Integral of sin8xcos3x dx

The solution

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$$\int\limits_{0}^{1} \sin{\left(8 x \right)} \cos{\left(3 x \right)}\, dx$$

Integral(sin(8*x)*cos(3*x), (x, 0, 1))

Detail solution

Rewrite the integrand:

$\sin{\left(8 x \right)} \cos{\left(3 x \right)} = - 512 \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)} + 384 \sin^{7}{\left(x \right)} \cos^{2}{\left(x \right)} + 768 \sin^{5}{\left(x \right)} \cos^{4}{\left(x \right)} - 576 \sin^{5}{\left(x \right)} \cos^{2}{\left(x \right)} - 320 \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)} + 240 \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)} + 32 \sin{\left(x \right)} \cos^{4}{\left(x \right)} - 24 \sin{\left(x \right)} \cos^{2}{\left(x \right)}$
Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 512 \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - 512 \int \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right)^{3} \sin{\left(x \right)} \cos^{4}{\left(x \right)}$
  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 $du$ :
      
      $\int \left(u^{10} - 3 u^{8} + 3 u^{6} - u^{4}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{10}\, du = \frac{u^{11}}{11}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 3 u^{8}\right)\, du = - 3 \int u^{8}\, du$
      
      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: $- \frac{u^{9}}{3}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 3 u^{6}\, du = 3 \int u^{6}\, du$
      
      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: $\frac{3 u^{7}}{7}$
      
      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: $\frac{u^{11}}{11} - \frac{u^{9}}{3} + \frac{3 u^{7}}{7} - \frac{u^{5}}{5}$
      
      Now substitute $u$ back in:
      
      $\frac{\cos^{11}{\left(x \right)}}{11} - \frac{\cos^{9}{\left(x \right)}}{3} + \frac{3 \cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}$
    Method #2
    1. Rewrite the integrand:
      
      $\left(1 - \cos^{2}{\left(x \right)}\right)^{3} \sin{\left(x \right)} \cos^{4}{\left(x \right)} = - \sin{\left(x \right)} \cos^{10}{\left(x \right)} + 3 \sin{\left(x \right)} \cos^{8}{\left(x \right)} - 3 \sin{\left(x \right)} \cos^{6}{\left(x \right)} + \sin{\left(x \right)} \cos^{4}{\left(x \right)}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin{\left(x \right)} \cos^{10}{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{10}{\left(x \right)}\, dx$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u^{10}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{10}\right)\, du = - \int u^{10}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{10}\, du = \frac{u^{11}}{11}$
      
      So, the result is: $- \frac{u^{11}}{11}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{11}{\left(x \right)}}{11}$
      
      So, the result is: $\frac{\cos^{11}{\left(x \right)}}{11}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 3 \sin{\left(x \right)} \cos^{8}{\left(x \right)}\, dx = 3 \int \sin{\left(x \right)} \cos^{8}{\left(x \right)}\, dx$
      
      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: $- \frac{\cos^{9}{\left(x \right)}}{3}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 3 \sin{\left(x \right)} \cos^{6}{\left(x \right)}\right)\, dx = - 3 \int \sin{\left(x \right)} \cos^{6}{\left(x \right)}\, dx$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u^{6}\, du$
      
      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:
      
      $- \frac{\cos^{7}{\left(x \right)}}{7}$
      
      So, the result is: $\frac{3 \cos^{7}{\left(x \right)}}{7}$
      
      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:
      
      $- \frac{\cos^{5}{\left(x \right)}}{5}$
      
      The result is: $\frac{\cos^{11}{\left(x \right)}}{11} - \frac{\cos^{9}{\left(x \right)}}{3} + \frac{3 \cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}$
    Method #3
    1. Rewrite the integrand:
      
      $\left(1 - \cos^{2}{\left(x \right)}\right)^{3} \sin{\left(x \right)} \cos^{4}{\left(x \right)} = - \sin{\left(x \right)} \cos^{10}{\left(x \right)} + 3 \sin{\left(x \right)} \cos^{8}{\left(x \right)} - 3 \sin{\left(x \right)} \cos^{6}{\left(x \right)} + \sin{\left(x \right)} \cos^{4}{\left(x \right)}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin{\left(x \right)} \cos^{10}{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{10}{\left(x \right)}\, dx$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u^{10}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{10}\right)\, du = - \int u^{10}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{10}\, du = \frac{u^{11}}{11}$
      
      So, the result is: $- \frac{u^{11}}{11}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{11}{\left(x \right)}}{11}$
      
      So, the result is: $\frac{\cos^{11}{\left(x \right)}}{11}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 3 \sin{\left(x \right)} \cos^{8}{\left(x \right)}\, dx = 3 \int \sin{\left(x \right)} \cos^{8}{\left(x \right)}\, dx$
      
      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: $- \frac{\cos^{9}{\left(x \right)}}{3}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 3 \sin{\left(x \right)} \cos^{6}{\left(x \right)}\right)\, dx = - 3 \int \sin{\left(x \right)} \cos^{6}{\left(x \right)}\, dx$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u^{6}\, du$
      
      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:
      
      $- \frac{\cos^{7}{\left(x \right)}}{7}$
      
      So, the result is: $\frac{3 \cos^{7}{\left(x \right)}}{7}$
      
      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:
      
      $- \frac{\cos^{5}{\left(x \right)}}{5}$
      
      The result is: $\frac{\cos^{11}{\left(x \right)}}{11} - \frac{\cos^{9}{\left(x \right)}}{3} + \frac{3 \cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}$
  So, the result is: $- \frac{512 \cos^{11}{\left(x \right)}}{11} + \frac{512 \cos^{9}{\left(x \right)}}{3} - \frac{1536 \cos^{7}{\left(x \right)}}{7} + \frac{512 \cos^{5}{\left(x \right)}}{5}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 384 \sin^{7}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 384 \int \sin^{7}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\sin^{7}{\left(x \right)} \cos^{2}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right)^{3} \sin{\left(x \right)} \cos^{2}{\left(x \right)}$
  2. Let $u = \cos{\left(x \right)}$ .
    
    Then let $du = - \sin{\left(x \right)} dx$ and substitute $du$ :
    
    $\int \left(u^{8} - 3 u^{6} + 3 u^{4} - u^{2}\right)\, du$
    1. Integrate term-by-term:
      1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
        
        $\int u^{8}\, du = \frac{u^{9}}{9}$
      1. The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \left(- 3 u^{6}\right)\, du = - 3 \int u^{6}\, du$
        
        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: $- \frac{3 u^{7}}{7}$
      1. The integral of a constant times a function is the constant times the integral of the function:
        
        $\int 3 u^{4}\, du = 3 \int u^{4}\, du$
        
        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: $\frac{3 u^{5}}{5}$
      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: $\frac{u^{9}}{9} - \frac{3 u^{7}}{7} + \frac{3 u^{5}}{5} - \frac{u^{3}}{3}$
    Now substitute $u$ back in:
    
    $\frac{\cos^{9}{\left(x \right)}}{9} - \frac{3 \cos^{7}{\left(x \right)}}{7} + \frac{3 \cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}$
  So, the result is: $\frac{128 \cos^{9}{\left(x \right)}}{3} - \frac{1152 \cos^{7}{\left(x \right)}}{7} + \frac{1152 \cos^{5}{\left(x \right)}}{5} - 128 \cos^{3}{\left(x \right)}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 768 \sin^{5}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx = 768 \int \sin^{5}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\sin^{5}{\left(x \right)} \cos^{4}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right)^{2} \sin{\left(x \right)} \cos^{4}{\left(x \right)}$
  2. Let $u = \cos{\left(x \right)}$ .
    
    Then let $du = - \sin{\left(x \right)} dx$ and substitute $du$ :
    
    $\int \left(- u^{8} + 2 u^{6} - u^{4}\right)\, du$
    1. Integrate term-by-term:
      1. 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: $- \frac{u^{9}}{9}$
      1. The integral of a constant times a function is the constant times the integral of the function:
        
        $\int 2 u^{6}\, du = 2 \int u^{6}\, du$
        
        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: $\frac{2 u^{7}}{7}$
      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: $- \frac{u^{9}}{9} + \frac{2 u^{7}}{7} - \frac{u^{5}}{5}$
    Now substitute $u$ back in:
    
    $- \frac{\cos^{9}{\left(x \right)}}{9} + \frac{2 \cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}$
  So, the result is: $- \frac{256 \cos^{9}{\left(x \right)}}{3} + \frac{1536 \cos^{7}{\left(x \right)}}{7} - \frac{768 \cos^{5}{\left(x \right)}}{5}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 576 \sin^{5}{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 576 \int \sin^{5}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
  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. Let $u = \cos{\left(x \right)}$ .
    
    Then let $du = - \sin{\left(x \right)} dx$ and substitute $du$ :
    
    $\int \left(- u^{6} + 2 u^{4} - u^{2}\right)\, du$
    1. Integrate term-by-term:
      1. The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \left(- u^{6}\right)\, du = - \int u^{6}\, du$
        
        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: $- \frac{u^{7}}{7}$
      1. 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: $\frac{2 u^{5}}{5}$
      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: $- \frac{u^{7}}{7} + \frac{2 u^{5}}{5} - \frac{u^{3}}{3}$
    Now substitute $u$ back in:
    
    $- \frac{\cos^{7}{\left(x \right)}}{7} + \frac{2 \cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}$
  So, the result is: $\frac{576 \cos^{7}{\left(x \right)}}{7} - \frac{1152 \cos^{5}{\left(x \right)}}{5} + 192 \cos^{3}{\left(x \right)}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 320 \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - 320 \int \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$
  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: $- \frac{320 \cos^{7}{\left(x \right)}}{7} + 64 \cos^{5}{\left(x \right)}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 240 \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 240 \int \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
  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: $48 \cos^{5}{\left(x \right)} - 80 \cos^{3}{\left(x \right)}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 32 \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx = 32 \int \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$
  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: $- \frac{32 \cos^{5}{\left(x \right)}}{5}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 24 \sin{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 24 \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
  1. Let $u = \cos{\left(x \right)}$ .
    
    Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
    
    $\int u^{2}\, du$
    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: $8 \cos^{3}{\left(x \right)}$
The result is: $- \frac{512 \cos^{11}{\left(x \right)}}{11} + 128 \cos^{9}{\left(x \right)} - 128 \cos^{7}{\left(x \right)} + \frac{272 \cos^{5}{\left(x \right)}}{5} - 8 \cos^{3}{\left(x \right)}$
Now simplify:

$\frac{8 \left(- 320 \cos^{8}{\left(x \right)} + 880 \cos^{6}{\left(x \right)} - 880 \cos^{4}{\left(x \right)} + 374 \cos^{2}{\left(x \right)} - 55\right) \cos^{3}{\left(x \right)}}{55}$
Add the constant of integration:

$\frac{8 \left(- 320 \cos^{8}{\left(x \right)} + 880 \cos^{6}{\left(x \right)} - 880 \cos^{4}{\left(x \right)} + 374 \cos^{2}{\left(x \right)} - 55\right) \cos^{3}{\left(x \right)}}{55}+ \mathrm{constant}$

The answer is:

$\frac{8 \left(- 320 \cos^{8}{\left(x \right)} + 880 \cos^{6}{\left(x \right)} - 880 \cos^{4}{\left(x \right)} + 374 \cos^{2}{\left(x \right)} - 55\right) \cos^{3}{\left(x \right)}}{55}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                          11             5   
 |                                   7           3             9      512*cos  (x)   272*cos (x)
 | sin(8*x)*cos(3*x) dx = C - 128*cos (x) - 8*cos (x) + 128*cos (x) - ------------ + -----------
 |                                                                         11             5     
/

$$-{{\cos \left(11\,x\right)}\over{22}}-{{\cos \left(5\,x\right) }\over{10}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

8    8*cos(3)*cos(8)   3*sin(3)*sin(8)
-- - --------------- - ---------------
55          55                55

$${{8}\over{55}}-{{5\,\cos 11+11\,\cos 5}\over{110}}$$

8    8*cos(3)*cos(8)   3*sin(3)*sin(8)
-- - --------------- - ---------------
55          55                55

$$- \frac{8 \cos{\left(3 \right)} \cos{\left(8 \right)}}{55} - \frac{3 \sin{\left(3 \right)} \sin{\left(8 \right)}}{55} + \frac{8}{55}$$

Упростить

Numerical answer [src]

0.116887158817857

0.116887158817857

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of sin8xcos3x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3