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Identical expressions
x*(six -x)*sin(x/ six)
x multiply by (6 minus x) multiply by sinus of (x divide by 6)
x multiply by (six minus x) multiply by sinus of (x divide by six)
x(6-x)sin(x/6)
x6-xsinx/6
x*(6-x)*sin(x divide by 6)
x*(6-x)*sin(x/6)dx
Similar expressions
x*(6+x)*sin(x/6)

Integral of x(6-x)sin(x/6) dx

The solution

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0

$$\int\limits_{0}^{6} x \left(6 - x\right) \sin{\left(\frac{x}{6} \right)}\, dx$$

Integral((x*(6 - x))*sin(x/6), (x, 0, 6))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = - x$ .
  
  Then let $du = - dx$ and substitute $du$ :
  
  $\int \left(- u^{2} \sin{\left(\frac{u}{6} \right)} - 6 u \sin{\left(\frac{u}{6} \right)}\right)\, du$
  1. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{2} \sin{\left(\frac{u}{6} \right)}\right)\, du = - \int u^{2} \sin{\left(\frac{u}{6} \right)}\, du$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u^{2}$ and let $\operatorname{dv}{\left(u \right)} = \sin{\left(\frac{u}{6} \right)}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 2 u$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = \frac{u}{6}$ .
      
      Then let $du = \frac{du}{6}$ and substitute $6 du$ :
      
      $\int 6 \sin{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = 6 \int \sin{\left(u \right)}\, du$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- 6 \cos{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- 6 \cos{\left(\frac{u}{6} \right)}$
      
      Now evaluate the sub-integral.
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = - 12 u$ and let $\operatorname{dv}{\left(u \right)} = \cos{\left(\frac{u}{6} \right)}$ .
      
      Then $\operatorname{du}{\left(u \right)} = -12$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = \frac{u}{6}$ .
      
      Then let $du = \frac{du}{6}$ and substitute $6 du$ :
      
      $\int 6 \cos{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = 6 \int \cos{\left(u \right)}\, du$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $6 \sin{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $6 \sin{\left(\frac{u}{6} \right)}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 72 \sin{\left(\frac{u}{6} \right)}\right)\, du = - 72 \int \sin{\left(\frac{u}{6} \right)}\, du$
      
      Let $u = \frac{u}{6}$ .
      
      Then let $du = \frac{du}{6}$ and substitute $6 du$ :
      
      $\int 6 \sin{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = 6 \int \sin{\left(u \right)}\, du$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- 6 \cos{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- 6 \cos{\left(\frac{u}{6} \right)}$
      
      So, the result is: $432 \cos{\left(\frac{u}{6} \right)}$
      
      So, the result is: $6 u^{2} \cos{\left(\frac{u}{6} \right)} - 72 u \sin{\left(\frac{u}{6} \right)} - 432 \cos{\left(\frac{u}{6} \right)}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 6 u \sin{\left(\frac{u}{6} \right)}\right)\, du = - 6 \int u \sin{\left(\frac{u}{6} \right)}\, du$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = \sin{\left(\frac{u}{6} \right)}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = \frac{u}{6}$ .
      
      Then let $du = \frac{du}{6}$ and substitute $6 du$ :
      
      $\int 6 \sin{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = 6 \int \sin{\left(u \right)}\, du$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- 6 \cos{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- 6 \cos{\left(\frac{u}{6} \right)}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 6 \cos{\left(\frac{u}{6} \right)}\right)\, du = - 6 \int \cos{\left(\frac{u}{6} \right)}\, du$
      
      Let $u = \frac{u}{6}$ .
      
      Then let $du = \frac{du}{6}$ and substitute $6 du$ :
      
      $\int 6 \cos{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = 6 \int \cos{\left(u \right)}\, du$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $6 \sin{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $6 \sin{\left(\frac{u}{6} \right)}$
      
      So, the result is: $- 36 \sin{\left(\frac{u}{6} \right)}$
      
      So, the result is: $36 u \cos{\left(\frac{u}{6} \right)} - 216 \sin{\left(\frac{u}{6} \right)}$
    The result is: $6 u^{2} \cos{\left(\frac{u}{6} \right)} - 72 u \sin{\left(\frac{u}{6} \right)} + 36 u \cos{\left(\frac{u}{6} \right)} - 216 \sin{\left(\frac{u}{6} \right)} - 432 \cos{\left(\frac{u}{6} \right)}$
  Now substitute $u$ back in:
  
  $6 x^{2} \cos{\left(\frac{x}{6} \right)} - 72 x \sin{\left(\frac{x}{6} \right)} - 36 x \cos{\left(\frac{x}{6} \right)} + 216 \sin{\left(\frac{x}{6} \right)} - 432 \cos{\left(\frac{x}{6} \right)}$
Method #2
1. Rewrite the integrand:
  
  $x \left(6 - x\right) \sin{\left(\frac{x}{6} \right)} = - x^{2} \sin{\left(\frac{x}{6} \right)} + 6 x \sin{\left(\frac{x}{6} \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- x^{2} \sin{\left(\frac{x}{6} \right)}\right)\, dx = - \int x^{2} \sin{\left(\frac{x}{6} \right)}\, dx$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = x^{2}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(\frac{x}{6} \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 2 x$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = \frac{x}{6}$ .
      
      Then let $du = \frac{dx}{6}$ and substitute $6 du$ :
      
      $\int 6 \sin{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = 6 \int \sin{\left(u \right)}\, du$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- 6 \cos{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- 6 \cos{\left(\frac{x}{6} \right)}$
      
      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 $u{\left(x \right)} = - 12 x$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(\frac{x}{6} \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = -12$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = \frac{x}{6}$ .
      
      Then let $du = \frac{dx}{6}$ and substitute $6 du$ :
      
      $\int 6 \cos{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = 6 \int \cos{\left(u \right)}\, du$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $6 \sin{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $6 \sin{\left(\frac{x}{6} \right)}$
      
      Now evaluate the sub-integral.
    3. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 72 \sin{\left(\frac{x}{6} \right)}\right)\, dx = - 72 \int \sin{\left(\frac{x}{6} \right)}\, dx$
      
      Let $u = \frac{x}{6}$ .
      
      Then let $du = \frac{dx}{6}$ and substitute $6 du$ :
      
      $\int 6 \sin{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = 6 \int \sin{\left(u \right)}\, du$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- 6 \cos{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- 6 \cos{\left(\frac{x}{6} \right)}$
      
      So, the result is: $432 \cos{\left(\frac{x}{6} \right)}$
    So, the result is: $6 x^{2} \cos{\left(\frac{x}{6} \right)} - 72 x \sin{\left(\frac{x}{6} \right)} - 432 \cos{\left(\frac{x}{6} \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 6 x \sin{\left(\frac{x}{6} \right)}\, dx = 6 \int x \sin{\left(\frac{x}{6} \right)}\, dx$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(\frac{x}{6} \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = \frac{x}{6}$ .
      
      Then let $du = \frac{dx}{6}$ and substitute $6 du$ :
      
      $\int 6 \sin{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = 6 \int \sin{\left(u \right)}\, du$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- 6 \cos{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- 6 \cos{\left(\frac{x}{6} \right)}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 6 \cos{\left(\frac{x}{6} \right)}\right)\, dx = - 6 \int \cos{\left(\frac{x}{6} \right)}\, dx$
      
      Let $u = \frac{x}{6}$ .
      
      Then let $du = \frac{dx}{6}$ and substitute $6 du$ :
      
      $\int 6 \cos{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = 6 \int \cos{\left(u \right)}\, du$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $6 \sin{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $6 \sin{\left(\frac{x}{6} \right)}$
      
      So, the result is: $- 36 \sin{\left(\frac{x}{6} \right)}$
    So, the result is: $- 36 x \cos{\left(\frac{x}{6} \right)} + 216 \sin{\left(\frac{x}{6} \right)}$
  The result is: $6 x^{2} \cos{\left(\frac{x}{6} \right)} - 72 x \sin{\left(\frac{x}{6} \right)} - 36 x \cos{\left(\frac{x}{6} \right)} + 216 \sin{\left(\frac{x}{6} \right)} - 432 \cos{\left(\frac{x}{6} \right)}$
Method #3
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = x \left(6 - x\right)$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(\frac{x}{6} \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = 6 - 2 x$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = \frac{x}{6}$ .
    
    Then let $du = \frac{dx}{6}$ and substitute $6 du$ :
    
    $\int 6 \sin{\left(u \right)}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = 6 \int \sin{\left(u \right)}\, du$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- 6 \cos{\left(u \right)}$
    Now substitute $u$ back in:
    
    $- 6 \cos{\left(\frac{x}{6} \right)}$
  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 $u{\left(x \right)} = 12 x - 36$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(\frac{x}{6} \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = 12$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = \frac{x}{6}$ .
    
    Then let $du = \frac{dx}{6}$ and substitute $6 du$ :
    
    $\int 6 \cos{\left(u \right)}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = 6 \int \cos{\left(u \right)}\, du$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $6 \sin{\left(u \right)}$
    Now substitute $u$ back in:
    
    $6 \sin{\left(\frac{x}{6} \right)}$
  Now evaluate the sub-integral.
3. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 72 \sin{\left(\frac{x}{6} \right)}\, dx = 72 \int \sin{\left(\frac{x}{6} \right)}\, dx$
  1. Let $u = \frac{x}{6}$ .
    
    Then let $du = \frac{dx}{6}$ and substitute $6 du$ :
    
    $\int 6 \sin{\left(u \right)}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = 6 \int \sin{\left(u \right)}\, du$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- 6 \cos{\left(u \right)}$
    Now substitute $u$ back in:
    
    $- 6 \cos{\left(\frac{x}{6} \right)}$
  So, the result is: $- 432 \cos{\left(\frac{x}{6} \right)}$
Method #4
1. Rewrite the integrand:
  
  $x \left(6 - x\right) \sin{\left(\frac{x}{6} \right)} = - x^{2} \sin{\left(\frac{x}{6} \right)} + 6 x \sin{\left(\frac{x}{6} \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- x^{2} \sin{\left(\frac{x}{6} \right)}\right)\, dx = - \int x^{2} \sin{\left(\frac{x}{6} \right)}\, dx$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = x^{2}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(\frac{x}{6} \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 2 x$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = \frac{x}{6}$ .
      
      Then let $du = \frac{dx}{6}$ and substitute $6 du$ :
      
      $\int 6 \sin{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = 6 \int \sin{\left(u \right)}\, du$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- 6 \cos{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- 6 \cos{\left(\frac{x}{6} \right)}$
      
      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 $u{\left(x \right)} = - 12 x$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(\frac{x}{6} \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = -12$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = \frac{x}{6}$ .
      
      Then let $du = \frac{dx}{6}$ and substitute $6 du$ :
      
      $\int 6 \cos{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = 6 \int \cos{\left(u \right)}\, du$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $6 \sin{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $6 \sin{\left(\frac{x}{6} \right)}$
      
      Now evaluate the sub-integral.
    3. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 72 \sin{\left(\frac{x}{6} \right)}\right)\, dx = - 72 \int \sin{\left(\frac{x}{6} \right)}\, dx$
      
      Let $u = \frac{x}{6}$ .
      
      Then let $du = \frac{dx}{6}$ and substitute $6 du$ :
      
      $\int 6 \sin{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = 6 \int \sin{\left(u \right)}\, du$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- 6 \cos{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- 6 \cos{\left(\frac{x}{6} \right)}$
      
      So, the result is: $432 \cos{\left(\frac{x}{6} \right)}$
    So, the result is: $6 x^{2} \cos{\left(\frac{x}{6} \right)} - 72 x \sin{\left(\frac{x}{6} \right)} - 432 \cos{\left(\frac{x}{6} \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 6 x \sin{\left(\frac{x}{6} \right)}\, dx = 6 \int x \sin{\left(\frac{x}{6} \right)}\, dx$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(\frac{x}{6} \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = \frac{x}{6}$ .
      
      Then let $du = \frac{dx}{6}$ and substitute $6 du$ :
      
      $\int 6 \sin{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = 6 \int \sin{\left(u \right)}\, du$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- 6 \cos{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- 6 \cos{\left(\frac{x}{6} \right)}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 6 \cos{\left(\frac{x}{6} \right)}\right)\, dx = - 6 \int \cos{\left(\frac{x}{6} \right)}\, dx$
      
      Let $u = \frac{x}{6}$ .
      
      Then let $du = \frac{dx}{6}$ and substitute $6 du$ :
      
      $\int 6 \cos{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = 6 \int \cos{\left(u \right)}\, du$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $6 \sin{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $6 \sin{\left(\frac{x}{6} \right)}$
      
      So, the result is: $- 36 \sin{\left(\frac{x}{6} \right)}$
    So, the result is: $- 36 x \cos{\left(\frac{x}{6} \right)} + 216 \sin{\left(\frac{x}{6} \right)}$
  The result is: $6 x^{2} \cos{\left(\frac{x}{6} \right)} - 72 x \sin{\left(\frac{x}{6} \right)} - 36 x \cos{\left(\frac{x}{6} \right)} + 216 \sin{\left(\frac{x}{6} \right)} - 432 \cos{\left(\frac{x}{6} \right)}$
Add the constant of integration:

$6 x^{2} \cos{\left(\frac{x}{6} \right)} - 72 x \sin{\left(\frac{x}{6} \right)} - 36 x \cos{\left(\frac{x}{6} \right)} + 216 \sin{\left(\frac{x}{6} \right)} - 432 \cos{\left(\frac{x}{6} \right)}+ \mathrm{constant}$

The answer is:

$6 x^{2} \cos{\left(\frac{x}{6} \right)} - 72 x \sin{\left(\frac{x}{6} \right)} - 36 x \cos{\left(\frac{x}{6} \right)} + 216 \sin{\left(\frac{x}{6} \right)} - 432 \cos{\left(\frac{x}{6} \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                           
 |                                                                                            
 |              /x\                 /x\          /x\           /x\           /x\      2    /x\
 | x*(6 - x)*sin|-| dx = C - 432*cos|-| + 216*sin|-| - 72*x*sin|-| - 36*x*cos|-| + 6*x *cos|-|
 |              \6/                 \6/          \6/           \6/           \6/           \6/
 |                                                                                            
/

$$\int x \left(6 - x\right) \sin{\left(\frac{x}{6} \right)}\, dx = C + 6 x^{2} \cos{\left(\frac{x}{6} \right)} - 72 x \sin{\left(\frac{x}{6} \right)} - 36 x \cos{\left(\frac{x}{6} \right)} + 216 \sin{\left(\frac{x}{6} \right)} - 432 \cos{\left(\frac{x}{6} \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

432 - 432*cos(1) - 216*sin(1)

$$- 432 \cos{\left(1 \right)} - 216 \sin{\left(1 \right)} + 432$$

432 - 432*cos(1) - 216*sin(1)

$$- 432 \cos{\left(1 \right)} - 216 \sin{\left(1 \right)} + 432$$

432 - 432*cos(1) - 216*sin(1)

Numerical answer [src]

16.831671146458

16.831671146458

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

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Integral of x(6-x)sin(x/6) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Method #4

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of x*(6-x)*sin(x/6) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Method #4

Integral of x(6-x)sin(x/6) dx