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Identical expressions
(one -6x)cos(2x+ one)
(1 minus 6x) co sinus of e of (2x plus 1)
(one minus 6x) co sinus of e of (2x plus one)
1-6xcos2x+1
(1-6x)cos(2x+1)dx
Similar expressions
(1-6x)cos(2x-1)
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Integral of (1-6x)cos(2x+1) dx

The solution

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 |  (1 - 6*x)*cos(2*x + 1) dx
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0

$$\int\limits_{0}^{0} \left(1 - 6 x\right) \cos{\left(2 x + 1 \right)}\, dx$$

Integral((1 - 6*x)*cos(2*x + 1), (x, 0, 0))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(1 - 6 x\right) \cos{\left(2 x + 1 \right)} = - 6 x \cos{\left(2 x + 1 \right)} + \cos{\left(2 x + 1 \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 6 x \cos{\left(2 x + 1 \right)}\right)\, dx = - 6 \int x \cos{\left(2 x + 1 \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \cos{\left(2 x + 1 \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 2 x + 1$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(2 x + 1 \right)}}{2}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(2 x + 1 \right)}}{2}\, dx = \frac{\int \sin{\left(2 x + 1 \right)}\, dx}{2}$
      
      Let $u = 2 x + 1$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\sin{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(2 x + 1 \right)}}{2}$
      
      So, the result is: $- \frac{\cos{\left(2 x + 1 \right)}}{4}$
    So, the result is: $- 3 x \sin{\left(2 x + 1 \right)} - \frac{3 \cos{\left(2 x + 1 \right)}}{2}$
  1. Let $u = 2 x + 1$ .
    
    Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{\cos{\left(u \right)}}{2}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
    Now substitute $u$ back in:
    
    $\frac{\sin{\left(2 x + 1 \right)}}{2}$
  The result is: $- 3 x \sin{\left(2 x + 1 \right)} + \frac{\sin{\left(2 x + 1 \right)}}{2} - \frac{3 \cos{\left(2 x + 1 \right)}}{2}$
Method #2
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = 1 - 6 x$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(2 x + 1 \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = -6$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 2 x + 1$ .
    
    Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{\cos{\left(u \right)}}{2}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
    Now substitute $u$ back in:
    
    $\frac{\sin{\left(2 x + 1 \right)}}{2}$
  Now evaluate the sub-integral.
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 3 \sin{\left(2 x + 1 \right)}\right)\, dx = - 3 \int \sin{\left(2 x + 1 \right)}\, dx$
  1. Let $u = 2 x + 1$ .
    
    Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{\sin{\left(u \right)}}{2}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{2}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(2 x + 1 \right)}}{2}$
  So, the result is: $\frac{3 \cos{\left(2 x + 1 \right)}}{2}$
Method #3
1. Rewrite the integrand:
  
  $\left(1 - 6 x\right) \cos{\left(2 x + 1 \right)} = - 6 x \cos{\left(2 x + 1 \right)} + \cos{\left(2 x + 1 \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 6 x \cos{\left(2 x + 1 \right)}\right)\, dx = - 6 \int x \cos{\left(2 x + 1 \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \cos{\left(2 x + 1 \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 2 x + 1$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(2 x + 1 \right)}}{2}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(2 x + 1 \right)}}{2}\, dx = \frac{\int \sin{\left(2 x + 1 \right)}\, dx}{2}$
      
      Let $u = 2 x + 1$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\sin{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(2 x + 1 \right)}}{2}$
      
      So, the result is: $- \frac{\cos{\left(2 x + 1 \right)}}{4}$
    So, the result is: $- 3 x \sin{\left(2 x + 1 \right)} - \frac{3 \cos{\left(2 x + 1 \right)}}{2}$
  1. Let $u = 2 x + 1$ .
    
    Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{\cos{\left(u \right)}}{2}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
    Now substitute $u$ back in:
    
    $\frac{\sin{\left(2 x + 1 \right)}}{2}$
  The result is: $- 3 x \sin{\left(2 x + 1 \right)} + \frac{\sin{\left(2 x + 1 \right)}}{2} - \frac{3 \cos{\left(2 x + 1 \right)}}{2}$
Add the constant of integration:

$- 3 x \sin{\left(2 x + 1 \right)} + \frac{\sin{\left(2 x + 1 \right)}}{2} - \frac{3 \cos{\left(2 x + 1 \right)}}{2}+ \mathrm{constant}$

The answer is:

$- 3 x \sin{\left(2 x + 1 \right)} + \frac{\sin{\left(2 x + 1 \right)}}{2} - \frac{3 \cos{\left(2 x + 1 \right)}}{2}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                
 |                                 sin(1 + 2*x)   3*cos(1 + 2*x)                   
 | (1 - 6*x)*cos(2*x + 1) dx = C + ------------ - -------------- - 3*x*sin(1 + 2*x)
 |                                      2               2                          
/

$$\int \left(1 - 6 x\right) \cos{\left(2 x + 1 \right)}\, dx = C - 3 x \sin{\left(2 x + 1 \right)} + \frac{\sin{\left(2 x + 1 \right)}}{2} - \frac{3 \cos{\left(2 x + 1 \right)}}{2}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

$$0$$

Numerical answer [src]

0.0

0.0

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

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Identical expressions

Similar expressions

Integral of (1-6x)cos(2x+1) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3