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Integral of d{x}:
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Integral of cos(7x)cos(3x)
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cos(7x)cos(3x)
co sinus of e of (7x) co sinus of e of (3x)
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cos(7x)cos(3x)dx

Solving integrals/ cos(7x)cos(3x)

Integral of cos(7x)cos(3x) dx

The solution

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$$\int\limits_{0}^{1} \cos{\left(3 x \right)} \cos{\left(7 x \right)}\, dx$$

Integral(cos(7*x)*cos(3*x), (x, 0, 1))

Detail solution

Rewrite the integrand:

$\cos{\left(3 x \right)} \cos{\left(7 x \right)} = 256 \cos^{10}{\left(x \right)} - 640 \cos^{8}{\left(x \right)} + 560 \cos^{6}{\left(x \right)} - 196 \cos^{4}{\left(x \right)} + 21 \cos^{2}{\left(x \right)}$
Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 256 \cos^{10}{\left(x \right)}\, dx = 256 \int \cos^{10}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\cos^{10}{\left(x \right)} = \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{5}$
  2. There are multiple ways to do this integral.
    Method #1
    1. Rewrite the integrand:
      
      $\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{5} = \frac{\cos^{5}{\left(2 x \right)}}{32} + \frac{5 \cos^{4}{\left(2 x \right)}}{32} + \frac{5 \cos^{3}{\left(2 x \right)}}{16} + \frac{5 \cos^{2}{\left(2 x \right)}}{16} + \frac{5 \cos{\left(2 x \right)}}{32} + \frac{1}{32}$
    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:
      
      $\cos^{5}{\left(2 x \right)} = \left(1 - \sin^{2}{\left(2 x \right)}\right)^{2} \cos{\left(2 x \right)}$
      
      There are multiple ways to do this integral.
      
      Method #1
      
      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:
      
      $\frac{\sin^{5}{\left(2 x \right)}}{10} - \frac{\sin^{3}{\left(2 x \right)}}{3} + \frac{\sin{\left(2 x \right)}}{2}$
      
      Method #2
      
      Rewrite the integrand:
      
      $\left(1 - \sin^{2}{\left(2 x \right)}\right)^{2} \cos{\left(2 x \right)} = \sin^{4}{\left(2 x \right)} \cos{\left(2 x \right)} - 2 \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)} + \cos{\left(2 x \right)}$
      
      Integrate term-by-term:
      
      Let $u = \sin{\left(2 x \right)}$ .
      
      Then let $du = 2 \cos{\left(2 x \right)} dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{u^{4}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{4}}{2}\, du = \frac{\int u^{4}\, du}{2}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
      
      So, the result is: $\frac{u^{5}}{10}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{5}{\left(2 x \right)}}{10}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)}\right)\, dx = - 2 \int \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)}\, dx$
      
      Let $u = \sin{\left(2 x \right)}$ .
      
      Then let $du = 2 \cos{\left(2 x \right)} dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{u^{2}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{2}}{2}\, du = \frac{\int u^{2}\, du}{2}$
      
      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: $\frac{u^{3}}{6}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{3}{\left(2 x \right)}}{6}$
      
      So, the result is: $- \frac{\sin^{3}{\left(2 x \right)}}{3}$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ 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:
      
      $\frac{\sin{\left(2 x \right)}}{2}$
      
      The result is: $\frac{\sin^{5}{\left(2 x \right)}}{10} - \frac{\sin^{3}{\left(2 x \right)}}{3} + \frac{\sin{\left(2 x \right)}}{2}$
      
      Method #3
      
      Rewrite the integrand:
      
      $\left(1 - \sin^{2}{\left(2 x \right)}\right)^{2} \cos{\left(2 x \right)} = \sin^{4}{\left(2 x \right)} \cos{\left(2 x \right)} - 2 \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)} + \cos{\left(2 x \right)}$
      
      Integrate term-by-term:
      
      Let $u = \sin{\left(2 x \right)}$ .
      
      Then let $du = 2 \cos{\left(2 x \right)} dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{u^{4}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{4}}{2}\, du = \frac{\int u^{4}\, du}{2}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
      
      So, the result is: $\frac{u^{5}}{10}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{5}{\left(2 x \right)}}{10}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)}\right)\, dx = - 2 \int \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)}\, dx$
      
      Let $u = \sin{\left(2 x \right)}$ .
      
      Then let $du = 2 \cos{\left(2 x \right)} dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{u^{2}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{2}}{2}\, du = \frac{\int u^{2}\, du}{2}$
      
      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: $\frac{u^{3}}{6}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{3}{\left(2 x \right)}}{6}$
      
      So, the result is: $- \frac{\sin^{3}{\left(2 x \right)}}{3}$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ 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:
      
      $\frac{\sin{\left(2 x \right)}}{2}$
      
      The result is: $\frac{\sin^{5}{\left(2 x \right)}}{10} - \frac{\sin^{3}{\left(2 x \right)}}{3} + \frac{\sin{\left(2 x \right)}}{2}$
      
      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:
      
      $\int \frac{5 \cos^{4}{\left(2 x \right)}}{32}\, dx = \frac{5 \int \cos^{4}{\left(2 x \right)}\, dx}{32}$
      
      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 $\frac{du}{8}$ :
      
      $\int \frac{\cos{\left(u \right)}}{64}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{8}\, du = \frac{\int \cos{\left(u \right)}\, du}{8}$
      
      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 $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      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:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      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: $\frac{15 x}{256} + \frac{5 \sin{\left(4 x \right)}}{256} + \frac{5 \sin{\left(8 x \right)}}{2048}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{5 \cos^{3}{\left(2 x \right)}}{16}\, dx = \frac{5 \int \cos^{3}{\left(2 x \right)}\, dx}{16}$
      
      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: $- \frac{5 \sin^{3}{\left(2 x \right)}}{96} + \frac{5 \sin{\left(2 x \right)}}{32}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{5 \cos^{2}{\left(2 x \right)}}{16}\, dx = \frac{5 \int \cos^{2}{\left(2 x \right)}\, dx}{16}$
      
      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 $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      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:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      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: $\frac{5 x}{32} + \frac{5 \sin{\left(4 x \right)}}{128}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{5 \cos{\left(2 x \right)}}{32}\, dx = \frac{5 \int \cos{\left(2 x \right)}\, dx}{32}$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      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:
      
      $\frac{\sin{\left(2 x \right)}}{2}$
      
      So, the result is: $\frac{5 \sin{\left(2 x \right)}}{64}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{32}\, dx = \frac{x}{32}$
      
      The result is: $\frac{63 x}{256} + \frac{\sin^{5}{\left(2 x \right)}}{320} - \frac{\sin^{3}{\left(2 x \right)}}{16} + \frac{\sin{\left(2 x \right)}}{4} + \frac{15 \sin{\left(4 x \right)}}{256} + \frac{5 \sin{\left(8 x \right)}}{2048}$
    Method #2
    1. Rewrite the integrand:
      
      $\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{5} = \frac{\cos^{5}{\left(2 x \right)}}{32} + \frac{5 \cos^{4}{\left(2 x \right)}}{32} + \frac{5 \cos^{3}{\left(2 x \right)}}{16} + \frac{5 \cos^{2}{\left(2 x \right)}}{16} + \frac{5 \cos{\left(2 x \right)}}{32} + \frac{1}{32}$
    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:
      
      $\int \frac{5 \cos^{4}{\left(2 x \right)}}{32}\, dx = \frac{5 \int \cos^{4}{\left(2 x \right)}\, dx}{32}$
      
      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 $\frac{du}{8}$ :
      
      $\int \frac{\cos{\left(u \right)}}{64}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{8}\, du = \frac{\int \cos{\left(u \right)}\, du}{8}$
      
      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 $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      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:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      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: $\frac{15 x}{256} + \frac{5 \sin{\left(4 x \right)}}{256} + \frac{5 \sin{\left(8 x \right)}}{2048}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{5 \cos^{3}{\left(2 x \right)}}{16}\, dx = \frac{5 \int \cos^{3}{\left(2 x \right)}\, dx}{16}$
      
      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: $- \frac{5 \sin^{3}{\left(2 x \right)}}{96} + \frac{5 \sin{\left(2 x \right)}}{32}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{5 \cos^{2}{\left(2 x \right)}}{16}\, dx = \frac{5 \int \cos^{2}{\left(2 x \right)}\, dx}{16}$
      
      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 $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      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:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      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: $\frac{5 x}{32} + \frac{5 \sin{\left(4 x \right)}}{128}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{5 \cos{\left(2 x \right)}}{32}\, dx = \frac{5 \int \cos{\left(2 x \right)}\, dx}{32}$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      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:
      
      $\frac{\sin{\left(2 x \right)}}{2}$
      
      So, the result is: $\frac{5 \sin{\left(2 x \right)}}{64}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{32}\, dx = \frac{x}{32}$
      
      The result is: $\frac{63 x}{256} + \frac{\sin^{5}{\left(2 x \right)}}{320} - \frac{\sin^{3}{\left(2 x \right)}}{16} + \frac{\sin{\left(2 x \right)}}{4} + \frac{15 \sin{\left(4 x \right)}}{256} + \frac{5 \sin{\left(8 x \right)}}{2048}$
  So, the result is: $63 x + \frac{4 \sin^{5}{\left(2 x \right)}}{5} - 16 \sin^{3}{\left(2 x \right)} + 64 \sin{\left(2 x \right)} + 15 \sin{\left(4 x \right)} + \frac{5 \sin{\left(8 x \right)}}{8}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 640 \cos^{8}{\left(x \right)}\right)\, dx = - 640 \int \cos^{8}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\cos^{8}{\left(x \right)} = \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{4}$
  2. Rewrite the integrand:
    
    $\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{4} = \frac{\cos^{4}{\left(2 x \right)}}{16} + \frac{\cos^{3}{\left(2 x \right)}}{4} + \frac{3 \cos^{2}{\left(2 x \right)}}{8} + \frac{\cos{\left(2 x \right)}}{4} + \frac{1}{16}$
  3. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{4}{\left(2 x \right)}}{16}\, dx = \frac{\int \cos^{4}{\left(2 x \right)}\, dx}{16}$
      1. Rewrite the integrand:
        
        $\cos^{4}{\left(2 x \right)} = \left(\frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}\right)^{2}$
      2. Rewrite the integrand:
        
        $\left(\frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}\right)^{2} = \frac{\cos^{2}{\left(4 x \right)}}{4} + \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{4}$
      3. 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 $\frac{du}{8}$ :
        
        $\int \frac{\cos{\left(u \right)}}{64}\, du$
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \frac{\cos{\left(u \right)}}{8}\, du = \frac{\int \cos{\left(u \right)}\, du}{8}$
        
        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 $\frac{du}{4}$ :
        
        $\int \frac{\cos{\left(u \right)}}{16}\, du$
        
        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:
        
        $\frac{\sin{\left(4 x \right)}}{4}$
        
        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: $\frac{3 x}{128} + \frac{\sin{\left(4 x \right)}}{128} + \frac{\sin{\left(8 x \right)}}{1024}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{3}{\left(2 x \right)}}{4}\, dx = \frac{\int \cos^{3}{\left(2 x \right)}\, dx}{4}$
      1. Rewrite the integrand:
        
        $\cos^{3}{\left(2 x \right)} = \left(1 - \sin^{2}{\left(2 x \right)}\right) \cos{\left(2 x \right)}$
      2. 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: $- \frac{\sin^{3}{\left(2 x \right)}}{24} + \frac{\sin{\left(2 x \right)}}{8}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{3 \cos^{2}{\left(2 x \right)}}{8}\, dx = \frac{3 \int \cos^{2}{\left(2 x \right)}\, dx}{8}$
      1. Rewrite the integrand:
        
        $\cos^{2}{\left(2 x \right)} = \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}$
      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 $\frac{du}{4}$ :
        
        $\int \frac{\cos{\left(u \right)}}{16}\, du$
        
        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:
        
        $\frac{\sin{\left(4 x \right)}}{4}$
        
        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: $\frac{3 x}{16} + \frac{3 \sin{\left(4 x \right)}}{64}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 x \right)}}{4}\, dx = \frac{\int \cos{\left(2 x \right)}\, dx}{4}$
      1. Let $u = 2 x$ .
        
        Then let $du = 2 dx$ 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:
        
        $\frac{\sin{\left(2 x \right)}}{2}$
      So, the result is: $\frac{\sin{\left(2 x \right)}}{8}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{16}\, dx = \frac{x}{16}$
    The result is: $\frac{35 x}{128} - \frac{\sin^{3}{\left(2 x \right)}}{24} + \frac{\sin{\left(2 x \right)}}{4} + \frac{7 \sin{\left(4 x \right)}}{128} + \frac{\sin{\left(8 x \right)}}{1024}$
  So, the result is: $- 175 x + \frac{80 \sin^{3}{\left(2 x \right)}}{3} - 160 \sin{\left(2 x \right)} - 35 \sin{\left(4 x \right)} - \frac{5 \sin{\left(8 x \right)}}{8}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 560 \cos^{6}{\left(x \right)}\, dx = 560 \int \cos^{6}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\cos^{6}{\left(x \right)} = \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{3}$
  2. Rewrite the integrand:
    
    $\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{3} = \frac{\cos^{3}{\left(2 x \right)}}{8} + \frac{3 \cos^{2}{\left(2 x \right)}}{8} + \frac{3 \cos{\left(2 x \right)}}{8} + \frac{1}{8}$
  3. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{3}{\left(2 x \right)}}{8}\, dx = \frac{\int \cos^{3}{\left(2 x \right)}\, dx}{8}$
      1. Rewrite the integrand:
        
        $\cos^{3}{\left(2 x \right)} = \left(1 - \sin^{2}{\left(2 x \right)}\right) \cos{\left(2 x \right)}$
      2. 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: $- \frac{\sin^{3}{\left(2 x \right)}}{48} + \frac{\sin{\left(2 x \right)}}{16}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{3 \cos^{2}{\left(2 x \right)}}{8}\, dx = \frac{3 \int \cos^{2}{\left(2 x \right)}\, dx}{8}$
      1. Rewrite the integrand:
        
        $\cos^{2}{\left(2 x \right)} = \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}$
      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 $\frac{du}{4}$ :
        
        $\int \frac{\cos{\left(u \right)}}{16}\, du$
        
        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:
        
        $\frac{\sin{\left(4 x \right)}}{4}$
        
        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: $\frac{3 x}{16} + \frac{3 \sin{\left(4 x \right)}}{64}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{3 \cos{\left(2 x \right)}}{8}\, dx = \frac{3 \int \cos{\left(2 x \right)}\, dx}{8}$
      1. Let $u = 2 x$ .
        
        Then let $du = 2 dx$ 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:
        
        $\frac{\sin{\left(2 x \right)}}{2}$
      So, the result is: $\frac{3 \sin{\left(2 x \right)}}{16}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{8}\, dx = \frac{x}{8}$
    The result is: $\frac{5 x}{16} - \frac{\sin^{3}{\left(2 x \right)}}{48} + \frac{\sin{\left(2 x \right)}}{4} + \frac{3 \sin{\left(4 x \right)}}{64}$
  So, the result is: $175 x - \frac{35 \sin^{3}{\left(2 x \right)}}{3} + 140 \sin{\left(2 x \right)} + \frac{105 \sin{\left(4 x \right)}}{4}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 196 \cos^{4}{\left(x \right)}\right)\, dx = - 196 \int \cos^{4}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\cos^{4}{\left(x \right)} = \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{2}$
  2. Rewrite the integrand:
    
    $\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{2} = \frac{\cos^{2}{\left(2 x \right)}}{4} + \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{4}$
  3. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{2}{\left(2 x \right)}}{4}\, dx = \frac{\int \cos^{2}{\left(2 x \right)}\, dx}{4}$
      1. Rewrite the integrand:
        
        $\cos^{2}{\left(2 x \right)} = \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}$
      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 $\frac{du}{4}$ :
        
        $\int \frac{\cos{\left(u \right)}}{16}\, du$
        
        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:
        
        $\frac{\sin{\left(4 x \right)}}{4}$
        
        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: $\frac{x}{8} + \frac{\sin{\left(4 x \right)}}{32}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 x \right)}}{2}\, dx = \frac{\int \cos{\left(2 x \right)}\, dx}{2}$
      1. Let $u = 2 x$ .
        
        Then let $du = 2 dx$ 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:
        
        $\frac{\sin{\left(2 x \right)}}{2}$
      So, the result is: $\frac{\sin{\left(2 x \right)}}{4}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{4}\, dx = \frac{x}{4}$
    The result is: $\frac{3 x}{8} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{32}$
  So, the result is: $- \frac{147 x}{2} - 49 \sin{\left(2 x \right)} - \frac{49 \sin{\left(4 x \right)}}{8}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 21 \cos^{2}{\left(x \right)}\, dx = 21 \int \cos^{2}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\cos^{2}{\left(x \right)} = \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 x \right)}}{2}\, dx = \frac{\int \cos{\left(2 x \right)}\, dx}{2}$
      1. Let $u = 2 x$ .
        
        Then let $du = 2 dx$ 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:
        
        $\frac{\sin{\left(2 x \right)}}{2}$
      So, the result is: $\frac{\sin{\left(2 x \right)}}{4}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
    The result is: $\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4}$
  So, the result is: $\frac{21 x}{2} + \frac{21 \sin{\left(2 x \right)}}{4}$
The result is: $\frac{4 \sin^{5}{\left(2 x \right)}}{5} - \sin^{3}{\left(2 x \right)} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{8}$
Add the constant of integration:

$\frac{4 \sin^{5}{\left(2 x \right)}}{5} - \sin^{3}{\left(2 x \right)} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{8}+ \mathrm{constant}$

The answer is:

$\frac{4 \sin^{5}{\left(2 x \right)}}{5} - \sin^{3}{\left(2 x \right)} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{8}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                  5     
 |                               3        sin(2*x)   sin(4*x)   4*sin (2*x)
 | cos(7*x)*cos(3*x) dx = C - sin (2*x) + -------- + -------- + -----------
 |                                           4          8            5     
/

$${{\sin \left(10\,x\right)}\over{20}}+{{\sin \left(4\,x\right) }\over{8}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

  3*cos(7)*sin(3)   7*cos(3)*sin(7)
- --------------- + ---------------
         40                40

$${{2\,\sin 10+5\,\sin 4}\over{40}}$$

  3*cos(7)*sin(3)   7*cos(3)*sin(7)
- --------------- + ---------------
         40                40

$$\frac{7 \sin{\left(7 \right)} \cos{\left(3 \right)}}{40} - \frac{3 \sin{\left(3 \right)} \cos{\left(7 \right)}}{40}$$

Simplify

Numerical answer [src]

-0.12180136745796

-0.12180136745796

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of cos(7x)cos(3x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #3

Method #2