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Identical expressions
((x^ two +4x+ three))*sin4x
((x squared plus 4x plus 3)) multiply by sinus of 4x
((x to the power of two plus 4x plus three)) multiply by sinus of 4x
((x2+4x+3))*sin4x
x2+4x+3*sin4x
((x²+4x+3))*sin4x
((x to the power of 2+4x+3))*sin4x
((x^2+4x+3))sin4x
((x2+4x+3))sin4x
x2+4x+3sin4x
x^2+4x+3sin4x
((x^2+4x+3))*sin4xdx
Similar expressions
((x^2-4x+3))*sin4x
((x^2+4x-3))*sin4x

Integral of ((x^2+4x+3))*sin4x dx

The solution

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  0                           
  /                           
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 |  / 2          \            
 |  \x  + 4*x + 3/*sin(4*x) dx
 |                            
/                             
-1

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

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

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(\left(x^{2} + 4 x\right) + 3\right) \sin{\left(4 x \right)} = x^{2} \sin{\left(4 x \right)} + 4 x \sin{\left(4 x \right)} + 3 \sin{\left(4 x \right)}$
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 $\operatorname{dv}{\left(x \right)} = \sin{\left(4 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 2 x$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(4 x \right)}}{4}$
    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 $u{\left(x \right)} = - \frac{x}{2}$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(4 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = - \frac{1}{2}$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
    Now evaluate the sub-integral.
  3. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\sin{\left(4 x \right)}}{8}\right)\, dx = - \frac{\int \sin{\left(4 x \right)}\, dx}{8}$
    1. Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(4 x \right)}}{4}$
    So, the result is: $\frac{\cos{\left(4 x \right)}}{32}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 4 x \sin{\left(4 x \right)}\, dx = 4 \int x \sin{\left(4 x \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \sin{\left(4 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(4 x \right)}}{4}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(4 x \right)}}{4}\right)\, dx = - \frac{\int \cos{\left(4 x \right)}\, dx}{4}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      So, the result is: $- \frac{\sin{\left(4 x \right)}}{16}$
    So, the result is: $- x \cos{\left(4 x \right)} + \frac{\sin{\left(4 x \right)}}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 3 \sin{\left(4 x \right)}\, dx = 3 \int \sin{\left(4 x \right)}\, dx$
    1. Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(4 x \right)}}{4}$
    So, the result is: $- \frac{3 \cos{\left(4 x \right)}}{4}$
  The result is: $- \frac{x^{2} \cos{\left(4 x \right)}}{4} + \frac{x \sin{\left(4 x \right)}}{8} - x \cos{\left(4 x \right)} + \frac{\sin{\left(4 x \right)}}{4} - \frac{23 \cos{\left(4 x \right)}}{32}$
Method #2
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = x^{2} + 4 x + 3$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(4 x \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = 2 x + 4$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 4 x$ .
    
    Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
    
    $\int \frac{\sin{\left(u \right)}}{4}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(4 x \right)}}{4}$
  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 $u{\left(x \right)} = - \frac{x}{2} - 1$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(4 x \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = - \frac{1}{2}$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 4 x$ .
    
    Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
    
    $\int \frac{\cos{\left(u \right)}}{4}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
    Now substitute $u$ back in:
    
    $\frac{\sin{\left(4 x \right)}}{4}$
  Now evaluate the sub-integral.
3. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{\sin{\left(4 x \right)}}{8}\right)\, dx = - \frac{\int \sin{\left(4 x \right)}\, dx}{8}$
  1. Let $u = 4 x$ .
    
    Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
    
    $\int \frac{\sin{\left(u \right)}}{4}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(4 x \right)}}{4}$
  So, the result is: $\frac{\cos{\left(4 x \right)}}{32}$
Method #3
1. Rewrite the integrand:
  
  $\left(\left(x^{2} + 4 x\right) + 3\right) \sin{\left(4 x \right)} = x^{2} \sin{\left(4 x \right)} + 4 x \sin{\left(4 x \right)} + 3 \sin{\left(4 x \right)}$
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 $\operatorname{dv}{\left(x \right)} = \sin{\left(4 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 2 x$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(4 x \right)}}{4}$
    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 $u{\left(x \right)} = - \frac{x}{2}$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(4 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = - \frac{1}{2}$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
    Now evaluate the sub-integral.
  3. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\sin{\left(4 x \right)}}{8}\right)\, dx = - \frac{\int \sin{\left(4 x \right)}\, dx}{8}$
    1. Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(4 x \right)}}{4}$
    So, the result is: $\frac{\cos{\left(4 x \right)}}{32}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 4 x \sin{\left(4 x \right)}\, dx = 4 \int x \sin{\left(4 x \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \sin{\left(4 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(4 x \right)}}{4}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(4 x \right)}}{4}\right)\, dx = - \frac{\int \cos{\left(4 x \right)}\, dx}{4}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      So, the result is: $- \frac{\sin{\left(4 x \right)}}{16}$
    So, the result is: $- x \cos{\left(4 x \right)} + \frac{\sin{\left(4 x \right)}}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 3 \sin{\left(4 x \right)}\, dx = 3 \int \sin{\left(4 x \right)}\, dx$
    1. Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(4 x \right)}}{4}$
    So, the result is: $- \frac{3 \cos{\left(4 x \right)}}{4}$
  The result is: $- \frac{x^{2} \cos{\left(4 x \right)}}{4} + \frac{x \sin{\left(4 x \right)}}{8} - x \cos{\left(4 x \right)} + \frac{\sin{\left(4 x \right)}}{4} - \frac{23 \cos{\left(4 x \right)}}{32}$
Add the constant of integration:

$- \frac{x^{2} \cos{\left(4 x \right)}}{4} + \frac{x \sin{\left(4 x \right)}}{8} - x \cos{\left(4 x \right)} + \frac{\sin{\left(4 x \right)}}{4} - \frac{23 \cos{\left(4 x \right)}}{32}+ \mathrm{constant}$

The answer is:

$- \frac{x^{2} \cos{\left(4 x \right)}}{4} + \frac{x \sin{\left(4 x \right)}}{8} - x \cos{\left(4 x \right)} + \frac{\sin{\left(4 x \right)}}{4} - \frac{23 \cos{\left(4 x \right)}}{32}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                               
 |                                                                         2                      
 | / 2          \                   23*cos(4*x)   sin(4*x)                x *cos(4*x)   x*sin(4*x)
 | \x  + 4*x + 3/*sin(4*x) dx = C - ----------- + -------- - x*cos(4*x) - ----------- + ----------
 |                                       32          4                         4            8     
/

$$\int \left(\left(x^{2} + 4 x\right) + 3\right) \sin{\left(4 x \right)}\, dx = C - \frac{x^{2} \cos{\left(4 x \right)}}{4} + \frac{x \sin{\left(4 x \right)}}{8} - x \cos{\left(4 x \right)} + \frac{\sin{\left(4 x \right)}}{4} - \frac{23 \cos{\left(4 x \right)}}{32}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

  23   cos(4)   sin(4)
- -- - ------ + ------
  32     32       8

$$- \frac{23}{32} + \frac{\sin{\left(4 \right)}}{8} - \frac{\cos{\left(4 \right)}}{32}$$

  23   cos(4)   sin(4)
- -- - ------ + ------
  32     32       8

$$- \frac{23}{32} + \frac{\sin{\left(4 \right)}}{8} - \frac{\cos{\left(4 \right)}}{32}$$

-23/32 - cos(4)/32 + sin(4)/8

Numerical answer [src]

-0.792923948761503

-0.792923948761503

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+4x+3))*sin4x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3