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(sin(two *x))^ four *cos(two *x)^ two
( sinus of (2 multiply by x)) to the power of 4 multiply by co sinus of e of (2 multiply by x) squared
( sinus of (two multiply by x)) to the power of four multiply by co sinus of e of (two multiply by x) to the power of two
(sin(2*x))4*cos(2*x)2
sin2*x4*cos2*x2
(sin(2*x))⁴*cos(2*x)²
(sin(2*x)) to the power of 4*cos(2*x) to the power of 2
(sin(2x))^4cos(2x)^2
(sin(2x))4cos(2x)2
sin2x4cos2x2
sin2x^4cos2x^2
(sin(2*x))^4*cos(2*x)^2dx

Solving integrals/ (sin(2*x))^4*cos(2*x)^2

Integral of (sin(2x))^4cos(2*x)^2 dx

The solution

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  1                       
  /                       
 |                        
 |     4         2        
 |  sin (2*x)*cos (2*x) dx
 |                        
/                         
0

$$\int\limits_{0}^{1} \sin^{4}{\left(2 x \right)} \cos^{2}{\left(2 x \right)}\, dx$$

Integral(sin(2*x)^4*cos(2*x)^2, (x, 0, 1))

Detail solution

Rewrite the integrand:

$\sin^{4}{\left(2 x \right)} \cos^{2}{\left(2 x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(4 x \right)}}{2}\right)^{2} \left(\frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}\right)$
There are multiple ways to do this integral.
Method #1
1. Let $u = 4 x$ .
  
  Then let $du = 4 dx$ and substitute $du$ :
  
  $\int \left(\frac{\cos^{3}{\left(u \right)}}{32} - \frac{\cos^{2}{\left(u \right)}}{32} - \frac{\cos{\left(u \right)}}{32} + \frac{1}{32}\right)\, du$
  1. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{3}{\left(u \right)}}{32}\, du = \frac{\int \cos^{3}{\left(u \right)}\, du}{32}$
      
      Rewrite the integrand:
      
      $\cos^{3}{\left(u \right)} = \left(1 - \sin^{2}{\left(u \right)}\right) \cos{\left(u \right)}$
      
      Let $u = \sin{\left(u \right)}$ .
      
      Then let $du = \cos{\left(u \right)} du$ and substitute $du$ :
      
      $\int \left(1 - u^{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{2}\right)\, du = - \int u^{2}\, du$
      
      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: $- \frac{u^{3}}{3}$
      
      The result is: $- \frac{u^{3}}{3} + u$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(u \right)}}{3} + \sin{\left(u \right)}$
      
      So, the result is: $- \frac{\sin^{3}{\left(u \right)}}{96} + \frac{\sin{\left(u \right)}}{32}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{2}{\left(u \right)}}{32}\right)\, du = - \frac{\int \cos^{2}{\left(u \right)}\, du}{32}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(u \right)} = \frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du$
      
      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:
      
      $\frac{\sin{\left(2 u \right)}}{2}$
      
      So, the result is: $\frac{\sin{\left(2 u \right)}}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The result is: $\frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}$
      
      So, the result is: $- \frac{u}{64} - \frac{\sin{\left(2 u \right)}}{128}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(u \right)}}{32}\right)\, du = - \frac{\int \cos{\left(u \right)}\, du}{32}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $- \frac{\sin{\left(u \right)}}{32}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{32}\, du = \frac{u}{32}$
    The result is: $\frac{u}{64} - \frac{\sin{\left(2 u \right)}}{128} - \frac{\sin^{3}{\left(u \right)}}{96}$
  Now substitute $u$ back in:
  
  $\frac{x}{16} - \frac{\sin^{3}{\left(4 x \right)}}{96} - \frac{\sin{\left(8 x \right)}}{128}$
Method #2
1. Rewrite the integrand:
  
  $\left(\frac{1}{2} - \frac{\cos{\left(4 x \right)}}{2}\right)^{2} \left(\frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}\right) = \frac{\cos^{3}{\left(4 x \right)}}{8} - \frac{\cos^{2}{\left(4 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{8} + \frac{1}{8}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\cos^{3}{\left(4 x \right)}}{8}\, dx = \frac{\int \cos^{3}{\left(4 x \right)}\, dx}{8}$
    1. Rewrite the integrand:
      
      $\cos^{3}{\left(4 x \right)} = \left(1 - \sin^{2}{\left(4 x \right)}\right) \cos{\left(4 x \right)}$
    2. Let $u = \sin{\left(4 x \right)}$ .
      
      Then let $du = 4 \cos{\left(4 x \right)} dx$ and substitute $du$ :
      
      $\int \left(\frac{1}{4} - \frac{u^{2}}{4}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{4}\, du = \frac{u}{4}$
      
      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:
      
      $- \frac{\sin^{3}{\left(4 x \right)}}{12} + \frac{\sin{\left(4 x \right)}}{4}$
    So, the result is: $- \frac{\sin^{3}{\left(4 x \right)}}{96} + \frac{\sin{\left(4 x \right)}}{32}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos^{2}{\left(4 x \right)}}{8}\right)\, dx = - \frac{\int \cos^{2}{\left(4 x \right)}\, dx}{8}$
    1. Rewrite the integrand:
      
      $\cos^{2}{\left(4 x \right)} = \frac{\cos{\left(8 x \right)}}{2} + \frac{1}{2}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(8 x \right)}}{2}\, dx = \frac{\int \cos{\left(8 x \right)}\, dx}{2}$
      
      Let $u = 8 x$ .
      
      Then let $du = 8 dx$ and substitute $\frac{du}{8}$ :
      
      $\int \frac{\cos{\left(u \right)}}{8}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{8}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{8}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(8 x \right)}}{8}$
      
      So, the result is: $\frac{\sin{\left(8 x \right)}}{16}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{x}{2} + \frac{\sin{\left(8 x \right)}}{16}$
    So, the result is: $- \frac{x}{16} - \frac{\sin{\left(8 x \right)}}{128}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos{\left(4 x \right)}}{8}\right)\, dx = - \frac{\int \cos{\left(4 x \right)}\, dx}{8}$
    1. Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      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:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
    So, the result is: $- \frac{\sin{\left(4 x \right)}}{32}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \frac{1}{8}\, dx = \frac{x}{8}$
  The result is: $\frac{x}{16} - \frac{\sin^{3}{\left(4 x \right)}}{96} - \frac{\sin{\left(8 x \right)}}{128}$
Method #3
1. Rewrite the integrand:
  
  $\left(\frac{1}{2} - \frac{\cos{\left(4 x \right)}}{2}\right)^{2} \left(\frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}\right) = \frac{\cos^{3}{\left(4 x \right)}}{8} - \frac{\cos^{2}{\left(4 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{8} + \frac{1}{8}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\cos^{3}{\left(4 x \right)}}{8}\, dx = \frac{\int \cos^{3}{\left(4 x \right)}\, dx}{8}$
    1. Rewrite the integrand:
      
      $\cos^{3}{\left(4 x \right)} = \left(1 - \sin^{2}{\left(4 x \right)}\right) \cos{\left(4 x \right)}$
    2. Let $u = \sin{\left(4 x \right)}$ .
      
      Then let $du = 4 \cos{\left(4 x \right)} dx$ and substitute $du$ :
      
      $\int \left(\frac{1}{4} - \frac{u^{2}}{4}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{4}\, du = \frac{u}{4}$
      
      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:
      
      $- \frac{\sin^{3}{\left(4 x \right)}}{12} + \frac{\sin{\left(4 x \right)}}{4}$
    So, the result is: $- \frac{\sin^{3}{\left(4 x \right)}}{96} + \frac{\sin{\left(4 x \right)}}{32}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos^{2}{\left(4 x \right)}}{8}\right)\, dx = - \frac{\int \cos^{2}{\left(4 x \right)}\, dx}{8}$
    1. Rewrite the integrand:
      
      $\cos^{2}{\left(4 x \right)} = \frac{\cos{\left(8 x \right)}}{2} + \frac{1}{2}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(8 x \right)}}{2}\, dx = \frac{\int \cos{\left(8 x \right)}\, dx}{2}$
      
      Let $u = 8 x$ .
      
      Then let $du = 8 dx$ and substitute $\frac{du}{8}$ :
      
      $\int \frac{\cos{\left(u \right)}}{8}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{8}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{8}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(8 x \right)}}{8}$
      
      So, the result is: $\frac{\sin{\left(8 x \right)}}{16}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{x}{2} + \frac{\sin{\left(8 x \right)}}{16}$
    So, the result is: $- \frac{x}{16} - \frac{\sin{\left(8 x \right)}}{128}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos{\left(4 x \right)}}{8}\right)\, dx = - \frac{\int \cos{\left(4 x \right)}\, dx}{8}$
    1. Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      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:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
    So, the result is: $- \frac{\sin{\left(4 x \right)}}{32}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \frac{1}{8}\, dx = \frac{x}{8}$
  The result is: $\frac{x}{16} - \frac{\sin^{3}{\left(4 x \right)}}{96} - \frac{\sin{\left(8 x \right)}}{128}$
Add the constant of integration:

$\frac{x}{16} - \frac{\sin^{3}{\left(4 x \right)}}{96} - \frac{\sin{\left(8 x \right)}}{128}+ \mathrm{constant}$

The answer is:

$\frac{x}{16} - \frac{\sin^{3}{\left(4 x \right)}}{96} - \frac{\sin{\left(8 x \right)}}{128}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                      
 |                                 3                     
 |    4         2               sin (4*x)   sin(8*x)   x 
 | sin (2*x)*cos (2*x) dx = C - --------- - -------- + --
 |                                  96        128      16
/

$$\int \sin^{4}{\left(2 x \right)} \cos^{2}{\left(2 x \right)}\, dx = C + \frac{x}{16} - \frac{\sin^{3}{\left(4 x \right)}}{96} - \frac{\sin{\left(8 x \right)}}{128}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                        3                5          
1    cos(2)*sin(2)   sin (2)*cos(2)   sin (2)*cos(2)
-- - ------------- - -------------- + --------------
16         32              48               12

$$\frac{\sin^{5}{\left(2 \right)} \cos{\left(2 \right)}}{12} - \frac{\sin^{3}{\left(2 \right)} \cos{\left(2 \right)}}{48} - \frac{\sin{\left(2 \right)} \cos{\left(2 \right)}}{32} + \frac{1}{16}$$

                        3                5          
1    cos(2)*sin(2)   sin (2)*cos(2)   sin (2)*cos(2)
-- - ------------- - -------------- + --------------
16         32              48               12

$$\frac{\sin^{5}{\left(2 \right)} \cos{\left(2 \right)}}{12} - \frac{\sin^{3}{\left(2 \right)} \cos{\left(2 \right)}}{48} - \frac{\sin{\left(2 \right)} \cos{\left(2 \right)}}{32} + \frac{1}{16}$$

1/16 - cos(2)*sin(2)/32 - sin(2)^3*cos(2)/48 + sin(2)^5*cos(2)/12

Numerical answer [src]

0.0592858328855552

0.0592858328855552

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

Other calculators

How to use it?

Identical expressions

Integral of (sin(2x))^4cos(2*x)^2 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Other calculators

How to use it?

Identical expressions

Integral of (sin(2*x))^4*cos(2*x)^2 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Integral of (sin(2x))^4cos(2*x)^2 dx