Integral of cosxcos4xdx dx

The solution

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

Simplify

Detail solution

Rewrite the integrand:

$\cos{\left(x \right)} \cos{\left(4 x \right)} 1 = 8 \cos^{5}{\left(x \right)} - 8 \cos^{3}{\left(x \right)} + \cos{\left(x \right)}$
Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 8 \cos^{5}{\left(x \right)}\, dx = 8 \int \cos^{5}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\cos^{5}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right)^{2} \cos{\left(x \right)}$
  2. There are multiple ways to do this integral.
    Method #1
    1. Rewrite the integrand:
      
      $\left(1 - \sin^{2}{\left(x \right)}\right)^{2} \cos{\left(x \right)} = \sin^{4}{\left(x \right)} \cos{\left(x \right)} - 2 \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos{\left(x \right)}$
    2. Integrate term-by-term:
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u^{4}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{5}{\left(x \right)}}{5}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 2 \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u^{2}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{3}{\left(x \right)}}{3}$
      
      So, the result is: $- \frac{2 \sin^{3}{\left(x \right)}}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$
      
      The result is: $\frac{\sin^{5}{\left(x \right)}}{5} - \frac{2 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}$
    Method #2
    1. Rewrite the integrand:
      
      $\left(1 - \sin^{2}{\left(x \right)}\right)^{2} \cos{\left(x \right)} = \sin^{4}{\left(x \right)} \cos{\left(x \right)} - 2 \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos{\left(x \right)}$
    2. Integrate term-by-term:
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u^{4}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{5}{\left(x \right)}}{5}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 2 \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u^{2}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{3}{\left(x \right)}}{3}$
      
      So, the result is: $- \frac{2 \sin^{3}{\left(x \right)}}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$
      
      The result is: $\frac{\sin^{5}{\left(x \right)}}{5} - \frac{2 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}$
  So, the result is: $\frac{8 \sin^{5}{\left(x \right)}}{5} - \frac{16 \sin^{3}{\left(x \right)}}{3} + 8 \sin{\left(x \right)}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 8 \cos^{3}{\left(x \right)}\right)\, dx = - 8 \int \cos^{3}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\cos^{3}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)}$
  2. Let $u = \sin{\left(x \right)}$ .
    
    Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
    
    $\int \left(1 - u^{2}\right)\, du$
    1. Integrate term-by-term:
      1. The integral of a constant is the constant times the variable of integration:
        
        $\int 1\, du = u$
      1. 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:
    
    $- \frac{\sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}$
  So, the result is: $\frac{8 \sin^{3}{\left(x \right)}}{3} - 8 \sin{\left(x \right)}$
1. The integral of cosine is sine:
  
  $\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$
The result is: $\frac{8 \sin^{5}{\left(x \right)}}{5} - \frac{8 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}$
Now simplify:

$\frac{\left(24 \sin^{4}{\left(x \right)} - 40 \sin^{2}{\left(x \right)} + 15\right) \sin{\left(x \right)}}{15}$
Add the constant of integration:

$\frac{\left(24 \sin^{4}{\left(x \right)} - 40 \sin^{2}{\left(x \right)} + 15\right) \sin{\left(x \right)}}{15}+ \mathrm{constant}$

The answer is:

$\frac{\left(24 \sin^{4}{\left(x \right)} - 40 \sin^{2}{\left(x \right)} + 15\right) \sin{\left(x \right)}}{15}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                3           5            
 |                            8*sin (x)   8*sin (x)         
 | cos(x)*cos(4*x)*1 dx = C - --------- + --------- + sin(x)
 |                                3           5             
/

$${{\sin \left(5\,x\right)}\over{10}}+{{\sin \left(3\,x\right)}\over{ 6}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

  cos(4)*sin(1)   4*cos(1)*sin(4)
- ------------- + ---------------
        15               15

$${{3\,\sin 5+5\,\sin 3}\over{30}}$$

  cos(4)*sin(1)   4*cos(1)*sin(4)
- ------------- + ---------------
        15               15

$$\frac{4 \sin{\left(4 \right)} \cos{\left(1 \right)}}{15} - \frac{\sin{\left(1 \right)} \cos{\left(4 \right)}}{15}$$

Simplify

Numerical answer [src]

-0.0723724261230026

-0.0723724261230026

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of cosxcos4xdx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2