Integral of sen3x(cos4x)dx dx

The solution

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$$\int\limits_{0}^{0} \sin{\left(3 x \right)} \cos{\left(4 x \right)}\, dx$$

Integral(sin(3*x)*cos(4*x), (x, 0, 0))

Detail solution

Rewrite the integrand:

$\sin{\left(3 x \right)} \cos{\left(4 x \right)} = - 32 \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)} + 32 \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)} - 4 \sin^{3}{\left(x \right)} + 24 \sin{\left(x \right)} \cos^{4}{\left(x \right)} - 24 \sin{\left(x \right)} \cos^{2}{\left(x \right)} + 3 \sin{\left(x \right)}$
Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 32 \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - 32 \int \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$
  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. 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 $du$ :
      
      $\int \left(u^{6} - u^{4}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      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:
      
      $\frac{\cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}$
    Method #2
    1. Rewrite the integrand:
      
      $\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{4}{\left(x \right)} = - \sin{\left(x \right)} \cos^{6}{\left(x \right)} + \sin{\left(x \right)} \cos^{4}{\left(x \right)}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin{\left(x \right)} \cos^{6}{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{6}{\left(x \right)}\, dx$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int \left(- u^{6}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{6}\, du = - \int u^{6}\, du$
      
      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:
      
      $- \frac{\cos^{7}{\left(x \right)}}{7}$
      
      So, the result is: $\frac{\cos^{7}{\left(x \right)}}{7}$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int \left(- u^{4}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{4}\, du = - \int u^{4}\, du$
      
      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:
      
      $- \frac{\cos^{5}{\left(x \right)}}{5}$
      
      The result is: $\frac{\cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}$
    Method #3
    1. Rewrite the integrand:
      
      $\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{4}{\left(x \right)} = - \sin{\left(x \right)} \cos^{6}{\left(x \right)} + \sin{\left(x \right)} \cos^{4}{\left(x \right)}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin{\left(x \right)} \cos^{6}{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{6}{\left(x \right)}\, dx$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int \left(- u^{6}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{6}\, du = - \int u^{6}\, du$
      
      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:
      
      $- \frac{\cos^{7}{\left(x \right)}}{7}$
      
      So, the result is: $\frac{\cos^{7}{\left(x \right)}}{7}$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int \left(- u^{4}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{4}\, du = - \int u^{4}\, du$
      
      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:
      
      $- \frac{\cos^{5}{\left(x \right)}}{5}$
      
      The result is: $\frac{\cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}$
  So, the result is: $- \frac{32 \cos^{7}{\left(x \right)}}{7} + \frac{32 \cos^{5}{\left(x \right)}}{5}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 32 \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 32 \int \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
  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: $\frac{32 \cos^{5}{\left(x \right)}}{5} - \frac{32 \cos^{3}{\left(x \right)}}{3}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 4 \sin^{3}{\left(x \right)}\right)\, dx = - 4 \int \sin^{3}{\left(x \right)}\, dx$
  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: $- \frac{4 \cos^{3}{\left(x \right)}}{3} + 4 \cos{\left(x \right)}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 24 \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx = 24 \int \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$
  1. Let $u = \cos{\left(x \right)}$ .
    
    Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
    
    $\int \left(- u^{4}\right)\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{4}\, du = - \int u^{4}\, du$
      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: $- \frac{24 \cos^{5}{\left(x \right)}}{5}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 24 \sin{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 24 \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
  1. Let $u = \cos{\left(x \right)}$ .
    
    Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
    
    $\int \left(- u^{2}\right)\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{2}\, du = - \int u^{2}\, du$
      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: $8 \cos^{3}{\left(x \right)}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 3 \sin{\left(x \right)}\, dx = 3 \int \sin{\left(x \right)}\, dx$
  1. The integral of sine is negative cosine:
    
    $\int \sin{\left(x \right)}\, dx = - \cos{\left(x \right)}$
  So, the result is: $- 3 \cos{\left(x \right)}$
The result is: $- \frac{32 \cos^{7}{\left(x \right)}}{7} + 8 \cos^{5}{\left(x \right)} - 4 \cos^{3}{\left(x \right)} + \cos{\left(x \right)}$
Add the constant of integration:

$- \frac{32 \cos^{7}{\left(x \right)}}{7} + 8 \cos^{5}{\left(x \right)} - 4 \cos^{3}{\left(x \right)} + \cos{\left(x \right)}+ \mathrm{constant}$

The answer is:

$- \frac{32 \cos^{7}{\left(x \right)}}{7} + 8 \cos^{5}{\left(x \right)} - 4 \cos^{3}{\left(x \right)} + \cos{\left(x \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                         7            
 |                                 3           5      32*cos (x)         
 | sin(3*x)*cos(4*x) dx = C - 4*cos (x) + 8*cos (x) - ---------- + cos(x)
 |                                                        7              
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$$\int \sin{\left(3 x \right)} \cos{\left(4 x \right)}\, dx = C - \frac{32 \cos^{7}{\left(x \right)}}{7} + 8 \cos^{5}{\left(x \right)} - 4 \cos^{3}{\left(x \right)} + \cos{\left(x \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

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Numerical answer [src]

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

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Identical expressions

Integral of sen3x(cos4x)dx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3