Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of 4
Integral of tg
Integral of sin^3(4x)
Integral of (2x^2-4x+5)dx
Derivative of:
sin^3(4x)
Identical expressions
sin^ three (4x)
sinus of cubed (4x)
sinus of to the power of three (4x)
sin3(4x)
sin34x
sin³(4x)
sin to the power of 3(4x)
sin^34x
sin^3(4x)dx
Similar expressions
sin^34x×cos7x
cos4x/sin^34x

Solving integrals/ sin^3(4x)

Integral of sin^3(4x) dx

The solution

You have entered [src]

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

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

Integral(sin(4*x)^3, (x, 0, 1))

Detail solution

Rewrite the integrand:

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

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

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

The answer is:

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

The answer (Indefinite) [src]

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

$$\int \sin^{3}{\left(4 x \right)}\, dx = C + \frac{\cos^{3}{\left(4 x \right)}}{12} - \frac{\cos{\left(4 x \right)}}{4}$$

Simplify

The answer [src]

                3   
1   cos(4)   cos (4)
- - ------ + -------
6     4         12

$$\frac{\cos^{3}{\left(4 \right)}}{12} - \frac{\cos{\left(4 \right)}}{4} + \frac{1}{6}$$

                3   
1   cos(4)   cos (4)
- - ------ + -------
6     4         12

$$\frac{\cos^{3}{\left(4 \right)}}{12} - \frac{\cos{\left(4 \right)}}{4} + \frac{1}{6}$$

1/6 - cos(4)/4 + cos(4)^3/12

Numerical answer [src]

0.306805136385521

0.306805136385521

The graph

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of sin^3(4x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3