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Similar expressions
(3+2x)cos(x/3)dx

Integral of (3-2x)cos(x/3)dx dx

The solution

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  1                    
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 |               /x\   
 |  (3 - 2*x)*cos|-| dx
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0

$$\int\limits_{0}^{1} \left(3 - 2 x\right) \cos{\left(\frac{x}{3} \right)}\, dx$$

Integral((3 - 2*x)*cos(x/3), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(3 - 2 x\right) \cos{\left(\frac{x}{3} \right)} = - 2 x \cos{\left(\frac{x}{3} \right)} + 3 \cos{\left(\frac{x}{3} \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 x \cos{\left(\frac{x}{3} \right)}\right)\, dx = - 2 \int x \cos{\left(\frac{x}{3} \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \cos{\left(\frac{x}{3} \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = \frac{x}{3}$ .
      
      Then let $du = \frac{dx}{3}$ and substitute $3 du$ :
      
      $\int 3 \cos{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = 3 \int \cos{\left(u \right)}\, du$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $3 \sin{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $3 \sin{\left(\frac{x}{3} \right)}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 3 \sin{\left(\frac{x}{3} \right)}\, dx = 3 \int \sin{\left(\frac{x}{3} \right)}\, dx$
      
      Let $u = \frac{x}{3}$ .
      
      Then let $du = \frac{dx}{3}$ and substitute $3 du$ :
      
      $\int 3 \sin{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = 3 \int \sin{\left(u \right)}\, du$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- 3 \cos{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- 3 \cos{\left(\frac{x}{3} \right)}$
      
      So, the result is: $- 9 \cos{\left(\frac{x}{3} \right)}$
    So, the result is: $- 6 x \sin{\left(\frac{x}{3} \right)} - 18 \cos{\left(\frac{x}{3} \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 3 \cos{\left(\frac{x}{3} \right)}\, dx = 3 \int \cos{\left(\frac{x}{3} \right)}\, dx$
    1. Let $u = \frac{x}{3}$ .
      
      Then let $du = \frac{dx}{3}$ and substitute $3 du$ :
      
      $\int 3 \cos{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = 3 \int \cos{\left(u \right)}\, du$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $3 \sin{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $3 \sin{\left(\frac{x}{3} \right)}$
    So, the result is: $9 \sin{\left(\frac{x}{3} \right)}$
  The result is: $- 6 x \sin{\left(\frac{x}{3} \right)} + 9 \sin{\left(\frac{x}{3} \right)} - 18 \cos{\left(\frac{x}{3} \right)}$
Method #2
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = 3 - 2 x$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(\frac{x}{3} \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = -2$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = \frac{x}{3}$ .
    
    Then let $du = \frac{dx}{3}$ and substitute $3 du$ :
    
    $\int 3 \cos{\left(u \right)}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = 3 \int \cos{\left(u \right)}\, du$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $3 \sin{\left(u \right)}$
    Now substitute $u$ back in:
    
    $3 \sin{\left(\frac{x}{3} \right)}$
  Now evaluate the sub-integral.
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 6 \sin{\left(\frac{x}{3} \right)}\right)\, dx = - 6 \int \sin{\left(\frac{x}{3} \right)}\, dx$
  1. Let $u = \frac{x}{3}$ .
    
    Then let $du = \frac{dx}{3}$ and substitute $3 du$ :
    
    $\int 3 \sin{\left(u \right)}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = 3 \int \sin{\left(u \right)}\, du$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- 3 \cos{\left(u \right)}$
    Now substitute $u$ back in:
    
    $- 3 \cos{\left(\frac{x}{3} \right)}$
  So, the result is: $18 \cos{\left(\frac{x}{3} \right)}$
Method #3
1. Rewrite the integrand:
  
  $\left(3 - 2 x\right) \cos{\left(\frac{x}{3} \right)} = - 2 x \cos{\left(\frac{x}{3} \right)} + 3 \cos{\left(\frac{x}{3} \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 x \cos{\left(\frac{x}{3} \right)}\right)\, dx = - 2 \int x \cos{\left(\frac{x}{3} \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \cos{\left(\frac{x}{3} \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = \frac{x}{3}$ .
      
      Then let $du = \frac{dx}{3}$ and substitute $3 du$ :
      
      $\int 3 \cos{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = 3 \int \cos{\left(u \right)}\, du$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $3 \sin{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $3 \sin{\left(\frac{x}{3} \right)}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 3 \sin{\left(\frac{x}{3} \right)}\, dx = 3 \int \sin{\left(\frac{x}{3} \right)}\, dx$
      
      Let $u = \frac{x}{3}$ .
      
      Then let $du = \frac{dx}{3}$ and substitute $3 du$ :
      
      $\int 3 \sin{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = 3 \int \sin{\left(u \right)}\, du$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- 3 \cos{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- 3 \cos{\left(\frac{x}{3} \right)}$
      
      So, the result is: $- 9 \cos{\left(\frac{x}{3} \right)}$
    So, the result is: $- 6 x \sin{\left(\frac{x}{3} \right)} - 18 \cos{\left(\frac{x}{3} \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 3 \cos{\left(\frac{x}{3} \right)}\, dx = 3 \int \cos{\left(\frac{x}{3} \right)}\, dx$
    1. Let $u = \frac{x}{3}$ .
      
      Then let $du = \frac{dx}{3}$ and substitute $3 du$ :
      
      $\int 3 \cos{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = 3 \int \cos{\left(u \right)}\, du$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $3 \sin{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $3 \sin{\left(\frac{x}{3} \right)}$
    So, the result is: $9 \sin{\left(\frac{x}{3} \right)}$
  The result is: $- 6 x \sin{\left(\frac{x}{3} \right)} + 9 \sin{\left(\frac{x}{3} \right)} - 18 \cos{\left(\frac{x}{3} \right)}$
Add the constant of integration:

$- 6 x \sin{\left(\frac{x}{3} \right)} + 9 \sin{\left(\frac{x}{3} \right)} - 18 \cos{\left(\frac{x}{3} \right)}+ \mathrm{constant}$

The answer is:

$- 6 x \sin{\left(\frac{x}{3} \right)} + 9 \sin{\left(\frac{x}{3} \right)} - 18 \cos{\left(\frac{x}{3} \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                           
 |                                                            
 |              /x\                /x\        /x\          /x\
 | (3 - 2*x)*cos|-| dx = C - 18*cos|-| + 9*sin|-| - 6*x*sin|-|
 |              \3/                \3/        \3/          \3/
 |                                                            
/

$$\int \left(3 - 2 x\right) \cos{\left(\frac{x}{3} \right)}\, dx = C - 6 x \sin{\left(\frac{x}{3} \right)} + 9 \sin{\left(\frac{x}{3} \right)} - 18 \cos{\left(\frac{x}{3} \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

18 - 18*cos(1/3) + 3*sin(1/3)

$$- 18 \cos{\left(\frac{1}{3} \right)} + 3 \sin{\left(\frac{1}{3} \right)} + 18$$

18 - 18*cos(1/3) + 3*sin(1/3)

$$- 18 \cos{\left(\frac{1}{3} \right)} + 3 \sin{\left(\frac{1}{3} \right)} + 18$$

18 - 18*cos(1/3) + 3*sin(1/3)

Numerical answer [src]

1.97235905672318

1.97235905672318

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 (3-2x)cos(x/3)dx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3