Integral of cosxcos5x dx

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 2*pi                  
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  |  cos(x)*cos(5*x) dx
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 0

$$\int\limits_{0}^{2 \pi} \cos{\left(x \right)} \cos{\left(5 x \right)}\, dx$$

Integral(cos(x)*cos(5*x), (x, 0, 2*pi))

Detail solution

Rewrite the integrand:

$\cos{\left(x \right)} \cos{\left(5 x \right)} = 16 \cos^{6}{\left(x \right)} - 20 \cos^{4}{\left(x \right)} + 5 \cos^{2}{\left(x \right)}$
Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 16 \cos^{6}{\left(x \right)}\, dx = 16 \int \cos^{6}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\cos^{6}{\left(x \right)} = \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{3}$
  2. There are multiple ways to do this integral.
    Method #1
    1. Rewrite the integrand:
      
      $\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{3} = \frac{\cos^{3}{\left(2 x \right)}}{8} + \frac{3 \cos^{2}{\left(2 x \right)}}{8} + \frac{3 \cos{\left(2 x \right)}}{8} + \frac{1}{8}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{3}{\left(2 x \right)}}{8}\, dx = \frac{\int \cos^{3}{\left(2 x \right)}\, dx}{8}$
      
      Rewrite the integrand:
      
      $\cos^{3}{\left(2 x \right)} = \left(1 - \sin^{2}{\left(2 x \right)}\right) \cos{\left(2 x \right)}$
      
      There are multiple ways to do this integral.
      
      Method #1
      
      Let $u = \sin{\left(2 x \right)}$ .
      
      Then let $du = 2 \cos{\left(2 x \right)} dx$ and substitute $du$ :
      
      $\int \left(\frac{1}{2} - \frac{u^{2}}{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{u^{2}}{2}\right)\, du = - \frac{\int u^{2}\, du}{2}$
      
      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: $- \frac{u^{3}}{6}$
      
      The result is: $- \frac{u^{3}}{6} + \frac{u}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(2 x \right)}}{2}$
      
      Method #2
      
      Rewrite the integrand:
      
      $\left(1 - \sin^{2}{\left(2 x \right)}\right) \cos{\left(2 x \right)} = - \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)} + \cos{\left(2 x \right)}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)}\right)\, dx = - \int \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)}\, dx$
      
      Let $u = \sin{\left(2 x \right)}$ .
      
      Then let $du = 2 \cos{\left(2 x \right)} dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{u^{2}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{2}}{2}\, du = \frac{\int u^{2}\, du}{2}$
      
      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: $\frac{u^{3}}{6}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{3}{\left(2 x \right)}}{6}$
      
      So, the result is: $- \frac{\sin^{3}{\left(2 x \right)}}{6}$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      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:
      
      $\frac{\sin{\left(2 x \right)}}{2}$
      
      The result is: $- \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(2 x \right)}}{2}$
      
      Method #3
      
      Rewrite the integrand:
      
      $\left(1 - \sin^{2}{\left(2 x \right)}\right) \cos{\left(2 x \right)} = - \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)} + \cos{\left(2 x \right)}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)}\right)\, dx = - \int \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)}\, dx$
      
      Let $u = \sin{\left(2 x \right)}$ .
      
      Then let $du = 2 \cos{\left(2 x \right)} dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{u^{2}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{2}}{2}\, du = \frac{\int u^{2}\, du}{2}$
      
      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: $\frac{u^{3}}{6}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{3}{\left(2 x \right)}}{6}$
      
      So, the result is: $- \frac{\sin^{3}{\left(2 x \right)}}{6}$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      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:
      
      $\frac{\sin{\left(2 x \right)}}{2}$
      
      The result is: $- \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(2 x \right)}}{2}$
      
      So, the result is: $- \frac{\sin^{3}{\left(2 x \right)}}{48} + \frac{\sin{\left(2 x \right)}}{16}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{3 \cos^{2}{\left(2 x \right)}}{8}\, dx = \frac{3 \int \cos^{2}{\left(2 x \right)}\, dx}{8}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(2 x \right)} = \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(4 x \right)}}{2}\, dx = \frac{\int \cos{\left(4 x \right)}\, dx}{2}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      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: $\frac{\sin{\left(4 x \right)}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{x}{2} + \frac{\sin{\left(4 x \right)}}{8}$
      
      So, the result is: $\frac{3 x}{16} + \frac{3 \sin{\left(4 x \right)}}{64}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{3 \cos{\left(2 x \right)}}{8}\, dx = \frac{3 \int \cos{\left(2 x \right)}\, dx}{8}$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      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:
      
      $\frac{\sin{\left(2 x \right)}}{2}$
      
      So, the result is: $\frac{3 \sin{\left(2 x \right)}}{16}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{8}\, dx = \frac{x}{8}$
      
      The result is: $\frac{5 x}{16} - \frac{\sin^{3}{\left(2 x \right)}}{48} + \frac{\sin{\left(2 x \right)}}{4} + \frac{3 \sin{\left(4 x \right)}}{64}$
    Method #2
    1. Rewrite the integrand:
      
      $\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{3} = \frac{\cos^{3}{\left(2 x \right)}}{8} + \frac{3 \cos^{2}{\left(2 x \right)}}{8} + \frac{3 \cos{\left(2 x \right)}}{8} + \frac{1}{8}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{3}{\left(2 x \right)}}{8}\, dx = \frac{\int \cos^{3}{\left(2 x \right)}\, dx}{8}$
      
      Rewrite the integrand:
      
      $\cos^{3}{\left(2 x \right)} = \left(1 - \sin^{2}{\left(2 x \right)}\right) \cos{\left(2 x \right)}$
      
      Let $u = \sin{\left(2 x \right)}$ .
      
      Then let $du = 2 \cos{\left(2 x \right)} dx$ and substitute $du$ :
      
      $\int \left(\frac{1}{2} - \frac{u^{2}}{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{u^{2}}{2}\right)\, du = - \frac{\int u^{2}\, du}{2}$
      
      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: $- \frac{u^{3}}{6}$
      
      The result is: $- \frac{u^{3}}{6} + \frac{u}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(2 x \right)}}{2}$
      
      So, the result is: $- \frac{\sin^{3}{\left(2 x \right)}}{48} + \frac{\sin{\left(2 x \right)}}{16}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{3 \cos^{2}{\left(2 x \right)}}{8}\, dx = \frac{3 \int \cos^{2}{\left(2 x \right)}\, dx}{8}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(2 x \right)} = \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(4 x \right)}}{2}\, dx = \frac{\int \cos{\left(4 x \right)}\, dx}{2}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      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: $\frac{\sin{\left(4 x \right)}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{x}{2} + \frac{\sin{\left(4 x \right)}}{8}$
      
      So, the result is: $\frac{3 x}{16} + \frac{3 \sin{\left(4 x \right)}}{64}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{3 \cos{\left(2 x \right)}}{8}\, dx = \frac{3 \int \cos{\left(2 x \right)}\, dx}{8}$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      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:
      
      $\frac{\sin{\left(2 x \right)}}{2}$
      
      So, the result is: $\frac{3 \sin{\left(2 x \right)}}{16}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{8}\, dx = \frac{x}{8}$
      
      The result is: $\frac{5 x}{16} - \frac{\sin^{3}{\left(2 x \right)}}{48} + \frac{\sin{\left(2 x \right)}}{4} + \frac{3 \sin{\left(4 x \right)}}{64}$
  So, the result is: $5 x - \frac{\sin^{3}{\left(2 x \right)}}{3} + 4 \sin{\left(2 x \right)} + \frac{3 \sin{\left(4 x \right)}}{4}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 20 \cos^{4}{\left(x \right)}\right)\, dx = - 20 \int \cos^{4}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\cos^{4}{\left(x \right)} = \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{2}$
  2. Rewrite the integrand:
    
    $\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{2} = \frac{\cos^{2}{\left(2 x \right)}}{4} + \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{4}$
  3. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{2}{\left(2 x \right)}}{4}\, dx = \frac{\int \cos^{2}{\left(2 x \right)}\, dx}{4}$
      1. Rewrite the integrand:
        
        $\cos^{2}{\left(2 x \right)} = \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}$
      2. Integrate term-by-term:
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \frac{\cos{\left(4 x \right)}}{2}\, dx = \frac{\int \cos{\left(4 x \right)}\, dx}{2}$
        
        Let $u = 4 x$ .
        
        Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
        
        $\int \frac{\cos{\left(u \right)}}{16}\, du$
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
        
        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: $\frac{\sin{\left(4 x \right)}}{8}$
        
        The integral of a constant is the constant times the variable of integration:
        
        $\int \frac{1}{2}\, dx = \frac{x}{2}$
        
        The result is: $\frac{x}{2} + \frac{\sin{\left(4 x \right)}}{8}$
      So, the result is: $\frac{x}{8} + \frac{\sin{\left(4 x \right)}}{32}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 x \right)}}{2}\, dx = \frac{\int \cos{\left(2 x \right)}\, dx}{2}$
      1. Let $u = 2 x$ .
        
        Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
        
        $\int \frac{\cos{\left(u \right)}}{4}\, du$
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
        
        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:
        
        $\frac{\sin{\left(2 x \right)}}{2}$
      So, the result is: $\frac{\sin{\left(2 x \right)}}{4}$
    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: $\frac{3 x}{8} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{32}$
  So, the result is: $- \frac{15 x}{2} - 5 \sin{\left(2 x \right)} - \frac{5 \sin{\left(4 x \right)}}{8}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 5 \cos^{2}{\left(x \right)}\, dx = 5 \int \cos^{2}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\cos^{2}{\left(x \right)} = \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 x \right)}}{2}\, dx = \frac{\int \cos{\left(2 x \right)}\, dx}{2}$
      1. Let $u = 2 x$ .
        
        Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
        
        $\int \frac{\cos{\left(u \right)}}{4}\, du$
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
        
        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:
        
        $\frac{\sin{\left(2 x \right)}}{2}$
      So, the result is: $\frac{\sin{\left(2 x \right)}}{4}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
    The result is: $\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4}$
  So, the result is: $\frac{5 x}{2} + \frac{5 \sin{\left(2 x \right)}}{4}$
The result is: $- \frac{\sin^{3}{\left(2 x \right)}}{3} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{8}$
Now simplify:

$\frac{\sin{\left(4 x \right)}}{8} + \frac{\sin{\left(6 x \right)}}{12}$
Add the constant of integration:

$\frac{\sin{\left(4 x \right)}}{8} + \frac{\sin{\left(6 x \right)}}{12}+ \mathrm{constant}$

The answer is:

$\frac{\sin{\left(4 x \right)}}{8} + \frac{\sin{\left(6 x \right)}}{12}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                            3                           
 |                          sin (2*x)   sin(2*x)   sin(4*x)
 | cos(x)*cos(5*x) dx = C - --------- + -------- + --------
 |                              3          4          8    
/

$${{\sin \left(6\,x\right)}\over{12}}+{{\sin \left(4\,x\right)}\over{ 8}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

$$0$$

=

$$0$$

Simplify

Numerical answer [src]

-2.46096520340058e-16

-2.46096520340058e-16

The graph

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Identical expressions

Similar expressions

Integral of cosxcos5x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #3

Method #2