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Identical expressions
(one -x)*sin4x
(1 minus x) multiply by sinus of 4x
(one minus x) multiply by sinus of 4x
(1-x)sin4x
1-xsin4x
(1-x)*sin4xdx
Similar expressions
(1+x)*sin4x

Integral of (1-x)*sin4x dx

The solution

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 oo                    
  /                    
 |                     
 |  (1 - x)*sin(4*x) dx
 |                     
/                      
-oo

$$\int\limits_{-\infty}^{\infty} \left(1 - x\right) \sin{\left(4 x \right)}\, dx$$

Integral((1 - x)*sin(4*x), (x, -oo, oo))

Detail solution

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

$\frac{x \cos{\left(4 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{16} - \frac{\cos{\left(4 x \right)}}{4}+ \mathrm{constant}$

The answer is:

$\frac{x \cos{\left(4 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{16} - \frac{\cos{\left(4 x \right)}}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                          
 |                           cos(4*x)   sin(4*x)   x*cos(4*x)
 | (1 - x)*sin(4*x) dx = C - -------- - -------- + ----------
 |                              4          16          4     
/

$$\int \left(1 - x\right) \sin{\left(4 x \right)}\, dx = C + \frac{x \cos{\left(4 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{16} - \frac{\cos{\left(4 x \right)}}{4}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

  oo                     
   /                     
  |                      
- |  (-1 + x)*sin(4*x) dx
  |                      
 /                       
 -oo

$$- \int\limits_{-\infty}^{\infty} \left(x - 1\right) \sin{\left(4 x \right)}\, dx$$

  oo                     
   /                     
  |                      
- |  (-1 + x)*sin(4*x) dx
  |                      
 /                       
 -oo

$$- \int\limits_{-\infty}^{\infty} \left(x - 1\right) \sin{\left(4 x \right)}\, dx$$

-Integral((-1 + x)*sin(4*x), (x, -oo, oo))

Numerical answer [src]

-1.06155338494956e+37

-1.06155338494956e+37

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 (1-x)*sin4x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Method #4