Integral of sin(5x)cos(2x)dx dx

The solution

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 |  sin(5*x)*cos(2*x)*1 dx
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0

$$\int\limits_{0}^{1} \sin{\left(5 x \right)} \cos{\left(2 x \right)} 1\, dx$$

Integral(sin(5*x)*cos(2*x)*1, (x, 0, 1))

Detail solution

Rewrite the integrand:

$\sin{\left(5 x \right)} \cos{\left(2 x \right)} 1 = 32 \sin^{5}{\left(x \right)} \cos^{2}{\left(x \right)} - 16 \sin^{5}{\left(x \right)} - 40 \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)} + 20 \sin^{3}{\left(x \right)} + 10 \sin{\left(x \right)} \cos^{2}{\left(x \right)} - 5 \sin{\left(x \right)}$
Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 32 \sin^{5}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 32 \int \sin^{5}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\sin^{5}{\left(x \right)} \cos^{2}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right)^{2} \sin{\left(x \right)} \cos^{2}{\left(x \right)}$
  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 $du$ :
      
      $\int \left(- u^{6} + 2 u^{4} - u^{2}\right)\, du$
      
      Integrate term-by-term:
      
      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 integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 u^{4}\, du = 2 \int u^{4}\, du$
      
      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: $\frac{2 u^{5}}{5}$
      
      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: $- \frac{u^{7}}{7} + \frac{2 u^{5}}{5} - \frac{u^{3}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{7}{\left(x \right)}}{7} + \frac{2 \cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}$
    Method #2
    1. Rewrite the integrand:
      
      $\left(1 - \cos^{2}{\left(x \right)}\right)^{2} \sin{\left(x \right)} \cos^{2}{\left(x \right)} = \sin{\left(x \right)} \cos^{6}{\left(x \right)} - 2 \sin{\left(x \right)} \cos^{4}{\left(x \right)} + \sin{\left(x \right)} \cos^{2}{\left(x \right)}$
    2. Integrate term-by-term:
      
      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 integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 \sin{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - 2 \int \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u^{4}\, du$
      
      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: $- \frac{u^{5}}{5}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{5}{\left(x \right)}}{5}$
      
      So, the result is: $\frac{2 \cos^{5}{\left(x \right)}}{5}$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u^{2}\, du$
      
      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: $- \frac{u^{3}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{3}{\left(x \right)}}{3}$
      
      The result is: $- \frac{\cos^{7}{\left(x \right)}}{7} + \frac{2 \cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}$
    Method #3
    1. Rewrite the integrand:
      
      $\left(1 - \cos^{2}{\left(x \right)}\right)^{2} \sin{\left(x \right)} \cos^{2}{\left(x \right)} = \sin{\left(x \right)} \cos^{6}{\left(x \right)} - 2 \sin{\left(x \right)} \cos^{4}{\left(x \right)} + \sin{\left(x \right)} \cos^{2}{\left(x \right)}$
    2. Integrate term-by-term:
      
      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 integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 \sin{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - 2 \int \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u^{4}\, du$
      
      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: $- \frac{u^{5}}{5}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{5}{\left(x \right)}}{5}$
      
      So, the result is: $\frac{2 \cos^{5}{\left(x \right)}}{5}$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u^{2}\, du$
      
      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: $- \frac{u^{3}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{3}{\left(x \right)}}{3}$
      
      The result is: $- \frac{\cos^{7}{\left(x \right)}}{7} + \frac{2 \cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}$
  So, the result is: $- \frac{32 \cos^{7}{\left(x \right)}}{7} + \frac{64 \cos^{5}{\left(x \right)}}{5} - \frac{32 \cos^{3}{\left(x \right)}}{3}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 16 \sin^{5}{\left(x \right)}\right)\, dx = - 16 \int \sin^{5}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\sin^{5}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right)^{2} \sin{\left(x \right)}$
  2. Rewrite the integrand:
    
    $\left(1 - \cos^{2}{\left(x \right)}\right)^{2} \sin{\left(x \right)} = \sin{\left(x \right)} \cos^{4}{\left(x \right)} - 2 \sin{\left(x \right)} \cos^{2}{\left(x \right)} + \sin{\left(x \right)}$
  3. Integrate term-by-term:
    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:
        
        $\int \left(- u^{4}\right)\, du = - \int u^{4}\, du$
        
        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:
      
      $- \frac{\cos^{5}{\left(x \right)}}{5}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 \sin{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 2 \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
      1. Let $u = \cos{\left(x \right)}$ .
        
        Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
        
        $\int u^{2}\, du$
        
        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: $- \frac{u^{3}}{3}$
        
        Now substitute $u$ back in:
        
        $- \frac{\cos^{3}{\left(x \right)}}{3}$
      So, the result is: $\frac{2 \cos^{3}{\left(x \right)}}{3}$
    1. The integral of sine is negative cosine:
      
      $\int \sin{\left(x \right)}\, dx = - \cos{\left(x \right)}$
    The result is: $- \frac{\cos^{5}{\left(x \right)}}{5} + \frac{2 \cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}$
  So, the result is: $\frac{16 \cos^{5}{\left(x \right)}}{5} - \frac{32 \cos^{3}{\left(x \right)}}{3} + 16 \cos{\left(x \right)}$
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:
  
  $\int 20 \sin^{3}{\left(x \right)}\, dx = 20 \int \sin^{3}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\sin^{3}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)}$
  2. Let $u = \cos{\left(x \right)}$ .
    
    Then let $du = - \sin{\left(x \right)} dx$ and substitute $du$ :
    
    $\int \left(u^{2} - 1\right)\, du$
    1. Integrate term-by-term:
      1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
        
        $\int u^{2}\, du = \frac{u^{3}}{3}$
      1. The integral of a constant is the constant times the variable of integration:
        
        $\int \left(-1\right)\, du = - u$
      The result is: $\frac{u^{3}}{3} - u$
    Now substitute $u$ back in:
    
    $\frac{\cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}$
  So, the result is: $\frac{20 \cos^{3}{\left(x \right)}}{3} - 20 \cos{\left(x \right)}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 10 \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 10 \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
  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: $- \frac{10 \cos^{3}{\left(x \right)}}{3}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 5 \sin{\left(x \right)}\right)\, dx = - 5 \int \sin{\left(x \right)}\, dx$
  1. The integral of sine is negative cosine:
    
    $\int \sin{\left(x \right)}\, dx = - \cos{\left(x \right)}$
  So, the result is: $5 \cos{\left(x \right)}$
The result is: $- \frac{32 \cos^{7}{\left(x \right)}}{7} + 8 \cos^{5}{\left(x \right)} - \frac{14 \cos^{3}{\left(x \right)}}{3} + \cos{\left(x \right)}$
Now simplify:

$\frac{\left(- 96 \cos^{6}{\left(x \right)} + 168 \cos^{4}{\left(x \right)} - 98 \cos^{2}{\left(x \right)} + 21\right) \cos{\left(x \right)}}{21}$
Add the constant of integration:

$\frac{\left(- 96 \cos^{6}{\left(x \right)} + 168 \cos^{4}{\left(x \right)} - 98 \cos^{2}{\left(x \right)} + 21\right) \cos{\left(x \right)}}{21}+ \mathrm{constant}$

The answer is:

$\frac{\left(- 96 \cos^{6}{\left(x \right)} + 168 \cos^{4}{\left(x \right)} - 98 \cos^{2}{\left(x \right)} + 21\right) \cos{\left(x \right)}}{21}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                               7            3            
 |                                   5      32*cos (x)   14*cos (x)         
 | sin(5*x)*cos(2*x)*1 dx = C + 8*cos (x) - ---------- - ---------- + cos(x)
 |                                              7            3              
/

$$-{{\cos \left(7\,x\right)}\over{14}}-{{\cos \left(3\,x\right) }\over{6}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

5    5*cos(2)*cos(5)   2*sin(2)*sin(5)
-- - --------------- - ---------------
21          21                21

$${{5}\over{21}}-{{3\,\cos 7+7\,\cos 3}\over{42}}$$

5    5*cos(2)*cos(5)   2*sin(2)*sin(5)
-- - --------------- - ---------------
21          21                21

$$- \frac{5 \cos{\left(2 \right)} \cos{\left(5 \right)}}{21} - \frac{2 \sin{\left(2 \right)} \sin{\left(5 \right)}}{21} + \frac{5}{21}$$

Simplify

Numerical answer [src]

0.349243826504124

0.349243826504124

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of sin(5x)cos(2x)dx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3