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(sin(x/ four))^ three
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(sin(x/4))^3dx

Integral of (sin(x/4))^3 dx

The solution

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 2*p          
  /           
 |            
 |     3/x\   
 |  sin |-| dx
 |      \4/   
 |            
/             
0

$$\int\limits_{0}^{2 p} \sin^{3}{\left(\frac{x}{4} \right)}\, dx$$

Integral(sin(x/4)^3, (x, 0, 2*p))

Detail solution

Rewrite the integrand:

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

$- 3 \cos{\left(\frac{x}{4} \right)} + \frac{\cos{\left(\frac{3 x}{4} \right)}}{3}$
Add the constant of integration:

$- 3 \cos{\left(\frac{x}{4} \right)} + \frac{\cos{\left(\frac{3 x}{4} \right)}}{3}+ \mathrm{constant}$

The answer is:

$- 3 \cos{\left(\frac{x}{4} \right)} + \frac{\cos{\left(\frac{3 x}{4} \right)}}{3}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                 3/x\
 |                             4*cos |-|
 |    3/x\               /x\         \4/
 | sin |-| dx = C - 4*cos|-| + ---------
 |     \4/               \4/       3    
 |                                      
/

$$\int \sin^{3}{\left(\frac{x}{4} \right)}\, dx = C + \frac{4 \cos^{3}{\left(\frac{x}{4} \right)}}{3} - 4 \cos{\left(\frac{x}{4} \right)}$$

Simplify

The answer [src]

                    3/p\
               4*cos |-|
8        /p\         \2/
- - 4*cos|-| + ---------
3        \2/       3

$$\frac{4 \cos^{3}{\left(\frac{p}{2} \right)}}{3} - 4 \cos{\left(\frac{p}{2} \right)} + \frac{8}{3}$$

                    3/p\
               4*cos |-|
8        /p\         \2/
- - 4*cos|-| + ---------
3        \2/       3

$$\frac{4 \cos^{3}{\left(\frac{p}{2} \right)}}{3} - 4 \cos{\left(\frac{p}{2} \right)} + \frac{8}{3}$$

8/3 - 4*cos(p/2) + 4*cos(p/2)^3/3

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

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How to use it?

Identical expressions

Integral of (sin(x/4))^3 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3