Integral of sin(3x)*sin(5x) dx

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$$\int\limits_{0}^{1} \sin{\left(3 x \right)} \sin{\left(5 x \right)}\, dx$$

Integral(sin(3*x)*sin(5*x), (x, 0, 1))

Detail solution

Rewrite the integrand:

$\sin{\left(3 x \right)} \sin{\left(5 x \right)} = - 64 \sin^{8}{\left(x \right)} + 128 \sin^{6}{\left(x \right)} - 80 \sin^{4}{\left(x \right)} + 15 \sin^{2}{\left(x \right)}$
Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 64 \sin^{8}{\left(x \right)}\right)\, dx = - 64 \int \sin^{8}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\sin^{8}{\left(x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{4}$
  2. There are multiple ways to do this integral.
    Method #1
    1. Rewrite the integrand:
      
      $\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{4} = \frac{\cos^{4}{\left(2 x \right)}}{16} - \frac{\cos^{3}{\left(2 x \right)}}{4} + \frac{3 \cos^{2}{\left(2 x \right)}}{8} - \frac{\cos{\left(2 x \right)}}{4} + \frac{1}{16}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{4}{\left(2 x \right)}}{16}\, dx = \frac{\int \cos^{4}{\left(2 x \right)}\, dx}{16}$
      
      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: $\frac{3 x}{128} + \frac{\sin{\left(4 x \right)}}{128} + \frac{\sin{\left(8 x \right)}}{1024}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{3}{\left(2 x \right)}}{4}\right)\, dx = - \frac{\int \cos^{3}{\left(2 x \right)}\, dx}{4}$
      
      Rewrite the integrand:
      
      $\cos^{3}{\left(2 x \right)} = \left(1 - \sin^{2}{\left(2 x \right)}\right) \cos{\left(2 x \right)}$
      
      There are multiple ways to do this integral.
      
      Method #1
      
      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:
      
      $- \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(2 x \right)}}{2}$
      
      Method #2
      
      Rewrite the integrand:
      
      $\left(1 - \sin^{2}{\left(2 x \right)}\right) \cos{\left(2 x \right)} = - \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)} + \cos{\left(2 x \right)}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)}\right)\, dx = - \int \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)}\, dx$
      
      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: $- \frac{\sin^{3}{\left(2 x \right)}}{6}$
      
      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: $- \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(2 x \right)}}{2}$
      
      Method #3
      
      Rewrite the integrand:
      
      $\left(1 - \sin^{2}{\left(2 x \right)}\right) \cos{\left(2 x \right)} = - \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)} + \cos{\left(2 x \right)}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)}\right)\, dx = - \int \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)}\, dx$
      
      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: $- \frac{\sin^{3}{\left(2 x \right)}}{6}$
      
      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: $- \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(2 x \right)}}{2}$
      
      So, the result is: $\frac{\sin^{3}{\left(2 x \right)}}{24} - \frac{\sin{\left(2 x \right)}}{8}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{3 \cos^{2}{\left(2 x \right)}}{8}\, dx = \frac{3 \int \cos^{2}{\left(2 x \right)}\, dx}{8}$
      
      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: $\frac{3 x}{16} + \frac{3 \sin{\left(4 x \right)}}{64}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(2 x \right)}}{4}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{4}$
      
      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: $- \frac{\sin{\left(2 x \right)}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{16}\, dx = \frac{x}{16}$
      
      The result is: $\frac{35 x}{128} + \frac{\sin^{3}{\left(2 x \right)}}{24} - \frac{\sin{\left(2 x \right)}}{4} + \frac{7 \sin{\left(4 x \right)}}{128} + \frac{\sin{\left(8 x \right)}}{1024}$
    Method #2
    1. Rewrite the integrand:
      
      $\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{4} = \frac{\cos^{4}{\left(2 x \right)}}{16} - \frac{\cos^{3}{\left(2 x \right)}}{4} + \frac{3 \cos^{2}{\left(2 x \right)}}{8} - \frac{\cos{\left(2 x \right)}}{4} + \frac{1}{16}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{4}{\left(2 x \right)}}{16}\, dx = \frac{\int \cos^{4}{\left(2 x \right)}\, dx}{16}$
      
      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: $\frac{3 x}{128} + \frac{\sin{\left(4 x \right)}}{128} + \frac{\sin{\left(8 x \right)}}{1024}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{3}{\left(2 x \right)}}{4}\right)\, dx = - \frac{\int \cos^{3}{\left(2 x \right)}\, dx}{4}$
      
      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: $\frac{\sin^{3}{\left(2 x \right)}}{24} - \frac{\sin{\left(2 x \right)}}{8}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{3 \cos^{2}{\left(2 x \right)}}{8}\, dx = \frac{3 \int \cos^{2}{\left(2 x \right)}\, dx}{8}$
      
      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: $\frac{3 x}{16} + \frac{3 \sin{\left(4 x \right)}}{64}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(2 x \right)}}{4}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{4}$
      
      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: $- \frac{\sin{\left(2 x \right)}}{8}$
      
      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: $- \frac{35 x}{2} - \frac{8 \sin^{3}{\left(2 x \right)}}{3} + 16 \sin{\left(2 x \right)} - \frac{7 \sin{\left(4 x \right)}}{2} - \frac{\sin{\left(8 x \right)}}{16}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 128 \sin^{6}{\left(x \right)}\, dx = 128 \int \sin^{6}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\sin^{6}{\left(x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{3}$
  2. Rewrite the integrand:
    
    $\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{3} = - \frac{\cos^{3}{\left(2 x \right)}}{8} + \frac{3 \cos^{2}{\left(2 x \right)}}{8} - \frac{3 \cos{\left(2 x \right)}}{8} + \frac{1}{8}$
  3. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{3}{\left(2 x \right)}}{8}\right)\, dx = - \frac{\int \cos^{3}{\left(2 x \right)}\, dx}{8}$
      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:
      
      $\int \left(- \frac{3 \cos{\left(2 x \right)}}{8}\right)\, dx = - \frac{3 \int \cos{\left(2 x \right)}\, dx}{8}$
      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: $40 x + \frac{8 \sin^{3}{\left(2 x \right)}}{3} - 32 \sin{\left(2 x \right)} + 6 \sin{\left(4 x \right)}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 80 \sin^{4}{\left(x \right)}\right)\, dx = - 80 \int \sin^{4}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\sin^{4}{\left(x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{2}$
  2. Rewrite the integrand:
    
    $\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{2} = \frac{\cos^{2}{\left(2 x \right)}}{4} - \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{4}$
  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:
      
      $\int \left(- \frac{\cos{\left(2 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{2}$
      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: $- 30 x + 20 \sin{\left(2 x \right)} - \frac{5 \sin{\left(4 x \right)}}{2}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 15 \sin^{2}{\left(x \right)}\, dx = 15 \int \sin^{2}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\sin^{2}{\left(x \right)} = \frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}$
  2. Integrate term-by-term:
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(2 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{2}$
      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: $- \frac{\sin{\left(2 x \right)}}{4}$
    The result is: $\frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4}$
  So, the result is: $\frac{15 x}{2} - \frac{15 \sin{\left(2 x \right)}}{4}$
The result is: $\frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(8 x \right)}}{16}$
Add the constant of integration:

$\frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(8 x \right)}}{16}+ \mathrm{constant}$

The answer is:

$\frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(8 x \right)}}{16}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                              
 |                            sin(8*x)   sin(2*x)
 | sin(3*x)*sin(5*x) dx = C - -------- + --------
 |                               16         4    
/

$${{\sin \left(2\,x\right)}\over{4}}-{{\sin \left(8\,x\right)}\over{ 16}}$$

Simplify

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The answer [src]

  5*cos(5)*sin(3)   3*cos(3)*sin(5)
- --------------- + ---------------
         16                16

$$-{{\sin 8-4\,\sin 2}\over{16}}$$

=

  5*cos(5)*sin(3)   3*cos(3)*sin(5)
- --------------- + ---------------
         16                16

$$- \frac{5 \sin{\left(3 \right)} \cos{\left(5 \right)}}{16} + \frac{3 \sin{\left(5 \right)} \cos{\left(3 \right)}}{16}$$

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Numerical answer [src]

0.165489466292459

0.165489466292459

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Integral of sin(3x)*sin(5x) dx

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Method #1

Method #1

Method #2

Method #3

Method #2