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sin^2(2x)cos^2(2x)dx

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

The solution

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  p                       
  -                       
  2                       
  /                       
 |                        
 |     2         2        
 |  sin (2*x)*cos (2*x) dx
 |                        
/                         
0

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

Integral(sin(2*x)^2*cos(2*x)^2, (x, 0, p/2))

Detail solution

Rewrite the integrand:

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

$\frac{x}{8} - \frac{\sin{\left(8 x \right)}}{64}+ \mathrm{constant}$

The answer is:

$\frac{x}{8} - \frac{\sin{\left(8 x \right)}}{64}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                         
 |                                          
 |    2         2               sin(8*x)   x
 | sin (2*x)*cos (2*x) dx = C - -------- + -
 |                                 64      8
/

$$\int \sin^{2}{\left(2 x \right)} \cos^{2}{\left(2 x \right)}\, dx = C + \frac{x}{8} - \frac{\sin{\left(8 x \right)}}{64}$$

Simplify

The answer [src]

p    cos(2*p)*sin(2*p)
-- - -----------------
16           32

$$\frac{p}{16} - \frac{\sin{\left(2 p \right)} \cos{\left(2 p \right)}}{32}$$

p    cos(2*p)*sin(2*p)
-- - -----------------
16           32

$$\frac{p}{16} - \frac{\sin{\left(2 p \right)} \cos{\left(2 p \right)}}{32}$$

p/16 - cos(2*p)*sin(2*p)/32

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

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Identical expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3