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Integral of d{x}:
Integral of tanx
Integral of x/sin(x)
Integral of atan(x)
Integral of tanh(x)
Identical expressions
(X^ two -3x+ two)*sin(3x)
(X squared minus 3x plus 2) multiply by sinus of (3x)
(X to the power of two minus 3x plus two) multiply by sinus of (3x)
(X2-3x+2)*sin(3x)
X2-3x+2*sin3x
(X²-3x+2)*sin(3x)
(X to the power of 2-3x+2)*sin(3x)
(X^2-3x+2)sin(3x)
(X2-3x+2)sin(3x)
X2-3x+2sin3x
X^2-3x+2sin3x
(X^2-3x+2)*sin(3x)dx
Similar expressions
(X^2+3x+2)*sin(3x)
(X^2-3x-2)*sin(3x)

Integral of (X^2-3x+2)*sin(3x) dx

The solution

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 |  \x  - 3*x + 2/*sin(3*x) dx
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0

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

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

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(\left(x^{2} - 3 x\right) + 2\right) \sin{\left(3 x \right)} = x^{2} \sin{\left(3 x \right)} - 3 x \sin{\left(3 x \right)} + 2 \sin{\left(3 x \right)}$
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 $u{\left(x \right)} = x^{2}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 2 x$ .
    
    To find $v{\left(x \right)}$ :
    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 $u{\left(x \right)} = - \frac{2 x}{3}$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(3 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = - \frac{2}{3}$ .
    
    To find $v{\left(x \right)}$ :
    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. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{2 \sin{\left(3 x \right)}}{9}\right)\, dx = - \frac{2 \int \sin{\left(3 x \right)}\, dx}{9}$
    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:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
    So, the result is: $\frac{2 \cos{\left(3 x \right)}}{27}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 3 x \sin{\left(3 x \right)}\right)\, dx = - 3 \int x \sin{\left(3 x \right)}\, dx$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      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:
      
      $\int \left(- \frac{\cos{\left(3 x \right)}}{3}\right)\, dx = - \frac{\int \cos{\left(3 x \right)}\, dx}{3}$
      
      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: $- \frac{\sin{\left(3 x \right)}}{9}$
    So, the result is: $x \cos{\left(3 x \right)} - \frac{\sin{\left(3 x \right)}}{3}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 \sin{\left(3 x \right)}\, dx = 2 \int \sin{\left(3 x \right)}\, dx$
    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:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
    So, the result is: $- \frac{2 \cos{\left(3 x \right)}}{3}$
  The result is: $- \frac{x^{2} \cos{\left(3 x \right)}}{3} + \frac{2 x \sin{\left(3 x \right)}}{9} + x \cos{\left(3 x \right)} - \frac{\sin{\left(3 x \right)}}{3} - \frac{16 \cos{\left(3 x \right)}}{27}$
Method #2
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = x^{2} - 3 x + 2$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = 2 x - 3$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 3 x$ .
    
    Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
    
    $\int \frac{\sin{\left(u \right)}}{3}\, du$
    1. 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 $u{\left(x \right)} = 1 - \frac{2 x}{3}$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(3 x \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = - \frac{2}{3}$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 3 x$ .
    
    Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
    
    $\int \frac{\cos{\left(u \right)}}{3}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      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:
  
  $\int \left(- \frac{2 \sin{\left(3 x \right)}}{9}\right)\, dx = - \frac{2 \int \sin{\left(3 x \right)}\, dx}{9}$
  1. Let $u = 3 x$ .
    
    Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
    
    $\int \frac{\sin{\left(u \right)}}{3}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
      
      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: $\frac{2 \cos{\left(3 x \right)}}{27}$
Method #3
1. Rewrite the integrand:
  
  $\left(\left(x^{2} - 3 x\right) + 2\right) \sin{\left(3 x \right)} = x^{2} \sin{\left(3 x \right)} - 3 x \sin{\left(3 x \right)} + 2 \sin{\left(3 x \right)}$
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 $u{\left(x \right)} = x^{2}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 2 x$ .
    
    To find $v{\left(x \right)}$ :
    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 $u{\left(x \right)} = - \frac{2 x}{3}$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(3 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = - \frac{2}{3}$ .
    
    To find $v{\left(x \right)}$ :
    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. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{2 \sin{\left(3 x \right)}}{9}\right)\, dx = - \frac{2 \int \sin{\left(3 x \right)}\, dx}{9}$
    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:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
    So, the result is: $\frac{2 \cos{\left(3 x \right)}}{27}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 3 x \sin{\left(3 x \right)}\right)\, dx = - 3 \int x \sin{\left(3 x \right)}\, dx$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      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:
      
      $\int \left(- \frac{\cos{\left(3 x \right)}}{3}\right)\, dx = - \frac{\int \cos{\left(3 x \right)}\, dx}{3}$
      
      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: $- \frac{\sin{\left(3 x \right)}}{9}$
    So, the result is: $x \cos{\left(3 x \right)} - \frac{\sin{\left(3 x \right)}}{3}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 \sin{\left(3 x \right)}\, dx = 2 \int \sin{\left(3 x \right)}\, dx$
    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:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
    So, the result is: $- \frac{2 \cos{\left(3 x \right)}}{3}$
  The result is: $- \frac{x^{2} \cos{\left(3 x \right)}}{3} + \frac{2 x \sin{\left(3 x \right)}}{9} + x \cos{\left(3 x \right)} - \frac{\sin{\left(3 x \right)}}{3} - \frac{16 \cos{\left(3 x \right)}}{27}$
Add the constant of integration:

$- \frac{x^{2} \cos{\left(3 x \right)}}{3} + \frac{2 x \sin{\left(3 x \right)}}{9} + x \cos{\left(3 x \right)} - \frac{\sin{\left(3 x \right)}}{3} - \frac{16 \cos{\left(3 x \right)}}{27}+ \mathrm{constant}$

The answer is:

$- \frac{x^{2} \cos{\left(3 x \right)}}{3} + \frac{2 x \sin{\left(3 x \right)}}{9} + x \cos{\left(3 x \right)} - \frac{\sin{\left(3 x \right)}}{3} - \frac{16 \cos{\left(3 x \right)}}{27}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                                 
 |                                                                         2                        
 | / 2          \                   16*cos(3*x)   sin(3*x)                x *cos(3*x)   2*x*sin(3*x)
 | \x  - 3*x + 2/*sin(3*x) dx = C - ----------- - -------- + x*cos(3*x) - ----------- + ------------
 |                                       27          3                         3             9      
/

$$\int \left(\left(x^{2} - 3 x\right) + 2\right) \sin{\left(3 x \right)}\, dx = C - \frac{x^{2} \cos{\left(3 x \right)}}{3} + \frac{2 x \sin{\left(3 x \right)}}{9} + x \cos{\left(3 x \right)} - \frac{\sin{\left(3 x \right)}}{3} - \frac{16 \cos{\left(3 x \right)}}{27}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

16   sin(3)   2*cos(3)
-- - ------ + --------
27     9         27

$$\frac{2 \cos{\left(3 \right)}}{27} - \frac{\sin{\left(3 \right)}}{9} + \frac{16}{27}$$

16   sin(3)   2*cos(3)
-- - ------ + --------
27     9         27

$$\frac{2 \cos{\left(3 \right)}}{27} - \frac{\sin{\left(3 \right)}}{9} + \frac{16}{27}$$

16/27 - sin(3)/9 + 2*cos(3)/27

Numerical answer [src]

0.503579814171093

0.503579814171093

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (X^2-3x+2)*sin(3x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3