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

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

The solution

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

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

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

Detail solution

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

$\frac{x \sin{\left(2 x - 1 \right)}}{2} + \frac{\sin{\left(2 x - 1 \right)}}{2} + \frac{\cos{\left(2 x - 1 \right)}}{4}+ \mathrm{constant}$

The answer is:

$\frac{x \sin{\left(2 x - 1 \right)}}{2} + \frac{\sin{\left(2 x - 1 \right)}}{2} + \frac{\cos{\left(2 x - 1 \right)}}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                             
 |                               sin(-1 + 2*x)   cos(-1 + 2*x)   x*sin(-1 + 2*x)
 | (x + 1)*cos(2*x - 1) dx = C + ------------- + ------------- + ---------------
 |                                     2               4                2       
/

$$\int \left(x + 1\right) \cos{\left(2 x - 1 \right)}\, dx = C + \frac{x \sin{\left(2 x - 1 \right)}}{2} + \frac{\sin{\left(2 x - 1 \right)}}{2} + \frac{\cos{\left(2 x - 1 \right)}}{4}$$

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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How to use it?

Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3