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Identical expressions
(6x- one)*cos(two *x)
(6x minus 1) multiply by co sinus of e of (2 multiply by x)
(6x minus one) multiply by co sinus of e of (two multiply by x)
(6x-1)cos(2x)
6x-1cos2x
(6x-1)*cos(2*x)dx
Similar expressions
(6x+1)*cos(2*x)

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

The solution

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

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

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

Detail solution

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

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

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

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

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

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

Numerical answer [src]

0.149023312243491

0.149023312243491

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

Other calculators

How to use it?

Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #1

Method #2

Method #3

Method #4

Other calculators

How to use it?

Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #1

Method #2

Method #3

Method #4

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