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Identical expressions
(3x+ two)*sin3x
(3x plus 2) multiply by sinus of 3x
(3x plus two) multiply by sinus of 3x
(3x+2)sin3x
3x+2sin3x
(3x+2)*sin3xdx
Similar expressions
(3x-2)*sin3x
(X^2-3x+2)*sin(3x)

Solving integrals/ (3x+2)*sin3x

Integral of (3x+2)*sin3x dx

The solution

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  1                      
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 |  (3*x + 2)*sin(3*x) dx
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0

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

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

Detail solution

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

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

The answer is:

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

The answer (Indefinite) [src]

  /                                                              
 |                             2*cos(3*x)   sin(3*x)             
 | (3*x + 2)*sin(3*x) dx = C - ---------- + -------- - x*cos(3*x)
 |                                 3           3                 
/

$${{\sin \left(3\,x\right)-3\,x\,\cos \left(3\,x\right)-2\,\cos \left(3\,x\right)}\over{3}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

2   5*cos(3)   sin(3)
- - -------- + ------
3      3         3

$${{\sin 3-5\,\cos 3}\over{3}}+{{2}\over{3}}$$

2   5*cos(3)   sin(3)
- - -------- + ------
3      3         3

$$\frac{\sin{\left(3 \right)}}{3} + \frac{2}{3} - \frac{5 \cos{\left(3 \right)}}{3}$$

Упростить

Numerical answer [src]

2.36369416368736

2.36369416368736

The graph

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

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Identical expressions

Similar expressions

Integral of (3x+2)*sin3x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Method #4