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Derivative of:
cos^3(2x)
Identical expressions
cos^ three (2x)
co sinus of e of cubed (2x)
co sinus of e of to the power of three (2x)
cos3(2x)
cos32x
cos³(2x)
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cos^32x
cos^3(2x)dx
Similar expressions
scrt(cos^32x)*sin2x

Solving integrals/ cos^3(2x)

Integral of cos^3(2x) dx

The solution

You have entered [src]

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

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

Integral(cos(2*x)^3, (x, 0, 1))

Detail solution

Rewrite the integrand:

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

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

            3   
sin(2)   sin (2)
------ - -------
  2         6

$$- \frac{\sin^{3}{\left(2 \right)}}{6} + \frac{\sin{\left(2 \right)}}{2}$$

            3   
sin(2)   sin (2)
------ - -------
  2         6

$$- \frac{\sin^{3}{\left(2 \right)}}{6} + \frac{\sin{\left(2 \right)}}{2}$$

sin(2)/2 - sin(2)^3/6

Numerical answer [src]

0.329344222634675

0.329344222634675

The graph

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

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Identical expressions

Similar expressions

Integral of cos^3(2x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3