Integral of (sin4x)(sin6x)dx dx

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$$\int\limits_{0}^{1} \sin{\left(4 x \right)} \sin{\left(6 x \right)}\, dx$$

Integral(sin(4*x)*sin(6*x), (x, 0, 1))

Detail solution

Rewrite the integrand:

$\sin{\left(4 x \right)} \sin{\left(6 x \right)} = - 256 \sin^{8}{\left(x \right)} \cos^{2}{\left(x \right)} + 384 \sin^{6}{\left(x \right)} \cos^{2}{\left(x \right)} - 176 \sin^{4}{\left(x \right)} \cos^{2}{\left(x \right)} + 24 \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}$
Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 256 \sin^{8}{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 256 \int \sin^{8}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
  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. There are multiple ways to do this integral.
    Method #1
    1. 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 $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      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 $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      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 $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      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 $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      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 $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      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:
      
      $\frac{7 x}{256} + \frac{\sin^{5}{\left(2 x \right)}}{320} - \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{256} - \frac{3 \sin{\left(8 x \right)}}{2048}$
    Method #2
    1. Rewrite the integrand:
      
      $\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{4} \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right) = \frac{\cos^{5}{\left(2 x \right)}}{32} - \frac{3 \cos^{4}{\left(2 x \right)}}{32} + \frac{\cos^{3}{\left(2 x \right)}}{16} + \frac{\cos^{2}{\left(2 x \right)}}{16} - \frac{3 \cos{\left(2 x \right)}}{32} + \frac{1}{32}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{5}{\left(2 x \right)}}{32}\, dx = \frac{\int \cos^{5}{\left(2 x \right)}\, dx}{32}$
      
      Rewrite the integrand:
      
      $\cos^{5}{\left(2 x \right)} = \left(1 - \sin^{2}{\left(2 x \right)}\right)^{2} \cos{\left(2 x \right)}$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $du$ :
      
      $\int \left(\frac{\sin^{4}{\left(u \right)} \cos{\left(u \right)}}{2} - \sin^{2}{\left(u \right)} \cos{\left(u \right)} + \frac{\cos{\left(u \right)}}{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin^{4}{\left(u \right)} \cos{\left(u \right)}}{2}\, du = \frac{\int \sin^{4}{\left(u \right)} \cos{\left(u \right)}\, du}{2}$
      
      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: $\frac{\sin^{5}{\left(u \right)}}{10}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin^{2}{\left(u \right)} \cos{\left(u \right)}\right)\, du = - \int \sin^{2}{\left(u \right)} \cos{\left(u \right)}\, du$
      
      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: $- \frac{\sin^{3}{\left(u \right)}}{3}$
      
      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: $\frac{\sin{\left(u \right)}}{2}$
      
      The result is: $\frac{\sin^{5}{\left(u \right)}}{10} - \frac{\sin^{3}{\left(u \right)}}{3} + \frac{\sin{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{5}{\left(2 x \right)}}{10} - \frac{\sin^{3}{\left(2 x \right)}}{3} + \frac{\sin{\left(2 x \right)}}{2}$
      
      So, the result is: $\frac{\sin^{5}{\left(2 x \right)}}{320} - \frac{\sin^{3}{\left(2 x \right)}}{96} + \frac{\sin{\left(2 x \right)}}{64}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{3 \cos^{4}{\left(2 x \right)}}{32}\right)\, dx = - \frac{3 \int \cos^{4}{\left(2 x \right)}\, dx}{32}$
      
      Rewrite the integrand:
      
      $\cos^{4}{\left(2 x \right)} = \left(\frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}\right)^{2}$
      
      Rewrite the integrand:
      
      $\left(\frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}\right)^{2} = \frac{\cos^{2}{\left(4 x \right)}}{4} + \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{4}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{2}{\left(4 x \right)}}{4}\, dx = \frac{\int \cos^{2}{\left(4 x \right)}\, dx}{4}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(4 x \right)} = \frac{\cos{\left(8 x \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(8 x \right)}}{2}\, dx = \frac{\int \cos{\left(8 x \right)}\, dx}{2}$
      
      Let $u = 8 x$ .
      
      Then let $du = 8 dx$ and substitute $\frac{du}{8}$ :
      
      $\int \frac{\cos{\left(u \right)}}{8}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{8}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{8}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(8 x \right)}}{8}$
      
      So, the result is: $\frac{\sin{\left(8 x \right)}}{16}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{x}{2} + \frac{\sin{\left(8 x \right)}}{16}$
      
      So, the result is: $\frac{x}{8} + \frac{\sin{\left(8 x \right)}}{64}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(4 x \right)}}{2}\, dx = \frac{\int \cos{\left(4 x \right)}\, dx}{2}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      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:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      So, the result is: $\frac{\sin{\left(4 x \right)}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{4}\, dx = \frac{x}{4}$
      
      The result is: $\frac{3 x}{8} + \frac{\sin{\left(4 x \right)}}{8} + \frac{\sin{\left(8 x \right)}}{64}$
      
      So, the result is: $- \frac{9 x}{256} - \frac{3 \sin{\left(4 x \right)}}{256} - \frac{3 \sin{\left(8 x \right)}}{2048}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{3}{\left(2 x \right)}}{16}\, dx = \frac{\int \cos^{3}{\left(2 x \right)}\, dx}{16}$
      
      Rewrite the integrand:
      
      $\cos^{3}{\left(2 x \right)} = \left(1 - \sin^{2}{\left(2 x \right)}\right) \cos{\left(2 x \right)}$
      
      Let $u = \sin{\left(2 x \right)}$ .
      
      Then let $du = 2 \cos{\left(2 x \right)} dx$ and substitute $du$ :
      
      $\int \left(\frac{1}{2} - \frac{u^{2}}{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{u^{2}}{2}\right)\, du = - \frac{\int u^{2}\, du}{2}$
      
      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: $- \frac{u^{3}}{6}$
      
      The result is: $- \frac{u^{3}}{6} + \frac{u}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(2 x \right)}}{2}$
      
      So, the result is: $- \frac{\sin^{3}{\left(2 x \right)}}{96} + \frac{\sin{\left(2 x \right)}}{32}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{2}{\left(2 x \right)}}{16}\, dx = \frac{\int \cos^{2}{\left(2 x \right)}\, dx}{16}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(2 x \right)} = \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(4 x \right)}}{2}\, dx = \frac{\int \cos{\left(4 x \right)}\, dx}{2}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      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:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      So, the result is: $\frac{\sin{\left(4 x \right)}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{x}{2} + \frac{\sin{\left(4 x \right)}}{8}$
      
      So, the result is: $\frac{x}{32} + \frac{\sin{\left(4 x \right)}}{128}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{3 \cos{\left(2 x \right)}}{32}\right)\, dx = - \frac{3 \int \cos{\left(2 x \right)}\, dx}{32}$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      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:
      
      $\frac{\sin{\left(2 x \right)}}{2}$
      
      So, the result is: $- \frac{3 \sin{\left(2 x \right)}}{64}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{32}\, dx = \frac{x}{32}$
      
      The result is: $\frac{7 x}{256} + \frac{\sin^{5}{\left(2 x \right)}}{320} - \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{256} - \frac{3 \sin{\left(8 x \right)}}{2048}$
    Method #3
    1. Rewrite the integrand:
      
      $\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{4} \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right) = \frac{\cos^{5}{\left(2 x \right)}}{32} - \frac{3 \cos^{4}{\left(2 x \right)}}{32} + \frac{\cos^{3}{\left(2 x \right)}}{16} + \frac{\cos^{2}{\left(2 x \right)}}{16} - \frac{3 \cos{\left(2 x \right)}}{32} + \frac{1}{32}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{5}{\left(2 x \right)}}{32}\, dx = \frac{\int \cos^{5}{\left(2 x \right)}\, dx}{32}$
      
      Rewrite the integrand:
      
      $\cos^{5}{\left(2 x \right)} = \left(1 - \sin^{2}{\left(2 x \right)}\right)^{2} \cos{\left(2 x \right)}$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $du$ :
      
      $\int \left(\frac{\sin^{4}{\left(u \right)} \cos{\left(u \right)}}{2} - \sin^{2}{\left(u \right)} \cos{\left(u \right)} + \frac{\cos{\left(u \right)}}{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin^{4}{\left(u \right)} \cos{\left(u \right)}}{2}\, du = \frac{\int \sin^{4}{\left(u \right)} \cos{\left(u \right)}\, du}{2}$
      
      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: $\frac{\sin^{5}{\left(u \right)}}{10}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin^{2}{\left(u \right)} \cos{\left(u \right)}\right)\, du = - \int \sin^{2}{\left(u \right)} \cos{\left(u \right)}\, du$
      
      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: $- \frac{\sin^{3}{\left(u \right)}}{3}$
      
      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: $\frac{\sin{\left(u \right)}}{2}$
      
      The result is: $\frac{\sin^{5}{\left(u \right)}}{10} - \frac{\sin^{3}{\left(u \right)}}{3} + \frac{\sin{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{5}{\left(2 x \right)}}{10} - \frac{\sin^{3}{\left(2 x \right)}}{3} + \frac{\sin{\left(2 x \right)}}{2}$
      
      So, the result is: $\frac{\sin^{5}{\left(2 x \right)}}{320} - \frac{\sin^{3}{\left(2 x \right)}}{96} + \frac{\sin{\left(2 x \right)}}{64}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{3 \cos^{4}{\left(2 x \right)}}{32}\right)\, dx = - \frac{3 \int \cos^{4}{\left(2 x \right)}\, dx}{32}$
      
      Rewrite the integrand:
      
      $\cos^{4}{\left(2 x \right)} = \left(\frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}\right)^{2}$
      
      Rewrite the integrand:
      
      $\left(\frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}\right)^{2} = \frac{\cos^{2}{\left(4 x \right)}}{4} + \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{4}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{2}{\left(4 x \right)}}{4}\, dx = \frac{\int \cos^{2}{\left(4 x \right)}\, dx}{4}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(4 x \right)} = \frac{\cos{\left(8 x \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(8 x \right)}}{2}\, dx = \frac{\int \cos{\left(8 x \right)}\, dx}{2}$
      
      Let $u = 8 x$ .
      
      Then let $du = 8 dx$ and substitute $\frac{du}{8}$ :
      
      $\int \frac{\cos{\left(u \right)}}{8}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{8}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{8}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(8 x \right)}}{8}$
      
      So, the result is: $\frac{\sin{\left(8 x \right)}}{16}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{x}{2} + \frac{\sin{\left(8 x \right)}}{16}$
      
      So, the result is: $\frac{x}{8} + \frac{\sin{\left(8 x \right)}}{64}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(4 x \right)}}{2}\, dx = \frac{\int \cos{\left(4 x \right)}\, dx}{2}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      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:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      So, the result is: $\frac{\sin{\left(4 x \right)}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{4}\, dx = \frac{x}{4}$
      
      The result is: $\frac{3 x}{8} + \frac{\sin{\left(4 x \right)}}{8} + \frac{\sin{\left(8 x \right)}}{64}$
      
      So, the result is: $- \frac{9 x}{256} - \frac{3 \sin{\left(4 x \right)}}{256} - \frac{3 \sin{\left(8 x \right)}}{2048}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{3}{\left(2 x \right)}}{16}\, dx = \frac{\int \cos^{3}{\left(2 x \right)}\, dx}{16}$
      
      Rewrite the integrand:
      
      $\cos^{3}{\left(2 x \right)} = \left(1 - \sin^{2}{\left(2 x \right)}\right) \cos{\left(2 x \right)}$
      
      Let $u = \sin{\left(2 x \right)}$ .
      
      Then let $du = 2 \cos{\left(2 x \right)} dx$ and substitute $du$ :
      
      $\int \left(\frac{1}{2} - \frac{u^{2}}{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{u^{2}}{2}\right)\, du = - \frac{\int u^{2}\, du}{2}$
      
      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: $- \frac{u^{3}}{6}$
      
      The result is: $- \frac{u^{3}}{6} + \frac{u}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(2 x \right)}}{2}$
      
      So, the result is: $- \frac{\sin^{3}{\left(2 x \right)}}{96} + \frac{\sin{\left(2 x \right)}}{32}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{2}{\left(2 x \right)}}{16}\, dx = \frac{\int \cos^{2}{\left(2 x \right)}\, dx}{16}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(2 x \right)} = \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(4 x \right)}}{2}\, dx = \frac{\int \cos{\left(4 x \right)}\, dx}{2}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      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:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      So, the result is: $\frac{\sin{\left(4 x \right)}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{x}{2} + \frac{\sin{\left(4 x \right)}}{8}$
      
      So, the result is: $\frac{x}{32} + \frac{\sin{\left(4 x \right)}}{128}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{3 \cos{\left(2 x \right)}}{32}\right)\, dx = - \frac{3 \int \cos{\left(2 x \right)}\, dx}{32}$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      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:
      
      $\frac{\sin{\left(2 x \right)}}{2}$
      
      So, the result is: $- \frac{3 \sin{\left(2 x \right)}}{64}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{32}\, dx = \frac{x}{32}$
      
      The result is: $\frac{7 x}{256} + \frac{\sin^{5}{\left(2 x \right)}}{320} - \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{256} - \frac{3 \sin{\left(8 x \right)}}{2048}$
  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 $du$ :
    
    $\int \left(- \frac{\cos^{4}{\left(u \right)}}{32} + \frac{\cos^{3}{\left(u \right)}}{16} - \frac{\cos{\left(u \right)}}{16} + \frac{1}{32}\right)\, du$
    1. Integrate term-by-term:
      1. The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \left(- \frac{\cos^{4}{\left(u \right)}}{32}\right)\, du = - \frac{\int \cos^{4}{\left(u \right)}\, du}{32}$
        
        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 $\frac{du}{4}$ :
        
        $\int \frac{\cos{\left(u \right)}}{4}\, du$
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
        
        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 $\frac{du}{2}$ :
        
        $\int \frac{\cos{\left(u \right)}}{2}\, du$
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
        
        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: $- \frac{3 u}{256} - \frac{\sin{\left(2 u \right)}}{128} - \frac{\sin{\left(4 u \right)}}{1024}$
      1. 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: $- \frac{\sin^{3}{\left(u \right)}}{48} + \frac{\sin{\left(u \right)}}{16}$
      1. 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: $- \frac{\sin{\left(u \right)}}{16}$
      1. 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:
  
  $\int \left(- 176 \sin^{4}{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 176 \int \sin^{4}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
  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 $du$ :
    
    $\int \left(\frac{\cos^{3}{\left(u \right)}}{16} - \frac{\cos^{2}{\left(u \right)}}{16} - \frac{\cos{\left(u \right)}}{16} + \frac{1}{16}\right)\, du$
    1. Integrate term-by-term:
      1. 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: $- \frac{\sin^{3}{\left(u \right)}}{48} + \frac{\sin{\left(u \right)}}{16}$
      1. 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 $\frac{du}{2}$ :
        
        $\int \frac{\cos{\left(u \right)}}{2}\, du$
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
        
        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: $- \frac{u}{32} - \frac{\sin{\left(2 u \right)}}{64}$
      1. 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: $- \frac{\sin{\left(u \right)}}{16}$
      1. 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: $- 11 x + \frac{11 \sin^{3}{\left(2 x \right)}}{3} + \frac{11 \sin{\left(4 x \right)}}{4}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 24 \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 24 \int \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
  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 $du$ :
    
    $\int \left(\frac{1}{8} - \frac{\cos^{2}{\left(u \right)}}{8}\right)\, du$
    1. Integrate term-by-term:
      1. The integral of a constant is the constant times the variable of integration:
        
        $\int \frac{1}{8}\, du = \frac{u}{8}$
      1. 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 $\frac{du}{2}$ :
        
        $\int \frac{\cos{\left(u \right)}}{2}\, du$
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
        
        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: $3 x - \frac{3 \sin{\left(4 x \right)}}{4}$
The result is: $- \frac{4 \sin^{5}{\left(2 x \right)}}{5} + \sin^{3}{\left(2 x \right)}$
Add the constant of integration:

$- \frac{4 \sin^{5}{\left(2 x \right)}}{5} + \sin^{3}{\left(2 x \right)}+ \mathrm{constant}$

The answer is:

$- \frac{4 \sin^{5}{\left(2 x \right)}}{5} + \sin^{3}{\left(2 x \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                            5     
 |                               3        4*sin (2*x)
 | sin(4*x)*sin(6*x) dx = C + sin (2*x) - -----------
 |                                             5     
/

$$\int \sin{\left(4 x \right)} \sin{\left(6 x \right)}\, dx = C - \frac{4 \sin^{5}{\left(2 x \right)}}{5} + \sin^{3}{\left(2 x \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

  3*cos(6)*sin(4)   cos(4)*sin(6)
- --------------- + -------------
         10               5

$$\frac{\sin{\left(6 \right)} \cos{\left(4 \right)}}{5} - \frac{3 \sin{\left(4 \right)} \cos{\left(6 \right)}}{10}$$

=

  3*cos(6)*sin(4)   cos(4)*sin(6)
- --------------- + -------------
         10               5

$$\frac{\sin{\left(6 \right)} \cos{\left(4 \right)}}{5} - \frac{3 \sin{\left(4 \right)} \cos{\left(6 \right)}}{10}$$

-3*cos(6)*sin(4)/10 + cos(4)*sin(6)/5

Numerical answer [src]

0.254525412250889

0.254525412250889

Other calculators

How to use it?

Identical expressions

Integral of (sin4x)(sin6x)dx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2

Method #3