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Similar expressions
(1+x)sin(x/4)dx

Integral of (1-x)sin(x/4)dx dx

The solution

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  1                  
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 |             /x\   
 |  (1 - x)*sin|-| dx
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0

$$\int\limits_{0}^{1} \left(1 - x\right) \sin{\left(\frac{x}{4} \right)}\, dx$$

Integral((1 - x)*sin(x/4), (x, 0, 1))

Detail solution

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

4 - 16*sin(1/4)

$$4 - 16 \sin{\left(\frac{1}{4} \right)}$$

4 - 16*sin(1/4)

$$4 - 16 \sin{\left(\frac{1}{4} \right)}$$

4 - 16*sin(1/4)

Numerical answer [src]

0.0415366519276331

0.0415366519276331

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

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Identical expressions

Similar expressions

Integral of (1-x)sin(x/4)dx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Method #4