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Integral of d{x}:
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Integral of cos^4(2x)
Identical expressions
cos^ four (2x)
co sinus of e of to the power of 4(2x)
co sinus of e of to the power of four (2x)
cos4(2x)
cos42x
cos⁴(2x)
cos^42x
cos^4(2x)dx
Similar expressions
sin2x/cos^42x

Solving integrals/ cos^4(2x)

Integral of cos^4(2x) dx

The solution

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

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

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

Detail solution

Rewrite the integrand:

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

$\frac{3 x}{8} + \frac{\sin{\left(4 x \right)}}{8} + \frac{\sin{\left(8 x \right)}}{64}+ \mathrm{constant}$

The answer is:

$\frac{3 x}{8} + \frac{\sin{\left(4 x \right)}}{8} + \frac{\sin{\left(8 x \right)}}{64}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                            
 |                                             
 |    4               sin(4*x)   sin(8*x)   3*x
 | cos (2*x) dx = C + -------- + -------- + ---
 |                       8          64       8 
/

$$\int \cos^{4}{\left(2 x \right)}\, dx = C + \frac{3 x}{8} + \frac{\sin{\left(4 x \right)}}{8} + \frac{\sin{\left(8 x \right)}}{64}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

       3                            
3   cos (2)*sin(2)   3*cos(2)*sin(2)
- + -------------- + ---------------
8         8                 16

$$\frac{3 \sin{\left(2 \right)} \cos{\left(2 \right)}}{16} + \frac{\sin{\left(2 \right)} \cos^{3}{\left(2 \right)}}{8} + \frac{3}{8}$$

       3                            
3   cos (2)*sin(2)   3*cos(2)*sin(2)
- + -------------- + ---------------
8         8                 16

$$\frac{3 \sin{\left(2 \right)} \cos{\left(2 \right)}}{16} + \frac{\sin{\left(2 \right)} \cos^{3}{\left(2 \right)}}{8} + \frac{3}{8}$$

3/8 + cos(2)^3*sin(2)/8 + 3*cos(2)*sin(2)/16

Numerical answer [src]

0.295858410689999

0.295858410689999

The graph

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

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Identical expressions

Similar expressions

Integral of cos^4(2x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2