$$

sin(-2 + x)/6 + cos(x)/12 + cos(-2 + x)/12 - x*cos(x)/6 + x*cos(-2 + x)/6

Use the examples entering the upper and lower limits of integration.

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Integral of (sin(x+2y)/3(x+y^2))dy dy

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2