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(x- one)^2sin(3x)x
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(x+1)^2sin(3x)x

Integral of (x-1)^2sin(3x)x dx

The solution

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0

$$\int\limits_{0}^{1} x \left(x - 1\right)^{2} \sin{\left(3 x \right)}\, dx$$

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

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $x \left(x - 1\right)^{2} \sin{\left(3 x \right)} = x^{3} \sin{\left(3 x \right)} - 2 x^{2} \sin{\left(3 x \right)} + x \sin{\left(3 x \right)}$
2. Integrate term-by-term:
  1. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(x \right)} = x^{3}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 3 x^{2}$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
    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 $u{\left(x \right)} = - x^{2}$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(3 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = - 2 x$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(3 x \right)}}{3}$
    Now evaluate the sub-integral.
  3. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(x \right)} = - \frac{2 x}{3}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = - \frac{2}{3}$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
    Now evaluate the sub-integral.
  4. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{2 \cos{\left(3 x \right)}}{9}\, dx = \frac{2 \int \cos{\left(3 x \right)}\, dx}{9}$
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(3 x \right)}}{3}$
    So, the result is: $\frac{2 \sin{\left(3 x \right)}}{27}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 x^{2} \sin{\left(3 x \right)}\right)\, dx = - 2 \int x^{2} \sin{\left(3 x \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 2 x$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
      
      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 $u{\left(x \right)} = - \frac{2 x}{3}$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(3 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = - \frac{2}{3}$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(3 x \right)}}{3}$
      
      Now evaluate the sub-integral.
    3. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{2 \sin{\left(3 x \right)}}{9}\right)\, dx = - \frac{2 \int \sin{\left(3 x \right)}\, dx}{9}$
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
      
      So, the result is: $\frac{2 \cos{\left(3 x \right)}}{27}$
    So, the result is: $\frac{2 x^{2} \cos{\left(3 x \right)}}{3} - \frac{4 x \sin{\left(3 x \right)}}{9} - \frac{4 \cos{\left(3 x \right)}}{27}$
  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 $\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 1$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
    Now evaluate the sub-integral.
  2. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos{\left(3 x \right)}}{3}\right)\, dx = - \frac{\int \cos{\left(3 x \right)}\, dx}{3}$
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(3 x \right)}}{3}$
    So, the result is: $- \frac{\sin{\left(3 x \right)}}{9}$
  The result is: $- \frac{x^{3} \cos{\left(3 x \right)}}{3} + \frac{x^{2} \sin{\left(3 x \right)}}{3} + \frac{2 x^{2} \cos{\left(3 x \right)}}{3} - \frac{4 x \sin{\left(3 x \right)}}{9} - \frac{x \cos{\left(3 x \right)}}{9} + \frac{\sin{\left(3 x \right)}}{27} - \frac{4 \cos{\left(3 x \right)}}{27}$
Method #2
1. Rewrite the integrand:
  
  $x \left(x - 1\right)^{2} \sin{\left(3 x \right)} = x^{3} \sin{\left(3 x \right)} - 2 x^{2} \sin{\left(3 x \right)} + x \sin{\left(3 x \right)}$
2. Integrate term-by-term:
  1. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(x \right)} = x^{3}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 3 x^{2}$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
    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 $u{\left(x \right)} = - x^{2}$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(3 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = - 2 x$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(3 x \right)}}{3}$
    Now evaluate the sub-integral.
  3. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(x \right)} = - \frac{2 x}{3}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = - \frac{2}{3}$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
    Now evaluate the sub-integral.
  4. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{2 \cos{\left(3 x \right)}}{9}\, dx = \frac{2 \int \cos{\left(3 x \right)}\, dx}{9}$
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(3 x \right)}}{3}$
    So, the result is: $\frac{2 \sin{\left(3 x \right)}}{27}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 x^{2} \sin{\left(3 x \right)}\right)\, dx = - 2 \int x^{2} \sin{\left(3 x \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 2 x$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
      
      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 $u{\left(x \right)} = - \frac{2 x}{3}$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(3 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = - \frac{2}{3}$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(3 x \right)}}{3}$
      
      Now evaluate the sub-integral.
    3. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{2 \sin{\left(3 x \right)}}{9}\right)\, dx = - \frac{2 \int \sin{\left(3 x \right)}\, dx}{9}$
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
      
      So, the result is: $\frac{2 \cos{\left(3 x \right)}}{27}$
    So, the result is: $\frac{2 x^{2} \cos{\left(3 x \right)}}{3} - \frac{4 x \sin{\left(3 x \right)}}{9} - \frac{4 \cos{\left(3 x \right)}}{27}$
  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 $\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 1$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
    Now evaluate the sub-integral.
  2. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos{\left(3 x \right)}}{3}\right)\, dx = - \frac{\int \cos{\left(3 x \right)}\, dx}{3}$
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(3 x \right)}}{3}$
    So, the result is: $- \frac{\sin{\left(3 x \right)}}{9}$
  The result is: $- \frac{x^{3} \cos{\left(3 x \right)}}{3} + \frac{x^{2} \sin{\left(3 x \right)}}{3} + \frac{2 x^{2} \cos{\left(3 x \right)}}{3} - \frac{4 x \sin{\left(3 x \right)}}{9} - \frac{x \cos{\left(3 x \right)}}{9} + \frac{\sin{\left(3 x \right)}}{27} - \frac{4 \cos{\left(3 x \right)}}{27}$
Add the constant of integration:

$- \frac{x^{3} \cos{\left(3 x \right)}}{3} + \frac{x^{2} \sin{\left(3 x \right)}}{3} + \frac{2 x^{2} \cos{\left(3 x \right)}}{3} - \frac{4 x \sin{\left(3 x \right)}}{9} - \frac{x \cos{\left(3 x \right)}}{9} + \frac{\sin{\left(3 x \right)}}{27} - \frac{4 \cos{\left(3 x \right)}}{27}+ \mathrm{constant}$

The answer is:

$- \frac{x^{3} \cos{\left(3 x \right)}}{3} + \frac{x^{2} \sin{\left(3 x \right)}}{3} + \frac{2 x^{2} \cos{\left(3 x \right)}}{3} - \frac{4 x \sin{\left(3 x \right)}}{9} - \frac{x \cos{\left(3 x \right)}}{9} + \frac{\sin{\left(3 x \right)}}{27} - \frac{4 \cos{\left(3 x \right)}}{27}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                                                          
 |                                                                      3                          2               2         
 |        2                     4*cos(3*x)   sin(3*x)   4*x*sin(3*x)   x *cos(3*x)   x*cos(3*x)   x *sin(3*x)   2*x *cos(3*x)
 | (x - 1) *sin(3*x)*x dx = C - ---------- + -------- - ------------ - ----------- - ---------- + ----------- + -------------
 |                                  27          27           9              3            9             3              3      
/

$$\int x \left(x - 1\right)^{2} \sin{\left(3 x \right)}\, dx = C - \frac{x^{3} \cos{\left(3 x \right)}}{3} + \frac{x^{2} \sin{\left(3 x \right)}}{3} + \frac{2 x^{2} \cos{\left(3 x \right)}}{3} - \frac{4 x \sin{\left(3 x \right)}}{9} - \frac{x \cos{\left(3 x \right)}}{9} + \frac{\sin{\left(3 x \right)}}{27} - \frac{4 \cos{\left(3 x \right)}}{27}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

4    2*sin(3)   2*cos(3)
-- - -------- + --------
27      27         27

$$\frac{2 \cos{\left(3 \right)}}{27} - \frac{2 \sin{\left(3 \right)}}{27} + \frac{4}{27}$$

4    2*sin(3)   2*cos(3)
-- - -------- + --------
27      27         27

$$\frac{2 \cos{\left(3 \right)}}{27} - \frac{2 \sin{\left(3 \right)}}{27} + \frac{4}{27}$$

4/27 - 2*sin(3)/27 + 2*cos(3)/27

Numerical answer [src]

0.0643620366918287

0.0643620366918287

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

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Identical expressions

Similar expressions

Integral of (x-1)^2sin(3x)x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2