Integral of sin(4x)cos(5x) dx

The solution

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$$\int\limits_{0}^{1} \sin{\left(4 x \right)} \cos{\left(5 x \right)}\, dx$$

Integral(sin(4*x)*cos(5*x), (x, 0, 1))

Detail solution

Rewrite the integrand:

$\sin{\left(4 x \right)} \cos{\left(5 x \right)} = - 128 \sin^{3}{\left(x \right)} \cos^{6}{\left(x \right)} + 160 \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)} - 40 \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)} + 64 \sin{\left(x \right)} \cos^{6}{\left(x \right)} - 80 \sin{\left(x \right)} \cos^{4}{\left(x \right)} + 20 \sin{\left(x \right)} \cos^{2}{\left(x \right)}$
Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 128 \sin^{3}{\left(x \right)} \cos^{6}{\left(x \right)}\right)\, dx = - 128 \int \sin^{3}{\left(x \right)} \cos^{6}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\sin^{3}{\left(x \right)} \cos^{6}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{6}{\left(x \right)}$
  2. There are multiple ways to do this integral.
    Method #1
    1. Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $du$ :
      
      $\int \left(u^{8} - u^{6}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{8}\, du = \frac{u^{9}}{9}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{6}\right)\, du = - \int u^{6}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      So, the result is: $- \frac{u^{7}}{7}$
      
      The result is: $\frac{u^{9}}{9} - \frac{u^{7}}{7}$
      
      Now substitute $u$ back in:
      
      $\frac{\cos^{9}{\left(x \right)}}{9} - \frac{\cos^{7}{\left(x \right)}}{7}$
    Method #2
    1. Rewrite the integrand:
      
      $\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{6}{\left(x \right)} = - \sin{\left(x \right)} \cos^{8}{\left(x \right)} + \sin{\left(x \right)} \cos^{6}{\left(x \right)}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin{\left(x \right)} \cos^{8}{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{8}{\left(x \right)}\, dx$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u^{8}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{8}\right)\, du = - \int u^{8}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{8}\, du = \frac{u^{9}}{9}$
      
      So, the result is: $- \frac{u^{9}}{9}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{9}{\left(x \right)}}{9}$
      
      So, the result is: $\frac{\cos^{9}{\left(x \right)}}{9}$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u^{6}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{6}\right)\, du = - \int u^{6}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      So, the result is: $- \frac{u^{7}}{7}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{7}{\left(x \right)}}{7}$
      
      The result is: $\frac{\cos^{9}{\left(x \right)}}{9} - \frac{\cos^{7}{\left(x \right)}}{7}$
    Method #3
    1. Rewrite the integrand:
      
      $\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{6}{\left(x \right)} = - \sin{\left(x \right)} \cos^{8}{\left(x \right)} + \sin{\left(x \right)} \cos^{6}{\left(x \right)}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin{\left(x \right)} \cos^{8}{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{8}{\left(x \right)}\, dx$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u^{8}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{8}\right)\, du = - \int u^{8}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{8}\, du = \frac{u^{9}}{9}$
      
      So, the result is: $- \frac{u^{9}}{9}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{9}{\left(x \right)}}{9}$
      
      So, the result is: $\frac{\cos^{9}{\left(x \right)}}{9}$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u^{6}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{6}\right)\, du = - \int u^{6}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      So, the result is: $- \frac{u^{7}}{7}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{7}{\left(x \right)}}{7}$
      
      The result is: $\frac{\cos^{9}{\left(x \right)}}{9} - \frac{\cos^{7}{\left(x \right)}}{7}$
  So, the result is: $- \frac{128 \cos^{9}{\left(x \right)}}{9} + \frac{128 \cos^{7}{\left(x \right)}}{7}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 160 \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx = 160 \int \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{4}{\left(x \right)}$
  2. Let $u = \cos{\left(x \right)}$ .
    
    Then let $du = - \sin{\left(x \right)} dx$ and substitute $du$ :
    
    $\int \left(u^{6} - u^{4}\right)\, du$
    1. Integrate term-by-term:
      1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
        
        $\int u^{6}\, du = \frac{u^{7}}{7}$
      1. The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \left(- u^{4}\right)\, du = - \int u^{4}\, du$
        
        The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
        
        $\int u^{4}\, du = \frac{u^{5}}{5}$
        
        So, the result is: $- \frac{u^{5}}{5}$
      The result is: $\frac{u^{7}}{7} - \frac{u^{5}}{5}$
    Now substitute $u$ back in:
    
    $\frac{\cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}$
  So, the result is: $\frac{160 \cos^{7}{\left(x \right)}}{7} - 32 \cos^{5}{\left(x \right)}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 40 \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 40 \int \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{2}{\left(x \right)}$
  2. Let $u = \cos{\left(x \right)}$ .
    
    Then let $du = - \sin{\left(x \right)} dx$ and substitute $du$ :
    
    $\int \left(u^{4} - u^{2}\right)\, du$
    1. Integrate term-by-term:
      1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
        
        $\int u^{4}\, du = \frac{u^{5}}{5}$
      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: $\frac{u^{5}}{5} - \frac{u^{3}}{3}$
    Now substitute $u$ back in:
    
    $\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}$
  So, the result is: $- 8 \cos^{5}{\left(x \right)} + \frac{40 \cos^{3}{\left(x \right)}}{3}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 64 \sin{\left(x \right)} \cos^{6}{\left(x \right)}\, dx = 64 \int \sin{\left(x \right)} \cos^{6}{\left(x \right)}\, dx$
  1. Let $u = \cos{\left(x \right)}$ .
    
    Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
    
    $\int u^{6}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{6}\right)\, du = - \int u^{6}\, du$
      1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
        
        $\int u^{6}\, du = \frac{u^{7}}{7}$
      So, the result is: $- \frac{u^{7}}{7}$
    Now substitute $u$ back in:
    
    $- \frac{\cos^{7}{\left(x \right)}}{7}$
  So, the result is: $- \frac{64 \cos^{7}{\left(x \right)}}{7}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 80 \sin{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - 80 \int \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$
  1. Let $u = \cos{\left(x \right)}$ .
    
    Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
    
    $\int u^{4}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{4}\right)\, du = - \int u^{4}\, du$
      1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
        
        $\int u^{4}\, du = \frac{u^{5}}{5}$
      So, the result is: $- \frac{u^{5}}{5}$
    Now substitute $u$ back in:
    
    $- \frac{\cos^{5}{\left(x \right)}}{5}$
  So, the result is: $16 \cos^{5}{\left(x \right)}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 20 \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 20 \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
  1. Let $u = \cos{\left(x \right)}$ .
    
    Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
    
    $\int u^{2}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{2}\right)\, du = - \int u^{2}\, du$
      1. 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: $- \frac{u^{3}}{3}$
    Now substitute $u$ back in:
    
    $- \frac{\cos^{3}{\left(x \right)}}{3}$
  So, the result is: $- \frac{20 \cos^{3}{\left(x \right)}}{3}$
The result is: $- \frac{128 \cos^{9}{\left(x \right)}}{9} + 32 \cos^{7}{\left(x \right)} - 24 \cos^{5}{\left(x \right)} + \frac{20 \cos^{3}{\left(x \right)}}{3}$
Now simplify:

$\frac{4 \left(- 32 \cos^{6}{\left(x \right)} + 72 \cos^{4}{\left(x \right)} - 54 \cos^{2}{\left(x \right)} + 15\right) \cos^{3}{\left(x \right)}}{9}$
Add the constant of integration:

$\frac{4 \left(- 32 \cos^{6}{\left(x \right)} + 72 \cos^{4}{\left(x \right)} - 54 \cos^{2}{\left(x \right)} + 15\right) \cos^{3}{\left(x \right)}}{9}+ \mathrm{constant}$

The answer is:

$\frac{4 \left(- 32 \cos^{6}{\left(x \right)} + 72 \cos^{4}{\left(x \right)} - 54 \cos^{2}{\left(x \right)} + 15\right) \cos^{3}{\left(x \right)}}{9}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                            9            3   
 |                                  5            7      128*cos (x)   20*cos (x)
 | sin(4*x)*cos(5*x) dx = C - 24*cos (x) + 32*cos (x) - ----------- + ----------
 |                                                           9            3     
/

$${{\cos x}\over{2}}-{{\cos \left(9\,x\right)}\over{18}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

  4   4*cos(4)*cos(5)   5*sin(4)*sin(5)
- - + --------------- + ---------------
  9          9                 9

$$-{{\cos 9-9\,\cos 1}\over{18}}-{{4}\over{9}}$$

  4   4*cos(4)*cos(5)   5*sin(4)*sin(5)
- - + --------------- + ---------------
  9          9                 9

$$- \frac{4}{9} + \frac{4 \cos{\left(4 \right)} \cos{\left(5 \right)}}{9} + \frac{5 \sin{\left(4 \right)} \sin{\left(5 \right)}}{9}$$

Simplify

Numerical answer [src]

-0.123674943627893

-0.123674943627893

The graph

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of sin(4x)cos(5x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3