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Integral of d{x}:
Integral of e^(-4x^2)
Integral of cos^3(4x)
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Integral of (4x^3-6x^2-4x+3)dx
Derivative of:
cos^3(4x)
Identical expressions
cos^ three (4x)
co sinus of e of cubed (4x)
co sinus of e of to the power of three (4x)
cos3(4x)
cos34x
cos³(4x)
cos to the power of 3(4x)
cos^34x
cos^3(4x)dx

Solving integrals/ cos^3(4x)

Integral of cos^3(4x) dx

The solution

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  1             
  /             
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 |     3        
 |  cos (4*x) dx
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0

$$\int\limits_{0}^{1} \cos^{3}{\left(4 x \right)}\, dx$$

Simplify

Detail solution

Rewrite the integrand:

$\cos^{3}{\left(4 x \right)} = \left(1 - \sin^{2}{\left(4 x \right)}\right) \cos{\left(4 x \right)}$
There are multiple ways to do this integral.
Method #1
1. Let $u = \sin{\left(4 x \right)}$ .
  
  Then let $du = 4 \cos{\left(4 x \right)} dx$ and substitute $du$ :
  
  $\int \left(\frac{1}{4} - \frac{u^{2}}{4}\right)\, du$
  1. Integrate term-by-term:
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{4}\, du = \frac{u}{4}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{u^{2}}{4}\right)\, du = - \frac{\int u^{2}\, du}{4}$
      
      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: $- \frac{u^{3}}{12}$
    The result is: $- \frac{u^{3}}{12} + \frac{u}{4}$
  Now substitute $u$ back in:
  
  $- \frac{\sin^{3}{\left(4 x \right)}}{12} + \frac{\sin{\left(4 x \right)}}{4}$
Method #2
1. Rewrite the integrand:
  
  $\left(1 - \sin^{2}{\left(4 x \right)}\right) \cos{\left(4 x \right)} = - \sin^{2}{\left(4 x \right)} \cos{\left(4 x \right)} + \cos{\left(4 x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \sin^{2}{\left(4 x \right)} \cos{\left(4 x \right)}\right)\, dx = - \int \sin^{2}{\left(4 x \right)} \cos{\left(4 x \right)}\, dx$
    1. Let $u = \sin{\left(4 x \right)}$ .
      
      Then let $du = 4 \cos{\left(4 x \right)} dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{u^{2}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{2}}{4}\, du = \frac{\int u^{2}\, du}{4}$
      
      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: $\frac{u^{3}}{12}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{3}{\left(4 x \right)}}{12}$
    So, the result is: $- \frac{\sin^{3}{\left(4 x \right)}}{12}$
  1. Let $u = 4 x$ .
    
    Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
    
    $\int \frac{\cos{\left(u \right)}}{16}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
    Now substitute $u$ back in:
    
    $\frac{\sin{\left(4 x \right)}}{4}$
  The result is: $- \frac{\sin^{3}{\left(4 x \right)}}{12} + \frac{\sin{\left(4 x \right)}}{4}$
Method #3
1. Rewrite the integrand:
  
  $\left(1 - \sin^{2}{\left(4 x \right)}\right) \cos{\left(4 x \right)} = - \sin^{2}{\left(4 x \right)} \cos{\left(4 x \right)} + \cos{\left(4 x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \sin^{2}{\left(4 x \right)} \cos{\left(4 x \right)}\right)\, dx = - \int \sin^{2}{\left(4 x \right)} \cos{\left(4 x \right)}\, dx$
    1. Let $u = \sin{\left(4 x \right)}$ .
      
      Then let $du = 4 \cos{\left(4 x \right)} dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{u^{2}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{2}}{4}\, du = \frac{\int u^{2}\, du}{4}$
      
      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: $\frac{u^{3}}{12}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{3}{\left(4 x \right)}}{12}$
    So, the result is: $- \frac{\sin^{3}{\left(4 x \right)}}{12}$
  1. Let $u = 4 x$ .
    
    Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
    
    $\int \frac{\cos{\left(u \right)}}{16}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
    Now substitute $u$ back in:
    
    $\frac{\sin{\left(4 x \right)}}{4}$
  The result is: $- \frac{\sin^{3}{\left(4 x \right)}}{12} + \frac{\sin{\left(4 x \right)}}{4}$
Now simplify:

$\frac{\left(3 - \sin^{2}{\left(4 x \right)}\right) \sin{\left(4 x \right)}}{12}$
Add the constant of integration:

$\frac{\left(3 - \sin^{2}{\left(4 x \right)}\right) \sin{\left(4 x \right)}}{12}+ \mathrm{constant}$

The answer is:

$\frac{\left(3 - \sin^{2}{\left(4 x \right)}\right) \sin{\left(4 x \right)}}{12}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                       
 |                       3                
 |    3               sin (4*x)   sin(4*x)
 | cos (4*x) dx = C - --------- + --------
 |                        12         4    
/

$${{\sin \left(4\,x\right)-{{\sin ^3\left(4\,x\right)}\over{3}} }\over{4}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

     3            
  sin (4)   sin(4)
- ------- + ------
     12       4

$$-{{\sin ^34-3\,\sin 4}\over{12}}$$

     3            
  sin (4)   sin(4)
- ------- + ------
     12       4

$$\frac{\sin{\left(4 \right)}}{4} - \frac{\sin^{3}{\left(4 \right)}}{12}$$

Simplify

Numerical answer [src]

-0.153079070328579

-0.153079070328579

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of cos^3(4x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3