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Integral of d{x}:
Integral of sin^3(6x)
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Identical expressions
sin^ three (6x)
sinus of cubed (6x)
sinus of to the power of three (6x)
sin3(6x)
sin36x
sin³(6x)
sin to the power of 3(6x)
sin^36x
sin^3(6x)dx
Similar expressions
(sin^3)*((6x)*cos6x)

Solving integrals/ sin^3(6x)

Integral of sin^3(6x) dx

The solution

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  1             
  /             
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 |     3        
 |  sin (6*x) dx
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0

$$\int\limits_{0}^{1} \sin^{3}{\left(6 x \right)}\, dx$$

Integral(sin(6*x)^3, (x, 0, 1))

Detail solution

Rewrite the integrand:

$\sin^{3}{\left(6 x \right)} = \left(1 - \cos^{2}{\left(6 x \right)}\right) \sin{\left(6 x \right)}$
There are multiple ways to do this integral.
Method #1
1. Let $u = \cos{\left(6 x \right)}$ .
  
  Then let $du = - 6 \sin{\left(6 x \right)} dx$ and substitute $du$ :
  
  $\int \left(\frac{u^{2}}{6} - \frac{1}{6}\right)\, du$
  1. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{2}}{6}\, du = \frac{\int u^{2}\, du}{6}$
      
      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: $\frac{u^{3}}{18}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \left(- \frac{1}{6}\right)\, du = - \frac{u}{6}$
    The result is: $\frac{u^{3}}{18} - \frac{u}{6}$
  Now substitute $u$ back in:
  
  $\frac{\cos^{3}{\left(6 x \right)}}{18} - \frac{\cos{\left(6 x \right)}}{6}$
Method #2
1. Rewrite the integrand:
  
  $\left(1 - \cos^{2}{\left(6 x \right)}\right) \sin{\left(6 x \right)} = - \sin{\left(6 x \right)} \cos^{2}{\left(6 x \right)} + \sin{\left(6 x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \sin{\left(6 x \right)} \cos^{2}{\left(6 x \right)}\right)\, dx = - \int \sin{\left(6 x \right)} \cos^{2}{\left(6 x \right)}\, dx$
    1. Let $u = \cos{\left(6 x \right)}$ .
      
      Then let $du = - 6 \sin{\left(6 x \right)} dx$ and substitute $- \frac{du}{6}$ :
      
      $\int \frac{u^{2}}{36}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{u^{2}}{6}\right)\, du = - \frac{\int u^{2}\, du}{6}$
      
      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: $- \frac{u^{3}}{18}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{3}{\left(6 x \right)}}{18}$
    So, the result is: $\frac{\cos^{3}{\left(6 x \right)}}{18}$
  1. Let $u = 6 x$ .
    
    Then let $du = 6 dx$ and substitute $\frac{du}{6}$ :
    
    $\int \frac{\sin{\left(u \right)}}{36}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{6}\, du = \frac{\int \sin{\left(u \right)}\, du}{6}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{6}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(6 x \right)}}{6}$
  The result is: $\frac{\cos^{3}{\left(6 x \right)}}{18} - \frac{\cos{\left(6 x \right)}}{6}$
Method #3
1. Rewrite the integrand:
  
  $\left(1 - \cos^{2}{\left(6 x \right)}\right) \sin{\left(6 x \right)} = - \sin{\left(6 x \right)} \cos^{2}{\left(6 x \right)} + \sin{\left(6 x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \sin{\left(6 x \right)} \cos^{2}{\left(6 x \right)}\right)\, dx = - \int \sin{\left(6 x \right)} \cos^{2}{\left(6 x \right)}\, dx$
    1. Let $u = \cos{\left(6 x \right)}$ .
      
      Then let $du = - 6 \sin{\left(6 x \right)} dx$ and substitute $- \frac{du}{6}$ :
      
      $\int \frac{u^{2}}{36}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{u^{2}}{6}\right)\, du = - \frac{\int u^{2}\, du}{6}$
      
      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: $- \frac{u^{3}}{18}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{3}{\left(6 x \right)}}{18}$
    So, the result is: $\frac{\cos^{3}{\left(6 x \right)}}{18}$
  1. Let $u = 6 x$ .
    
    Then let $du = 6 dx$ and substitute $\frac{du}{6}$ :
    
    $\int \frac{\sin{\left(u \right)}}{36}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{6}\, du = \frac{\int \sin{\left(u \right)}\, du}{6}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{6}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(6 x \right)}}{6}$
  The result is: $\frac{\cos^{3}{\left(6 x \right)}}{18} - \frac{\cos{\left(6 x \right)}}{6}$
Now simplify:

$\frac{\left(\cos^{2}{\left(6 x \right)} - 3\right) \cos{\left(6 x \right)}}{18}$
Add the constant of integration:

$\frac{\left(\cos^{2}{\left(6 x \right)} - 3\right) \cos{\left(6 x \right)}}{18}+ \mathrm{constant}$

The answer is:

$\frac{\left(\cos^{2}{\left(6 x \right)} - 3\right) \cos{\left(6 x \right)}}{18}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

$${{{{\cos ^3\left(6\,x\right)}\over{3}}-\cos \left(6\,x\right) }\over{6}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                3   
1   cos(6)   cos (6)
- - ------ + -------
9     6         18

$${{\cos ^36-3\,\cos 6}\over{18}}+{{1}\over{9}}$$

                3   
1   cos(6)   cos (6)
- - ------ + -------
9     6         18

$$- \frac{\cos{\left(6 \right)}}{6} + \frac{\cos^{3}{\left(6 \right)}}{18} + \frac{1}{9}$$

Simplify

Numerical answer [src]

0.000260890672094249

0.000260890672094249

The graph

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

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Identical expressions

Similar expressions

Integral of sin^3(6x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3