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(sin^ three)*((6x)*cos6x)
( sinus of cubed ) multiply by ((6x) multiply by co sinus of e of 6x)
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(sin^3)((6x)cos6x)
(sin3)((6x)cos6x)
sin36xcos6x
sin^36xcos6x
(sin^3)*((6x)*cos6x)dx

Integral of (sin^3)((6x)cos6x) dx

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$$\int\limits_{0}^{0} \sin^{3}{\left(x \right)} 6 x \cos{\left(6 x \right)}\, dx$$

Integral(sin(x)^3*6*x*cos(6*x), (x, 0, 0))

Detail solution

The integral of a constant times a function is the constant times the integral of the function:

$\int \sin^{3}{\left(x \right)} 6 x \cos{\left(6 x \right)}\, dx = 6 \int x \sin^{3}{\left(x \right)} \cos{\left(6 x \right)}\, dx$
1. There are multiple ways to do this integral.
  Method #1
  1. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = \sin^{3}{\left(x \right)} \cos{\left(6 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 1$ .
    
    To find $v{\left(x \right)}$ :
    1. Rewrite the integrand:
      
      $\sin^{3}{\left(x \right)} \cos{\left(6 x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos{\left(6 x \right)}$
    2. Rewrite the integrand:
      
      $\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos{\left(6 x \right)} = - \sin{\left(x \right)} \cos^{2}{\left(x \right)} \cos{\left(6 x \right)} + \sin{\left(x \right)} \cos{\left(6 x \right)}$
    3. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin{\left(x \right)} \cos^{2}{\left(x \right)} \cos{\left(6 x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{2}{\left(x \right)} \cos{\left(6 x \right)}\, dx$
      
      Rewrite the integrand:
      
      $\sin{\left(x \right)} \cos^{2}{\left(x \right)} \cos{\left(6 x \right)} = 32 \sin{\left(x \right)} \cos^{8}{\left(x \right)} - 48 \sin{\left(x \right)} \cos^{6}{\left(x \right)} + 18 \sin{\left(x \right)} \cos^{4}{\left(x \right)} - \sin{\left(x \right)} \cos^{2}{\left(x \right)}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 32 \sin{\left(x \right)} \cos^{8}{\left(x \right)}\, dx = 32 \int \sin{\left(x \right)} \cos^{8}{\left(x \right)}\, dx$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u^{8}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{8}\right)\, du = - \int u^{8}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{8}\, du = \frac{u^{9}}{9}$
      
      So, the result is: $- \frac{u^{9}}{9}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{9}{\left(x \right)}}{9}$
      
      So, the result is: $- \frac{32 \cos^{9}{\left(x \right)}}{9}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 48 \sin{\left(x \right)} \cos^{6}{\left(x \right)}\right)\, dx = - 48 \int \sin{\left(x \right)} \cos^{6}{\left(x \right)}\, dx$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u^{6}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{6}\right)\, du = - \int u^{6}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      So, the result is: $- \frac{u^{7}}{7}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{7}{\left(x \right)}}{7}$
      
      So, the result is: $\frac{48 \cos^{7}{\left(x \right)}}{7}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 18 \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx = 18 \int \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u^{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{4}\right)\, du = - \int u^{4}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
      
      So, the result is: $- \frac{u^{5}}{5}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{5}{\left(x \right)}}{5}$
      
      So, the result is: $- \frac{18 \cos^{5}{\left(x \right)}}{5}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u^{2}\, du$
      
      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: $- \frac{u^{3}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{3}{\left(x \right)}}{3}$
      
      So, the result is: $\frac{\cos^{3}{\left(x \right)}}{3}$
      
      The result is: $- \frac{32 \cos^{9}{\left(x \right)}}{9} + \frac{48 \cos^{7}{\left(x \right)}}{7} - \frac{18 \cos^{5}{\left(x \right)}}{5} + \frac{\cos^{3}{\left(x \right)}}{3}$
      
      So, the result is: $\frac{32 \cos^{9}{\left(x \right)}}{9} - \frac{48 \cos^{7}{\left(x \right)}}{7} + \frac{18 \cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}$
      
      Rewrite the integrand:
      
      $\sin{\left(x \right)} \cos{\left(6 x \right)} = 32 \sin{\left(x \right)} \cos^{6}{\left(x \right)} - 48 \sin{\left(x \right)} \cos^{4}{\left(x \right)} + 18 \sin{\left(x \right)} \cos^{2}{\left(x \right)} - \sin{\left(x \right)}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 32 \sin{\left(x \right)} \cos^{6}{\left(x \right)}\, dx = 32 \int \sin{\left(x \right)} \cos^{6}{\left(x \right)}\, dx$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u^{6}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{6}\right)\, du = - \int u^{6}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      So, the result is: $- \frac{u^{7}}{7}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{7}{\left(x \right)}}{7}$
      
      So, the result is: $- \frac{32 \cos^{7}{\left(x \right)}}{7}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 48 \sin{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - 48 \int \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u^{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{4}\right)\, du = - \int u^{4}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
      
      So, the result is: $- \frac{u^{5}}{5}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{5}{\left(x \right)}}{5}$
      
      So, the result is: $\frac{48 \cos^{5}{\left(x \right)}}{5}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 18 \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 18 \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u^{2}\, du$
      
      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: $- \frac{u^{3}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{3}{\left(x \right)}}{3}$
      
      So, the result is: $- 6 \cos^{3}{\left(x \right)}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)}\, dx$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(x \right)}\, dx = - \cos{\left(x \right)}$
      
      So, the result is: $\cos{\left(x \right)}$
      
      The result is: $- \frac{32 \cos^{7}{\left(x \right)}}{7} + \frac{48 \cos^{5}{\left(x \right)}}{5} - 6 \cos^{3}{\left(x \right)} + \cos{\left(x \right)}$
      
      The result is: $\frac{32 \cos^{9}{\left(x \right)}}{9} - \frac{80 \cos^{7}{\left(x \right)}}{7} + \frac{66 \cos^{5}{\left(x \right)}}{5} - \frac{19 \cos^{3}{\left(x \right)}}{3} + \cos{\left(x \right)}$
    Now evaluate the sub-integral.
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{32 \cos^{9}{\left(x \right)}}{9}\, dx = \frac{32 \int \cos^{9}{\left(x \right)}\, dx}{9}$
      
      Rewrite the integrand:
      
      $\cos^{9}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right)^{4} \cos{\left(x \right)}$
      
      Rewrite the integrand:
      
      $\left(1 - \sin^{2}{\left(x \right)}\right)^{4} \cos{\left(x \right)} = \sin^{8}{\left(x \right)} \cos{\left(x \right)} - 4 \sin^{6}{\left(x \right)} \cos{\left(x \right)} + 6 \sin^{4}{\left(x \right)} \cos{\left(x \right)} - 4 \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos{\left(x \right)}$
      
      Integrate term-by-term:
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u^{8}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{8}\, du = \frac{u^{9}}{9}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{9}{\left(x \right)}}{9}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 4 \sin^{6}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 4 \int \sin^{6}{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u^{6}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{7}{\left(x \right)}}{7}$
      
      So, the result is: $- \frac{4 \sin^{7}{\left(x \right)}}{7}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 6 \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx = 6 \int \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ 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:
      
      $\frac{\sin^{5}{\left(x \right)}}{5}$
      
      So, the result is: $\frac{6 \sin^{5}{\left(x \right)}}{5}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 4 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 4 \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ 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:
      
      $\frac{\sin^{3}{\left(x \right)}}{3}$
      
      So, the result is: $- \frac{4 \sin^{3}{\left(x \right)}}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$
      
      The result is: $\frac{\sin^{9}{\left(x \right)}}{9} - \frac{4 \sin^{7}{\left(x \right)}}{7} + \frac{6 \sin^{5}{\left(x \right)}}{5} - \frac{4 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}$
      
      So, the result is: $\frac{32 \sin^{9}{\left(x \right)}}{81} - \frac{128 \sin^{7}{\left(x \right)}}{63} + \frac{64 \sin^{5}{\left(x \right)}}{15} - \frac{128 \sin^{3}{\left(x \right)}}{27} + \frac{32 \sin{\left(x \right)}}{9}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{80 \cos^{7}{\left(x \right)}}{7}\right)\, dx = - \frac{80 \int \cos^{7}{\left(x \right)}\, dx}{7}$
      
      Rewrite the integrand:
      
      $\cos^{7}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right)^{3} \cos{\left(x \right)}$
      
      Rewrite the integrand:
      
      $\left(1 - \sin^{2}{\left(x \right)}\right)^{3} \cos{\left(x \right)} = - \sin^{6}{\left(x \right)} \cos{\left(x \right)} + 3 \sin^{4}{\left(x \right)} \cos{\left(x \right)} - 3 \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos{\left(x \right)}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin^{6}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - \int \sin^{6}{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u^{6}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{7}{\left(x \right)}}{7}$
      
      So, the result is: $- \frac{\sin^{7}{\left(x \right)}}{7}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 3 \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx = 3 \int \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ 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:
      
      $\frac{\sin^{5}{\left(x \right)}}{5}$
      
      So, the result is: $\frac{3 \sin^{5}{\left(x \right)}}{5}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 3 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 3 \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ 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:
      
      $\frac{\sin^{3}{\left(x \right)}}{3}$
      
      So, the result is: $- \sin^{3}{\left(x \right)}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$
      
      The result is: $- \frac{\sin^{7}{\left(x \right)}}{7} + \frac{3 \sin^{5}{\left(x \right)}}{5} - \sin^{3}{\left(x \right)} + \sin{\left(x \right)}$
      
      So, the result is: $\frac{80 \sin^{7}{\left(x \right)}}{49} - \frac{48 \sin^{5}{\left(x \right)}}{7} + \frac{80 \sin^{3}{\left(x \right)}}{7} - \frac{80 \sin{\left(x \right)}}{7}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{66 \cos^{5}{\left(x \right)}}{5}\, dx = \frac{66 \int \cos^{5}{\left(x \right)}\, dx}{5}$
      
      Rewrite the integrand:
      
      $\cos^{5}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right)^{2} \cos{\left(x \right)}$
      
      Rewrite the integrand:
      
      $\left(1 - \sin^{2}{\left(x \right)}\right)^{2} \cos{\left(x \right)} = \sin^{4}{\left(x \right)} \cos{\left(x \right)} - 2 \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos{\left(x \right)}$
      
      Integrate term-by-term:
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ 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:
      
      $\frac{\sin^{5}{\left(x \right)}}{5}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 2 \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ 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:
      
      $\frac{\sin^{3}{\left(x \right)}}{3}$
      
      So, the result is: $- \frac{2 \sin^{3}{\left(x \right)}}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$
      
      The result is: $\frac{\sin^{5}{\left(x \right)}}{5} - \frac{2 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}$
      
      So, the result is: $\frac{66 \sin^{5}{\left(x \right)}}{25} - \frac{44 \sin^{3}{\left(x \right)}}{5} + \frac{66 \sin{\left(x \right)}}{5}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{19 \cos^{3}{\left(x \right)}}{3}\right)\, dx = - \frac{19 \int \cos^{3}{\left(x \right)}\, dx}{3}$
      
      Rewrite the integrand:
      
      $\cos^{3}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)}$
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ 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:
      
      $- \frac{\sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}$
      
      So, the result is: $\frac{19 \sin^{3}{\left(x \right)}}{9} - \frac{19 \sin{\left(x \right)}}{3}$
    1. The integral of cosine is sine:
      
      $\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$
    The result is: $\frac{32 \sin^{9}{\left(x \right)}}{81} - \frac{176 \sin^{7}{\left(x \right)}}{441} + \frac{26 \sin^{5}{\left(x \right)}}{525} - \frac{\sin^{3}{\left(x \right)}}{945} - \frac{2 \sin{\left(x \right)}}{315}$
  Method #2
  1. Rewrite the integrand:
    
    $x \sin^{3}{\left(x \right)} \cos{\left(6 x \right)} = 32 x \sin^{3}{\left(x \right)} \cos^{6}{\left(x \right)} - 48 x \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)} + 18 x \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)} - x \sin^{3}{\left(x \right)}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 32 x \sin^{3}{\left(x \right)} \cos^{6}{\left(x \right)}\, dx = 32 \int x \sin^{3}{\left(x \right)} \cos^{6}{\left(x \right)}\, dx$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = \sin^{3}{\left(x \right)} \cos^{6}{\left(x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Rewrite the integrand:
      
      $\sin^{3}{\left(x \right)} \cos^{6}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{6}{\left(x \right)}$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $du$ :
      
      $\int \left(u^{8} - u^{6}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{8}\, du = \frac{u^{9}}{9}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{6}\right)\, du = - \int u^{6}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      So, the result is: $- \frac{u^{7}}{7}$
      
      The result is: $\frac{u^{9}}{9} - \frac{u^{7}}{7}$
      
      Now substitute $u$ back in:
      
      $\frac{\cos^{9}{\left(x \right)}}{9} - \frac{\cos^{7}{\left(x \right)}}{7}$
      
      Now evaluate the sub-integral.
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{9}{\left(x \right)}}{9}\, dx = \frac{\int \cos^{9}{\left(x \right)}\, dx}{9}$
      
      Rewrite the integrand:
      
      $\cos^{9}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right)^{4} \cos{\left(x \right)}$
      
      Rewrite the integrand:
      
      $\left(1 - \sin^{2}{\left(x \right)}\right)^{4} \cos{\left(x \right)} = \sin^{8}{\left(x \right)} \cos{\left(x \right)} - 4 \sin^{6}{\left(x \right)} \cos{\left(x \right)} + 6 \sin^{4}{\left(x \right)} \cos{\left(x \right)} - 4 \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos{\left(x \right)}$
      
      Integrate term-by-term:
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u^{8}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{8}\, du = \frac{u^{9}}{9}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{9}{\left(x \right)}}{9}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 4 \sin^{6}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 4 \int \sin^{6}{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u^{6}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{7}{\left(x \right)}}{7}$
      
      So, the result is: $- \frac{4 \sin^{7}{\left(x \right)}}{7}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 6 \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx = 6 \int \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ 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:
      
      $\frac{\sin^{5}{\left(x \right)}}{5}$
      
      So, the result is: $\frac{6 \sin^{5}{\left(x \right)}}{5}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 4 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 4 \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ 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:
      
      $\frac{\sin^{3}{\left(x \right)}}{3}$
      
      So, the result is: $- \frac{4 \sin^{3}{\left(x \right)}}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$
      
      The result is: $\frac{\sin^{9}{\left(x \right)}}{9} - \frac{4 \sin^{7}{\left(x \right)}}{7} + \frac{6 \sin^{5}{\left(x \right)}}{5} - \frac{4 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}$
      
      So, the result is: $\frac{\sin^{9}{\left(x \right)}}{81} - \frac{4 \sin^{7}{\left(x \right)}}{63} + \frac{2 \sin^{5}{\left(x \right)}}{15} - \frac{4 \sin^{3}{\left(x \right)}}{27} + \frac{\sin{\left(x \right)}}{9}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{7}{\left(x \right)}}{7}\right)\, dx = - \frac{\int \cos^{7}{\left(x \right)}\, dx}{7}$
      
      Rewrite the integrand:
      
      $\cos^{7}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right)^{3} \cos{\left(x \right)}$
      
      Rewrite the integrand:
      
      $\left(1 - \sin^{2}{\left(x \right)}\right)^{3} \cos{\left(x \right)} = - \sin^{6}{\left(x \right)} \cos{\left(x \right)} + 3 \sin^{4}{\left(x \right)} \cos{\left(x \right)} - 3 \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos{\left(x \right)}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin^{6}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - \int \sin^{6}{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u^{6}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{7}{\left(x \right)}}{7}$
      
      So, the result is: $- \frac{\sin^{7}{\left(x \right)}}{7}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 3 \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx = 3 \int \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ 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:
      
      $\frac{\sin^{5}{\left(x \right)}}{5}$
      
      So, the result is: $\frac{3 \sin^{5}{\left(x \right)}}{5}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 3 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 3 \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ 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:
      
      $\frac{\sin^{3}{\left(x \right)}}{3}$
      
      So, the result is: $- \sin^{3}{\left(x \right)}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$
      
      The result is: $- \frac{\sin^{7}{\left(x \right)}}{7} + \frac{3 \sin^{5}{\left(x \right)}}{5} - \sin^{3}{\left(x \right)} + \sin{\left(x \right)}$
      
      So, the result is: $\frac{\sin^{7}{\left(x \right)}}{49} - \frac{3 \sin^{5}{\left(x \right)}}{35} + \frac{\sin^{3}{\left(x \right)}}{7} - \frac{\sin{\left(x \right)}}{7}$
      
      The result is: $\frac{\sin^{9}{\left(x \right)}}{81} - \frac{19 \sin^{7}{\left(x \right)}}{441} + \frac{\sin^{5}{\left(x \right)}}{21} - \frac{\sin^{3}{\left(x \right)}}{189} - \frac{2 \sin{\left(x \right)}}{63}$
      
      So, the result is: $32 x \left(\frac{\cos^{9}{\left(x \right)}}{9} - \frac{\cos^{7}{\left(x \right)}}{7}\right) - \frac{32 \sin^{9}{\left(x \right)}}{81} + \frac{608 \sin^{7}{\left(x \right)}}{441} - \frac{32 \sin^{5}{\left(x \right)}}{21} + \frac{32 \sin^{3}{\left(x \right)}}{189} + \frac{64 \sin{\left(x \right)}}{63}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 48 x \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - 48 \int x \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Rewrite the integrand:
      
      $\sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{4}{\left(x \right)}$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $du$ :
      
      $\int \left(u^{6} - u^{4}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{4}\right)\, du = - \int u^{4}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
      
      So, the result is: $- \frac{u^{5}}{5}$
      
      The result is: $\frac{u^{7}}{7} - \frac{u^{5}}{5}$
      
      Now substitute $u$ back in:
      
      $\frac{\cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}$
      
      Now evaluate the sub-integral.
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{7}{\left(x \right)}}{7}\, dx = \frac{\int \cos^{7}{\left(x \right)}\, dx}{7}$
      
      Rewrite the integrand:
      
      $\cos^{7}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right)^{3} \cos{\left(x \right)}$
      
      Rewrite the integrand:
      
      $\left(1 - \sin^{2}{\left(x \right)}\right)^{3} \cos{\left(x \right)} = - \sin^{6}{\left(x \right)} \cos{\left(x \right)} + 3 \sin^{4}{\left(x \right)} \cos{\left(x \right)} - 3 \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos{\left(x \right)}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin^{6}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - \int \sin^{6}{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u^{6}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{7}{\left(x \right)}}{7}$
      
      So, the result is: $- \frac{\sin^{7}{\left(x \right)}}{7}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 3 \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx = 3 \int \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ 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:
      
      $\frac{\sin^{5}{\left(x \right)}}{5}$
      
      So, the result is: $\frac{3 \sin^{5}{\left(x \right)}}{5}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 3 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 3 \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ 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:
      
      $\frac{\sin^{3}{\left(x \right)}}{3}$
      
      So, the result is: $- \sin^{3}{\left(x \right)}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$
      
      The result is: $- \frac{\sin^{7}{\left(x \right)}}{7} + \frac{3 \sin^{5}{\left(x \right)}}{5} - \sin^{3}{\left(x \right)} + \sin{\left(x \right)}$
      
      So, the result is: $- \frac{\sin^{7}{\left(x \right)}}{49} + \frac{3 \sin^{5}{\left(x \right)}}{35} - \frac{\sin^{3}{\left(x \right)}}{7} + \frac{\sin{\left(x \right)}}{7}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{5}{\left(x \right)}}{5}\right)\, dx = - \frac{\int \cos^{5}{\left(x \right)}\, dx}{5}$
      
      Rewrite the integrand:
      
      $\cos^{5}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right)^{2} \cos{\left(x \right)}$
      
      Rewrite the integrand:
      
      $\left(1 - \sin^{2}{\left(x \right)}\right)^{2} \cos{\left(x \right)} = \sin^{4}{\left(x \right)} \cos{\left(x \right)} - 2 \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos{\left(x \right)}$
      
      Integrate term-by-term:
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ 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:
      
      $\frac{\sin^{5}{\left(x \right)}}{5}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 2 \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ 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:
      
      $\frac{\sin^{3}{\left(x \right)}}{3}$
      
      So, the result is: $- \frac{2 \sin^{3}{\left(x \right)}}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$
      
      The result is: $\frac{\sin^{5}{\left(x \right)}}{5} - \frac{2 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}$
      
      So, the result is: $- \frac{\sin^{5}{\left(x \right)}}{25} + \frac{2 \sin^{3}{\left(x \right)}}{15} - \frac{\sin{\left(x \right)}}{5}$
      
      The result is: $- \frac{\sin^{7}{\left(x \right)}}{49} + \frac{8 \sin^{5}{\left(x \right)}}{175} - \frac{\sin^{3}{\left(x \right)}}{105} - \frac{2 \sin{\left(x \right)}}{35}$
      
      So, the result is: $- 48 x \left(\frac{\cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}\right) - \frac{48 \sin^{7}{\left(x \right)}}{49} + \frac{384 \sin^{5}{\left(x \right)}}{175} - \frac{16 \sin^{3}{\left(x \right)}}{35} - \frac{96 \sin{\left(x \right)}}{35}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 18 x \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 18 \int x \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Rewrite the integrand:
      
      $\sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{2}{\left(x \right)}$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $du$ :
      
      $\int \left(u^{4} - u^{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{2}\right)\, du = - \int u^{2}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      So, the result is: $- \frac{u^{3}}{3}$
      
      The result is: $\frac{u^{5}}{5} - \frac{u^{3}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}$
      
      Now evaluate the sub-integral.
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{5}{\left(x \right)}}{5}\, dx = \frac{\int \cos^{5}{\left(x \right)}\, dx}{5}$
      
      Rewrite the integrand:
      
      $\cos^{5}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right)^{2} \cos{\left(x \right)}$
      
      Rewrite the integrand:
      
      $\left(1 - \sin^{2}{\left(x \right)}\right)^{2} \cos{\left(x \right)} = \sin^{4}{\left(x \right)} \cos{\left(x \right)} - 2 \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos{\left(x \right)}$
      
      Integrate term-by-term:
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ 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:
      
      $\frac{\sin^{5}{\left(x \right)}}{5}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 2 \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ 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:
      
      $\frac{\sin^{3}{\left(x \right)}}{3}$
      
      So, the result is: $- \frac{2 \sin^{3}{\left(x \right)}}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$
      
      The result is: $\frac{\sin^{5}{\left(x \right)}}{5} - \frac{2 \sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}$
      
      So, the result is: $\frac{\sin^{5}{\left(x \right)}}{25} - \frac{2 \sin^{3}{\left(x \right)}}{15} + \frac{\sin{\left(x \right)}}{5}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{3}{\left(x \right)}}{3}\right)\, dx = - \frac{\int \cos^{3}{\left(x \right)}\, dx}{3}$
      
      Rewrite the integrand:
      
      $\cos^{3}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)}$
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ 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:
      
      $- \frac{\sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}$
      
      So, the result is: $\frac{\sin^{3}{\left(x \right)}}{9} - \frac{\sin{\left(x \right)}}{3}$
      
      The result is: $\frac{\sin^{5}{\left(x \right)}}{25} - \frac{\sin^{3}{\left(x \right)}}{45} - \frac{2 \sin{\left(x \right)}}{15}$
      
      So, the result is: $18 x \left(\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}\right) - \frac{18 \sin^{5}{\left(x \right)}}{25} + \frac{2 \sin^{3}{\left(x \right)}}{5} + \frac{12 \sin{\left(x \right)}}{5}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- x \sin^{3}{\left(x \right)}\right)\, dx = - \int x \sin^{3}{\left(x \right)}\, dx$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = \sin^{3}{\left(x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Rewrite the integrand:
      
      $\sin^{3}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)}$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $du$ :
      
      $\int \left(u^{2} - 1\right)\, du$
      
      Integrate term-by-term:
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \left(-1\right)\, du = - u$
      
      The result is: $\frac{u^{3}}{3} - u$
      
      Now substitute $u$ back in:
      
      $\frac{\cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}$
      
      Now evaluate the sub-integral.
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{3}{\left(x \right)}}{3}\, dx = \frac{\int \cos^{3}{\left(x \right)}\, dx}{3}$
      
      Rewrite the integrand:
      
      $\cos^{3}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)}$
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ 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:
      
      $- \frac{\sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}$
      
      So, the result is: $- \frac{\sin^{3}{\left(x \right)}}{9} + \frac{\sin{\left(x \right)}}{3}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \cos{\left(x \right)}\right)\, dx = - \int \cos{\left(x \right)}\, dx$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$
      
      So, the result is: $- \sin{\left(x \right)}$
      
      The result is: $- \frac{\sin^{3}{\left(x \right)}}{9} - \frac{2 \sin{\left(x \right)}}{3}$
      
      So, the result is: $- x \left(\frac{\cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}\right) - \frac{\sin^{3}{\left(x \right)}}{9} - \frac{2 \sin{\left(x \right)}}{3}$
    The result is: $- x \left(\frac{\cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}\right) + 18 x \left(\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}\right) - 48 x \left(\frac{\cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}\right) + 32 x \left(\frac{\cos^{9}{\left(x \right)}}{9} - \frac{\cos^{7}{\left(x \right)}}{7}\right) - \frac{32 \sin^{9}{\left(x \right)}}{81} + \frac{176 \sin^{7}{\left(x \right)}}{441} - \frac{26 \sin^{5}{\left(x \right)}}{525} + \frac{\sin^{3}{\left(x \right)}}{945} + \frac{2 \sin{\left(x \right)}}{315}$
So, the result is: $6 x \left(\frac{32 \cos^{9}{\left(x \right)}}{9} - \frac{80 \cos^{7}{\left(x \right)}}{7} + \frac{66 \cos^{5}{\left(x \right)}}{5} - \frac{19 \cos^{3}{\left(x \right)}}{3} + \cos{\left(x \right)}\right) - \frac{64 \sin^{9}{\left(x \right)}}{27} + \frac{352 \sin^{7}{\left(x \right)}}{147} - \frac{52 \sin^{5}{\left(x \right)}}{175} + \frac{2 \sin^{3}{\left(x \right)}}{315} + \frac{4 \sin{\left(x \right)}}{105}$
Now simplify:

$\frac{2 x \left(1120 \cos^{8}{\left(x \right)} - 3600 \cos^{6}{\left(x \right)} + 4158 \cos^{4}{\left(x \right)} - 1995 \cos^{2}{\left(x \right)} + 315\right) \cos{\left(x \right)}}{105} - \frac{64 \sin^{9}{\left(x \right)}}{27} + \frac{352 \sin^{7}{\left(x \right)}}{147} - \frac{52 \sin^{5}{\left(x \right)}}{175} + \frac{2 \sin^{3}{\left(x \right)}}{315} + \frac{4 \sin{\left(x \right)}}{105}$
Add the constant of integration:

$\frac{2 x \left(1120 \cos^{8}{\left(x \right)} - 3600 \cos^{6}{\left(x \right)} + 4158 \cos^{4}{\left(x \right)} - 1995 \cos^{2}{\left(x \right)} + 315\right) \cos{\left(x \right)}}{105} - \frac{64 \sin^{9}{\left(x \right)}}{27} + \frac{352 \sin^{7}{\left(x \right)}}{147} - \frac{52 \sin^{5}{\left(x \right)}}{175} + \frac{2 \sin^{3}{\left(x \right)}}{315} + \frac{4 \sin{\left(x \right)}}{105}+ \mathrm{constant}$

The answer is:

$\frac{2 x \left(1120 \cos^{8}{\left(x \right)} - 3600 \cos^{6}{\left(x \right)} + 4158 \cos^{4}{\left(x \right)} - 1995 \cos^{2}{\left(x \right)} + 315\right) \cos{\left(x \right)}}{105} - \frac{64 \sin^{9}{\left(x \right)}}{27} + \frac{352 \sin^{7}{\left(x \right)}}{147} - \frac{52 \sin^{5}{\left(x \right)}}{175} + \frac{2 \sin^{3}{\left(x \right)}}{315} + \frac{4 \sin{\left(x \right)}}{105}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                                                                                               
 |                                     9            5           3                        7          /        7            3            9            5            \
 |    3                          64*sin (x)   52*sin (x)   2*sin (x)   4*sin(x)   352*sin (x)       |  80*cos (x)   19*cos (x)   32*cos (x)   66*cos (x)         |
 | sin (x)*6*x*cos(6*x) dx = C - ---------- - ---------- + --------- + -------- + ----------- + 6*x*|- ---------- - ---------- + ---------- + ---------- + cos(x)|
 |                                   27          175          315        105          147           \      7            3            9            5              /
/

$$-{{1225\,\sin \left(9\,x\right)-11025\,x\,\cos \left(9\,x\right)- 6075\,\sin \left(7\,x\right)+42525\,x\,\cos \left(7\,x\right)+11907 \,\sin \left(5\,x\right)-59535\,x\,\cos \left(5\,x\right)-11025\, \sin \left(3\,x\right)+33075\,x\,\cos \left(3\,x\right)}\over{132300 }}$$

Simplify

The answer [src]

$$0$$

Упростить

Numerical answer [src]

0.0

0.0

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

Other calculators

How to use it?

Identical expressions

Integral of (sin^3)((6x)cos6x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Other calculators

How to use it?

Identical expressions

Integral of (sin^3)*((6x)*cos6x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Integral of (sin^3)((6x)cos6x) dx