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Identical expressions
(2x+ four)sin3x
(2x plus 4) sinus of 3x
(2x plus four) sinus of 3x
2x+4sin3x
(2x+4)sin3xdx
Similar expressions
(2x-4)sin3x

Integral of (2x+4)sin3x dx

The solution

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 |  (2*x + 4)*sin(3*x) dx
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$$\int\limits_{0}^{1} \left(2 x + 4\right) \sin{\left(3 x \right)}\, dx$$

Integral((2*x + 4)*sin(3*x), (x, 0, 1))

Detail solution

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

4              2*sin(3)
- - 2*cos(3) + --------
3                 9

$$\frac{2 \sin{\left(3 \right)}}{9} + \frac{4}{3} - 2 \cos{\left(3 \right)}$$

4              2*sin(3)
- - 2*cos(3) + --------
3                 9

$$\frac{2 \sin{\left(3 \right)}}{9} + \frac{4}{3} - 2 \cos{\left(3 \right)}$$

4/3 - 2*cos(3) + 2*sin(3)/9

Numerical answer [src]

3.34467832832531

3.34467832832531

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

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How to use it?

Identical expressions

Similar expressions

Integral of (2x+4)sin3x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3