Integral of sin(2x)sin(8x) dx

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$$\int\limits_{0}^{1} \sin{\left(2 x \right)} \sin{\left(8 x \right)}\, dx$$

Integral(sin(2*x)*sin(8*x), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = 2 x$ .
  
  Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
  
  $\int \frac{\sin{\left(u \right)} \sin{\left(4 u \right)}}{4}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\sin{\left(4 u \right)} \sin{\left(u \right)}}{2}\, du = \frac{\int \sin{\left(4 u \right)} \sin{\left(u \right)}\, du}{2}$
    1. Rewrite the integrand:
      
      $\sin{\left(4 u \right)} \sin{\left(u \right)} = - 8 \sin^{4}{\left(u \right)} \cos{\left(u \right)} + 4 \sin^{2}{\left(u \right)} \cos{\left(u \right)}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 8 \sin^{4}{\left(u \right)} \cos{\left(u \right)}\right)\, du = - 8 \int \sin^{4}{\left(u \right)} \cos{\left(u \right)}\, du$
      
      Let $u = \sin{\left(u \right)}$ .
      
      Then let $du = \cos{\left(u \right)} du$ and substitute $du$ :
      
      $\int u^{4}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{5}{\left(u \right)}}{5}$
      
      So, the result is: $- \frac{8 \sin^{5}{\left(u \right)}}{5}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 4 \sin^{2}{\left(u \right)} \cos{\left(u \right)}\, du = 4 \int \sin^{2}{\left(u \right)} \cos{\left(u \right)}\, du$
      
      Let $u = \sin{\left(u \right)}$ .
      
      Then let $du = \cos{\left(u \right)} du$ and substitute $du$ :
      
      $\int u^{2}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{3}{\left(u \right)}}{3}$
      
      So, the result is: $\frac{4 \sin^{3}{\left(u \right)}}{3}$
      
      The result is: $- \frac{8 \sin^{5}{\left(u \right)}}{5} + \frac{4 \sin^{3}{\left(u \right)}}{3}$
    So, the result is: $- \frac{4 \sin^{5}{\left(u \right)}}{5} + \frac{2 \sin^{3}{\left(u \right)}}{3}$
  Now substitute $u$ back in:
  
  $- \frac{4 \sin^{5}{\left(2 x \right)}}{5} + \frac{2 \sin^{3}{\left(2 x \right)}}{3}$
Method #2
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 2 \sin{\left(x \right)} \sin{\left(8 x \right)} \cos{\left(x \right)}\, dx = 2 \int \sin{\left(x \right)} \sin{\left(8 x \right)} \cos{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\sin{\left(x \right)} \sin{\left(8 x \right)} \cos{\left(x \right)} = - 128 \sin^{8}{\left(x \right)} \cos^{2}{\left(x \right)} + 192 \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)} - 80 \sin^{4}{\left(x \right)} \cos^{2}{\left(x \right)} + 8 \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 128 \sin^{8}{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 128 \int \sin^{8}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
      
      Rewrite the integrand:
      
      $\sin^{8}{\left(x \right)} \cos^{2}{\left(x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{4} \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $du$ :
      
      $\int \left(\frac{\cos^{5}{\left(u \right)}}{64} - \frac{3 \cos^{4}{\left(u \right)}}{64} + \frac{\cos^{3}{\left(u \right)}}{32} + \frac{\cos^{2}{\left(u \right)}}{32} - \frac{3 \cos{\left(u \right)}}{64} + \frac{1}{64}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{5}{\left(u \right)}}{64}\, du = \frac{\int \cos^{5}{\left(u \right)}\, du}{64}$
      
      Rewrite the integrand:
      
      $\cos^{5}{\left(u \right)} = \left(1 - \sin^{2}{\left(u \right)}\right)^{2} \cos{\left(u \right)}$
      
      Rewrite the integrand:
      
      $\left(1 - \sin^{2}{\left(u \right)}\right)^{2} \cos{\left(u \right)} = \sin^{4}{\left(u \right)} \cos{\left(u \right)} - 2 \sin^{2}{\left(u \right)} \cos{\left(u \right)} + \cos{\left(u \right)}$
      
      Integrate term-by-term:
      
      Let $u = \sin{\left(u \right)}$ .
      
      Then let $du = \cos{\left(u \right)} du$ and substitute $du$ :
      
      $\int u^{4}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{5}{\left(u \right)}}{5}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 \sin^{2}{\left(u \right)} \cos{\left(u \right)}\right)\, du = - 2 \int \sin^{2}{\left(u \right)} \cos{\left(u \right)}\, du$
      
      Let $u = \sin{\left(u \right)}$ .
      
      Then let $du = \cos{\left(u \right)} du$ and substitute $du$ :
      
      $\int u^{2}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{3}{\left(u \right)}}{3}$
      
      So, the result is: $- \frac{2 \sin^{3}{\left(u \right)}}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      The result is: $\frac{\sin^{5}{\left(u \right)}}{5} - \frac{2 \sin^{3}{\left(u \right)}}{3} + \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin^{5}{\left(u \right)}}{320} - \frac{\sin^{3}{\left(u \right)}}{96} + \frac{\sin{\left(u \right)}}{64}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{3 \cos^{4}{\left(u \right)}}{64}\right)\, du = - \frac{3 \int \cos^{4}{\left(u \right)}\, du}{64}$
      
      Rewrite the integrand:
      
      $\cos^{4}{\left(u \right)} = \left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)^{2}$
      
      There are multiple ways to do this integral.
      
      Method #1
      
      Rewrite the integrand:
      
      $\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)^{2} = \frac{\cos^{2}{\left(2 u \right)}}{4} + \frac{\cos{\left(2 u \right)}}{2} + \frac{1}{4}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{2}{\left(2 u \right)}}{4}\, du = \frac{\int \cos^{2}{\left(2 u \right)}\, du}{4}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(2 u \right)} = \frac{\cos{\left(4 u \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(4 u \right)}}{2}\, du = \frac{\int \cos{\left(4 u \right)}\, du}{2}$
      
      Let $u = 4 u$ .
      
      Then let $du = 4 du$ 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:
      
      $\frac{\sin{\left(4 u \right)}}{4}$
      
      So, the result is: $\frac{\sin{\left(4 u \right)}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The result is: $\frac{u}{2} + \frac{\sin{\left(4 u \right)}}{8}$
      
      So, the result is: $\frac{u}{8} + \frac{\sin{\left(4 u \right)}}{32}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ 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:
      
      $\frac{\sin{\left(2 u \right)}}{2}$
      
      So, the result is: $\frac{\sin{\left(2 u \right)}}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{4}\, du = \frac{u}{4}$
      
      The result is: $\frac{3 u}{8} + \frac{\sin{\left(2 u \right)}}{4} + \frac{\sin{\left(4 u \right)}}{32}$
      
      Method #2
      
      Rewrite the integrand:
      
      $\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)^{2} = \frac{\cos^{2}{\left(2 u \right)}}{4} + \frac{\cos{\left(2 u \right)}}{2} + \frac{1}{4}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{2}{\left(2 u \right)}}{4}\, du = \frac{\int \cos^{2}{\left(2 u \right)}\, du}{4}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(2 u \right)} = \frac{\cos{\left(4 u \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(4 u \right)}}{2}\, du = \frac{\int \cos{\left(4 u \right)}\, du}{2}$
      
      Let $u = 4 u$ .
      
      Then let $du = 4 du$ 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:
      
      $\frac{\sin{\left(4 u \right)}}{4}$
      
      So, the result is: $\frac{\sin{\left(4 u \right)}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The result is: $\frac{u}{2} + \frac{\sin{\left(4 u \right)}}{8}$
      
      So, the result is: $\frac{u}{8} + \frac{\sin{\left(4 u \right)}}{32}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ 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:
      
      $\frac{\sin{\left(2 u \right)}}{2}$
      
      So, the result is: $\frac{\sin{\left(2 u \right)}}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{4}\, du = \frac{u}{4}$
      
      The result is: $\frac{3 u}{8} + \frac{\sin{\left(2 u \right)}}{4} + \frac{\sin{\left(4 u \right)}}{32}$
      
      So, the result is: $- \frac{9 u}{512} - \frac{3 \sin{\left(2 u \right)}}{256} - \frac{3 \sin{\left(4 u \right)}}{2048}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{3}{\left(u \right)}}{32}\, du = \frac{\int \cos^{3}{\left(u \right)}\, du}{32}$
      
      Rewrite the integrand:
      
      $\cos^{3}{\left(u \right)} = \left(1 - \sin^{2}{\left(u \right)}\right) \cos{\left(u \right)}$
      
      Let $u = \sin{\left(u \right)}$ .
      
      Then let $du = \cos{\left(u \right)} du$ and substitute $du$ :
      
      $\int \left(1 - u^{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      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: $- \frac{u^{3}}{3} + u$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(u \right)}}{3} + \sin{\left(u \right)}$
      
      So, the result is: $- \frac{\sin^{3}{\left(u \right)}}{96} + \frac{\sin{\left(u \right)}}{32}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{2}{\left(u \right)}}{32}\, du = \frac{\int \cos^{2}{\left(u \right)}\, du}{32}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(u \right)} = \frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ 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:
      
      $\frac{\sin{\left(2 u \right)}}{2}$
      
      So, the result is: $\frac{\sin{\left(2 u \right)}}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The result is: $\frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}$
      
      So, the result is: $\frac{u}{64} + \frac{\sin{\left(2 u \right)}}{128}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{3 \cos{\left(u \right)}}{64}\right)\, du = - \frac{3 \int \cos{\left(u \right)}\, du}{64}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $- \frac{3 \sin{\left(u \right)}}{64}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{64}\, du = \frac{u}{64}$
      
      The result is: $\frac{7 u}{512} - \frac{\sin{\left(2 u \right)}}{256} - \frac{3 \sin{\left(4 u \right)}}{2048} + \frac{\sin^{5}{\left(u \right)}}{320} - \frac{\sin^{3}{\left(u \right)}}{48}$
      
      Now substitute $u$ back in:
      
      $\frac{7 x}{256} + \frac{\sin^{5}{\left(2 x \right)}}{320} - \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{256} - \frac{3 \sin{\left(8 x \right)}}{2048}$
      
      So, the result is: $- \frac{7 x}{2} - \frac{2 \sin^{5}{\left(2 x \right)}}{5} + \frac{8 \sin^{3}{\left(2 x \right)}}{3} + \frac{\sin{\left(4 x \right)}}{2} + \frac{3 \sin{\left(8 x \right)}}{16}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 192 \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 192 \int \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
      
      Rewrite the integrand:
      
      $\sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{3} \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $du$ :
      
      $\int \left(- \frac{\cos^{4}{\left(u \right)}}{32} + \frac{\cos^{3}{\left(u \right)}}{16} - \frac{\cos{\left(u \right)}}{16} + \frac{1}{32}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{4}{\left(u \right)}}{32}\right)\, du = - \frac{\int \cos^{4}{\left(u \right)}\, du}{32}$
      
      Rewrite the integrand:
      
      $\cos^{4}{\left(u \right)} = \left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)^{2}$
      
      Rewrite the integrand:
      
      $\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)^{2} = \frac{\cos^{2}{\left(2 u \right)}}{4} + \frac{\cos{\left(2 u \right)}}{2} + \frac{1}{4}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{2}{\left(2 u \right)}}{4}\, du = \frac{\int \cos^{2}{\left(2 u \right)}\, du}{4}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(2 u \right)} = \frac{\cos{\left(4 u \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(4 u \right)}}{2}\, du = \frac{\int \cos{\left(4 u \right)}\, du}{2}$
      
      Let $u = 4 u$ .
      
      Then let $du = 4 du$ 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:
      
      $\frac{\sin{\left(4 u \right)}}{4}$
      
      So, the result is: $\frac{\sin{\left(4 u \right)}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The result is: $\frac{u}{2} + \frac{\sin{\left(4 u \right)}}{8}$
      
      So, the result is: $\frac{u}{8} + \frac{\sin{\left(4 u \right)}}{32}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ 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:
      
      $\frac{\sin{\left(2 u \right)}}{2}$
      
      So, the result is: $\frac{\sin{\left(2 u \right)}}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{4}\, du = \frac{u}{4}$
      
      The result is: $\frac{3 u}{8} + \frac{\sin{\left(2 u \right)}}{4} + \frac{\sin{\left(4 u \right)}}{32}$
      
      So, the result is: $- \frac{3 u}{256} - \frac{\sin{\left(2 u \right)}}{128} - \frac{\sin{\left(4 u \right)}}{1024}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{3}{\left(u \right)}}{16}\, du = \frac{\int \cos^{3}{\left(u \right)}\, du}{16}$
      
      Rewrite the integrand:
      
      $\cos^{3}{\left(u \right)} = \left(1 - \sin^{2}{\left(u \right)}\right) \cos{\left(u \right)}$
      
      Let $u = \sin{\left(u \right)}$ .
      
      Then let $du = \cos{\left(u \right)} du$ and substitute $du$ :
      
      $\int \left(1 - u^{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      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: $- \frac{u^{3}}{3} + u$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(u \right)}}{3} + \sin{\left(u \right)}$
      
      So, the result is: $- \frac{\sin^{3}{\left(u \right)}}{48} + \frac{\sin{\left(u \right)}}{16}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(u \right)}}{16}\right)\, du = - \frac{\int \cos{\left(u \right)}\, du}{16}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $- \frac{\sin{\left(u \right)}}{16}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{32}\, du = \frac{u}{32}$
      
      The result is: $\frac{5 u}{256} - \frac{\sin{\left(2 u \right)}}{128} - \frac{\sin{\left(4 u \right)}}{1024} - \frac{\sin^{3}{\left(u \right)}}{48}$
      
      Now substitute $u$ back in:
      
      $\frac{5 x}{128} - \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{128} - \frac{\sin{\left(8 x \right)}}{1024}$
      
      So, the result is: $\frac{15 x}{2} - 4 \sin^{3}{\left(2 x \right)} - \frac{3 \sin{\left(4 x \right)}}{2} - \frac{3 \sin{\left(8 x \right)}}{16}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 80 \sin^{4}{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 80 \int \sin^{4}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
      
      Rewrite the integrand:
      
      $\sin^{4}{\left(x \right)} \cos^{2}{\left(x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{2} \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $du$ :
      
      $\int \left(\frac{\cos^{3}{\left(u \right)}}{16} - \frac{\cos^{2}{\left(u \right)}}{16} - \frac{\cos{\left(u \right)}}{16} + \frac{1}{16}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{3}{\left(u \right)}}{16}\, du = \frac{\int \cos^{3}{\left(u \right)}\, du}{16}$
      
      Rewrite the integrand:
      
      $\cos^{3}{\left(u \right)} = \left(1 - \sin^{2}{\left(u \right)}\right) \cos{\left(u \right)}$
      
      Let $u = \sin{\left(u \right)}$ .
      
      Then let $du = \cos{\left(u \right)} du$ and substitute $du$ :
      
      $\int \left(1 - u^{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      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: $- \frac{u^{3}}{3} + u$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(u \right)}}{3} + \sin{\left(u \right)}$
      
      So, the result is: $- \frac{\sin^{3}{\left(u \right)}}{48} + \frac{\sin{\left(u \right)}}{16}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{2}{\left(u \right)}}{16}\right)\, du = - \frac{\int \cos^{2}{\left(u \right)}\, du}{16}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(u \right)} = \frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ 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:
      
      $\frac{\sin{\left(2 u \right)}}{2}$
      
      So, the result is: $\frac{\sin{\left(2 u \right)}}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The result is: $\frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}$
      
      So, the result is: $- \frac{u}{32} - \frac{\sin{\left(2 u \right)}}{64}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(u \right)}}{16}\right)\, du = - \frac{\int \cos{\left(u \right)}\, du}{16}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $- \frac{\sin{\left(u \right)}}{16}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{16}\, du = \frac{u}{16}$
      
      The result is: $\frac{u}{32} - \frac{\sin{\left(2 u \right)}}{64} - \frac{\sin^{3}{\left(u \right)}}{48}$
      
      Now substitute $u$ back in:
      
      $\frac{x}{16} - \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{64}$
      
      So, the result is: $- 5 x + \frac{5 \sin^{3}{\left(2 x \right)}}{3} + \frac{5 \sin{\left(4 x \right)}}{4}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 8 \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 8 \int \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
      
      Rewrite the integrand:
      
      $\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right) \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $du$ :
      
      $\int \left(\frac{1}{8} - \frac{\cos^{2}{\left(u \right)}}{8}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{8}\, du = \frac{u}{8}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{2}{\left(u \right)}}{8}\right)\, du = - \frac{\int \cos^{2}{\left(u \right)}\, du}{8}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(u \right)} = \frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ 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:
      
      $\frac{\sin{\left(2 u \right)}}{2}$
      
      So, the result is: $\frac{\sin{\left(2 u \right)}}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The result is: $\frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}$
      
      So, the result is: $- \frac{u}{16} - \frac{\sin{\left(2 u \right)}}{32}$
      
      The result is: $\frac{u}{16} - \frac{\sin{\left(2 u \right)}}{32}$
      
      Now substitute $u$ back in:
      
      $\frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32}$
      
      So, the result is: $x - \frac{\sin{\left(4 x \right)}}{4}$
    The result is: $- \frac{2 \sin^{5}{\left(2 x \right)}}{5} + \frac{\sin^{3}{\left(2 x \right)}}{3}$
  So, the result is: $- \frac{4 \sin^{5}{\left(2 x \right)}}{5} + \frac{2 \sin^{3}{\left(2 x \right)}}{3}$
Method #3
1. Rewrite the integrand:
  
  $\sin{\left(2 x \right)} \sin{\left(8 x \right)} = - 256 \sin^{8}{\left(x \right)} \cos^{2}{\left(x \right)} + 384 \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)} - 160 \sin^{4}{\left(x \right)} \cos^{2}{\left(x \right)} + 16 \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 256 \sin^{8}{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 256 \int \sin^{8}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
    1. Rewrite the integrand:
      
      $\sin^{8}{\left(x \right)} \cos^{2}{\left(x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{4} \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)$
    2. Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $du$ :
      
      $\int \left(\frac{\cos^{5}{\left(u \right)}}{64} - \frac{3 \cos^{4}{\left(u \right)}}{64} + \frac{\cos^{3}{\left(u \right)}}{32} + \frac{\cos^{2}{\left(u \right)}}{32} - \frac{3 \cos{\left(u \right)}}{64} + \frac{1}{64}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{5}{\left(u \right)}}{64}\, du = \frac{\int \cos^{5}{\left(u \right)}\, du}{64}$
      
      Rewrite the integrand:
      
      $\cos^{5}{\left(u \right)} = \left(1 - \sin^{2}{\left(u \right)}\right)^{2} \cos{\left(u \right)}$
      
      Rewrite the integrand:
      
      $\left(1 - \sin^{2}{\left(u \right)}\right)^{2} \cos{\left(u \right)} = \sin^{4}{\left(u \right)} \cos{\left(u \right)} - 2 \sin^{2}{\left(u \right)} \cos{\left(u \right)} + \cos{\left(u \right)}$
      
      Integrate term-by-term:
      
      Let $u = \sin{\left(u \right)}$ .
      
      Then let $du = \cos{\left(u \right)} du$ and substitute $du$ :
      
      $\int u^{4}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{5}{\left(u \right)}}{5}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 \sin^{2}{\left(u \right)} \cos{\left(u \right)}\right)\, du = - 2 \int \sin^{2}{\left(u \right)} \cos{\left(u \right)}\, du$
      
      Let $u = \sin{\left(u \right)}$ .
      
      Then let $du = \cos{\left(u \right)} du$ and substitute $du$ :
      
      $\int u^{2}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{3}{\left(u \right)}}{3}$
      
      So, the result is: $- \frac{2 \sin^{3}{\left(u \right)}}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      The result is: $\frac{\sin^{5}{\left(u \right)}}{5} - \frac{2 \sin^{3}{\left(u \right)}}{3} + \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin^{5}{\left(u \right)}}{320} - \frac{\sin^{3}{\left(u \right)}}{96} + \frac{\sin{\left(u \right)}}{64}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{3 \cos^{4}{\left(u \right)}}{64}\right)\, du = - \frac{3 \int \cos^{4}{\left(u \right)}\, du}{64}$
      
      Rewrite the integrand:
      
      $\cos^{4}{\left(u \right)} = \left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)^{2}$
      
      Rewrite the integrand:
      
      $\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)^{2} = \frac{\cos^{2}{\left(2 u \right)}}{4} + \frac{\cos{\left(2 u \right)}}{2} + \frac{1}{4}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{2}{\left(2 u \right)}}{4}\, du = \frac{\int \cos^{2}{\left(2 u \right)}\, du}{4}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(2 u \right)} = \frac{\cos{\left(4 u \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(4 u \right)}}{2}\, du = \frac{\int \cos{\left(4 u \right)}\, du}{2}$
      
      Let $u = 4 u$ .
      
      Then let $du = 4 du$ 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:
      
      $\frac{\sin{\left(4 u \right)}}{4}$
      
      So, the result is: $\frac{\sin{\left(4 u \right)}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The result is: $\frac{u}{2} + \frac{\sin{\left(4 u \right)}}{8}$
      
      So, the result is: $\frac{u}{8} + \frac{\sin{\left(4 u \right)}}{32}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ 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:
      
      $\frac{\sin{\left(2 u \right)}}{2}$
      
      So, the result is: $\frac{\sin{\left(2 u \right)}}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{4}\, du = \frac{u}{4}$
      
      The result is: $\frac{3 u}{8} + \frac{\sin{\left(2 u \right)}}{4} + \frac{\sin{\left(4 u \right)}}{32}$
      
      So, the result is: $- \frac{9 u}{512} - \frac{3 \sin{\left(2 u \right)}}{256} - \frac{3 \sin{\left(4 u \right)}}{2048}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{3}{\left(u \right)}}{32}\, du = \frac{\int \cos^{3}{\left(u \right)}\, du}{32}$
      
      Rewrite the integrand:
      
      $\cos^{3}{\left(u \right)} = \left(1 - \sin^{2}{\left(u \right)}\right) \cos{\left(u \right)}$
      
      Let $u = \sin{\left(u \right)}$ .
      
      Then let $du = \cos{\left(u \right)} du$ and substitute $du$ :
      
      $\int \left(1 - u^{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      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: $- \frac{u^{3}}{3} + u$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(u \right)}}{3} + \sin{\left(u \right)}$
      
      So, the result is: $- \frac{\sin^{3}{\left(u \right)}}{96} + \frac{\sin{\left(u \right)}}{32}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{2}{\left(u \right)}}{32}\, du = \frac{\int \cos^{2}{\left(u \right)}\, du}{32}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(u \right)} = \frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ 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:
      
      $\frac{\sin{\left(2 u \right)}}{2}$
      
      So, the result is: $\frac{\sin{\left(2 u \right)}}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The result is: $\frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}$
      
      So, the result is: $\frac{u}{64} + \frac{\sin{\left(2 u \right)}}{128}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{3 \cos{\left(u \right)}}{64}\right)\, du = - \frac{3 \int \cos{\left(u \right)}\, du}{64}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $- \frac{3 \sin{\left(u \right)}}{64}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{64}\, du = \frac{u}{64}$
      
      The result is: $\frac{7 u}{512} - \frac{\sin{\left(2 u \right)}}{256} - \frac{3 \sin{\left(4 u \right)}}{2048} + \frac{\sin^{5}{\left(u \right)}}{320} - \frac{\sin^{3}{\left(u \right)}}{48}$
      
      Now substitute $u$ back in:
      
      $\frac{7 x}{256} + \frac{\sin^{5}{\left(2 x \right)}}{320} - \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{256} - \frac{3 \sin{\left(8 x \right)}}{2048}$
    So, the result is: $- 7 x - \frac{4 \sin^{5}{\left(2 x \right)}}{5} + \frac{16 \sin^{3}{\left(2 x \right)}}{3} + \sin{\left(4 x \right)} + \frac{3 \sin{\left(8 x \right)}}{8}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 384 \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 384 \int \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
    1. Rewrite the integrand:
      
      $\sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{3} \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)$
    2. Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $du$ :
      
      $\int \left(- \frac{\cos^{4}{\left(u \right)}}{32} + \frac{\cos^{3}{\left(u \right)}}{16} - \frac{\cos{\left(u \right)}}{16} + \frac{1}{32}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{4}{\left(u \right)}}{32}\right)\, du = - \frac{\int \cos^{4}{\left(u \right)}\, du}{32}$
      
      Rewrite the integrand:
      
      $\cos^{4}{\left(u \right)} = \left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)^{2}$
      
      Rewrite the integrand:
      
      $\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)^{2} = \frac{\cos^{2}{\left(2 u \right)}}{4} + \frac{\cos{\left(2 u \right)}}{2} + \frac{1}{4}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{2}{\left(2 u \right)}}{4}\, du = \frac{\int \cos^{2}{\left(2 u \right)}\, du}{4}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(2 u \right)} = \frac{\cos{\left(4 u \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(4 u \right)}}{2}\, du = \frac{\int \cos{\left(4 u \right)}\, du}{2}$
      
      Let $u = 4 u$ .
      
      Then let $du = 4 du$ 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:
      
      $\frac{\sin{\left(4 u \right)}}{4}$
      
      So, the result is: $\frac{\sin{\left(4 u \right)}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The result is: $\frac{u}{2} + \frac{\sin{\left(4 u \right)}}{8}$
      
      So, the result is: $\frac{u}{8} + \frac{\sin{\left(4 u \right)}}{32}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ 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:
      
      $\frac{\sin{\left(2 u \right)}}{2}$
      
      So, the result is: $\frac{\sin{\left(2 u \right)}}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{4}\, du = \frac{u}{4}$
      
      The result is: $\frac{3 u}{8} + \frac{\sin{\left(2 u \right)}}{4} + \frac{\sin{\left(4 u \right)}}{32}$
      
      So, the result is: $- \frac{3 u}{256} - \frac{\sin{\left(2 u \right)}}{128} - \frac{\sin{\left(4 u \right)}}{1024}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{3}{\left(u \right)}}{16}\, du = \frac{\int \cos^{3}{\left(u \right)}\, du}{16}$
      
      Rewrite the integrand:
      
      $\cos^{3}{\left(u \right)} = \left(1 - \sin^{2}{\left(u \right)}\right) \cos{\left(u \right)}$
      
      Let $u = \sin{\left(u \right)}$ .
      
      Then let $du = \cos{\left(u \right)} du$ and substitute $du$ :
      
      $\int \left(1 - u^{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      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: $- \frac{u^{3}}{3} + u$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(u \right)}}{3} + \sin{\left(u \right)}$
      
      So, the result is: $- \frac{\sin^{3}{\left(u \right)}}{48} + \frac{\sin{\left(u \right)}}{16}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(u \right)}}{16}\right)\, du = - \frac{\int \cos{\left(u \right)}\, du}{16}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $- \frac{\sin{\left(u \right)}}{16}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{32}\, du = \frac{u}{32}$
      
      The result is: $\frac{5 u}{256} - \frac{\sin{\left(2 u \right)}}{128} - \frac{\sin{\left(4 u \right)}}{1024} - \frac{\sin^{3}{\left(u \right)}}{48}$
      
      Now substitute $u$ back in:
      
      $\frac{5 x}{128} - \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{128} - \frac{\sin{\left(8 x \right)}}{1024}$
    So, the result is: $15 x - 8 \sin^{3}{\left(2 x \right)} - 3 \sin{\left(4 x \right)} - \frac{3 \sin{\left(8 x \right)}}{8}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 160 \sin^{4}{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 160 \int \sin^{4}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
    1. Rewrite the integrand:
      
      $\sin^{4}{\left(x \right)} \cos^{2}{\left(x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{2} \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)$
    2. Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $du$ :
      
      $\int \left(\frac{\cos^{3}{\left(u \right)}}{16} - \frac{\cos^{2}{\left(u \right)}}{16} - \frac{\cos{\left(u \right)}}{16} + \frac{1}{16}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{3}{\left(u \right)}}{16}\, du = \frac{\int \cos^{3}{\left(u \right)}\, du}{16}$
      
      Rewrite the integrand:
      
      $\cos^{3}{\left(u \right)} = \left(1 - \sin^{2}{\left(u \right)}\right) \cos{\left(u \right)}$
      
      Let $u = \sin{\left(u \right)}$ .
      
      Then let $du = \cos{\left(u \right)} du$ and substitute $du$ :
      
      $\int \left(1 - u^{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      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: $- \frac{u^{3}}{3} + u$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(u \right)}}{3} + \sin{\left(u \right)}$
      
      So, the result is: $- \frac{\sin^{3}{\left(u \right)}}{48} + \frac{\sin{\left(u \right)}}{16}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{2}{\left(u \right)}}{16}\right)\, du = - \frac{\int \cos^{2}{\left(u \right)}\, du}{16}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(u \right)} = \frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ 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:
      
      $\frac{\sin{\left(2 u \right)}}{2}$
      
      So, the result is: $\frac{\sin{\left(2 u \right)}}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The result is: $\frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}$
      
      So, the result is: $- \frac{u}{32} - \frac{\sin{\left(2 u \right)}}{64}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(u \right)}}{16}\right)\, du = - \frac{\int \cos{\left(u \right)}\, du}{16}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $- \frac{\sin{\left(u \right)}}{16}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{16}\, du = \frac{u}{16}$
      
      The result is: $\frac{u}{32} - \frac{\sin{\left(2 u \right)}}{64} - \frac{\sin^{3}{\left(u \right)}}{48}$
      
      Now substitute $u$ back in:
      
      $\frac{x}{16} - \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{64}$
    So, the result is: $- 10 x + \frac{10 \sin^{3}{\left(2 x \right)}}{3} + \frac{5 \sin{\left(4 x \right)}}{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 16 \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 16 \int \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
    1. Rewrite the integrand:
      
      $\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right) \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)$
    2. Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $du$ :
      
      $\int \left(\frac{1}{8} - \frac{\cos^{2}{\left(u \right)}}{8}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{8}\, du = \frac{u}{8}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{2}{\left(u \right)}}{8}\right)\, du = - \frac{\int \cos^{2}{\left(u \right)}\, du}{8}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(u \right)} = \frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ 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:
      
      $\frac{\sin{\left(2 u \right)}}{2}$
      
      So, the result is: $\frac{\sin{\left(2 u \right)}}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The result is: $\frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}$
      
      So, the result is: $- \frac{u}{16} - \frac{\sin{\left(2 u \right)}}{32}$
      
      The result is: $\frac{u}{16} - \frac{\sin{\left(2 u \right)}}{32}$
      
      Now substitute $u$ back in:
      
      $\frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32}$
    So, the result is: $2 x - \frac{\sin{\left(4 x \right)}}{2}$
  The result is: $- \frac{4 \sin^{5}{\left(2 x \right)}}{5} + \frac{2 \sin^{3}{\left(2 x \right)}}{3}$
Now simplify:

$\frac{2 \cdot \left(5 - 6 \sin^{2}{\left(2 x \right)}\right) \sin^{3}{\left(2 x \right)}}{15}$
Add the constant of integration:

$\frac{2 \cdot \left(5 - 6 \sin^{2}{\left(2 x \right)}\right) \sin^{3}{\left(2 x \right)}}{15}+ \mathrm{constant}$

The answer is:

$\frac{2 \cdot \left(5 - 6 \sin^{2}{\left(2 x \right)}\right) \sin^{3}{\left(2 x \right)}}{15}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

$$\int \sin{\left(2 x \right)} \sin{\left(8 x \right)}\, dx = C - \frac{4 \sin^{5}{\left(2 x \right)}}{5} + \frac{2 \sin^{3}{\left(2 x \right)}}{3}$$

Simplify

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The answer [src]

  2*cos(8)*sin(2)   cos(2)*sin(8)
- --------------- + -------------
         15               30

$$\frac{\sin{\left(8 \right)} \cos{\left(2 \right)}}{30} - \frac{2 \sin{\left(2 \right)} \cos{\left(8 \right)}}{15}$$

=

  2*cos(8)*sin(2)   cos(2)*sin(8)
- --------------- + -------------
         15               30

$$\frac{\sin{\left(8 \right)} \cos{\left(2 \right)}}{30} - \frac{2 \sin{\left(2 \right)} \cos{\left(8 \right)}}{15}$$

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Numerical answer [src]

0.003916430694558

0.003916430694558

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Identical expressions

Integral of sin(2x)sin(8x) dx

Limits of integration:

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Piecewise:

The solution

Method #1

Method #2

Method #1

Method #2

Method #3