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Integral of d{x}:
Integral of sqrt(2x+4)
Integral of (sin^4)x×(cos^4)x
Integral of (3x^3-4x^2+3x)/(x^2+1)
Integral of sin^5(2x)
Identical expressions
(sin^ four)x×(cos^ four)x
( sinus of to the power of 4)x×( co sinus of e of to the power of 4)x
( sinus of to the power of four)x×( co sinus of e of to the power of four)x
(sin4)x×(cos4)x
sin4x×cos4x
(sin⁴)x×(cos⁴)x
sin^4x×cos^4x
(sin^4)x×(cos^4)xdx

Integral of (sin^4)x×(cos^4)x dx

The solution

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0

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

Integral(sin(x)^4*x*cos(x)^4*x, (x, 0, 1))

Detail solution

Use integration by parts:

$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$

Let $u{\left(x \right)} = x^{2}$ and let $\operatorname{dv}{\left(x \right)} = \sin^{4}{\left(x \right)} \cos^{4}{\left(x \right)}$ .

Then $\operatorname{du}{\left(x \right)} = 2 x$ .

To find $v{\left(x \right)}$ :
1. Rewrite the integrand:
  
  $\sin^{4}{\left(x \right)} \cos^{4}{\left(x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{2} \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{2}$
2. There are multiple ways to do this integral.
  Method #1
  1. Rewrite the integrand:
    
    $\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{2} \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{2} = \frac{\cos^{4}{\left(2 x \right)}}{16} - \frac{\cos^{2}{\left(2 x \right)}}{8} + \frac{1}{16}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{4}{\left(2 x \right)}}{16}\, dx = \frac{\int \cos^{4}{\left(2 x \right)}\, dx}{16}$
      
      Rewrite the integrand:
      
      $\cos^{4}{\left(2 x \right)} = \left(\frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}\right)^{2}$
      
      There are multiple ways to do this integral.
      
      Method #1
      
      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: $\frac{3 x}{8} + \frac{\sin{\left(4 x \right)}}{8} + \frac{\sin{\left(8 x \right)}}{64}$
      
      Method #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: $\frac{3 x}{128} + \frac{\sin{\left(4 x \right)}}{128} + \frac{\sin{\left(8 x \right)}}{1024}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{2}{\left(2 x \right)}}{8}\right)\, dx = - \frac{\int \cos^{2}{\left(2 x \right)}\, dx}{8}$
      
      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: $- \frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64}$
    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: $\frac{3 x}{128} - \frac{\sin{\left(4 x \right)}}{128} + \frac{\sin{\left(8 x \right)}}{1024}$
  Method #2
  1. Rewrite the integrand:
    
    $\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{2} \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{2} = \frac{\cos^{4}{\left(2 x \right)}}{16} - \frac{\cos^{2}{\left(2 x \right)}}{8} + \frac{1}{16}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{4}{\left(2 x \right)}}{16}\, dx = \frac{\int \cos^{4}{\left(2 x \right)}\, dx}{16}$
      
      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: $\frac{3 x}{128} + \frac{\sin{\left(4 x \right)}}{128} + \frac{\sin{\left(8 x \right)}}{1024}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{2}{\left(2 x \right)}}{8}\right)\, dx = - \frac{\int \cos^{2}{\left(2 x \right)}\, dx}{8}$
      
      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: $- \frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64}$
    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: $\frac{3 x}{128} - \frac{\sin{\left(4 x \right)}}{128} + \frac{\sin{\left(8 x \right)}}{1024}$
Now evaluate the sub-integral.
Use integration by parts:

$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$

Let $u{\left(x \right)} = \frac{x}{512}$ and let $\operatorname{dv}{\left(x \right)} = 24 x - 8 \sin{\left(4 x \right)} + \sin{\left(8 x \right)}$ .

Then $\operatorname{du}{\left(x \right)} = \frac{1}{512}$ .

To find $v{\left(x \right)}$ :
1. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 24 x\, dx = 24 \int x\, dx$
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
    So, the result is: $12 x^{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 8 \sin{\left(4 x \right)}\right)\, dx = - 8 \int \sin{\left(4 x \right)}\, dx$
    1. Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{16}\, du$
      1. The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \frac{\sin{\left(u \right)}}{4}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
        
        The integral of sine is negative cosine:
        
        $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
        
        So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(4 x \right)}}{4}$
    So, the result is: $2 \cos{\left(4 x \right)}$
  1. Let $u = 8 x$ .
    
    Then let $du = 8 dx$ and substitute $\frac{du}{8}$ :
    
    $\int \frac{\sin{\left(u \right)}}{64}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{8}\, du = \frac{\int \sin{\left(u \right)}\, du}{8}$
      1. The integral of sine is negative cosine:
        
        $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      So, the result is: $- \frac{\cos{\left(u \right)}}{8}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(8 x \right)}}{8}$
  The result is: $12 x^{2} + 2 \cos{\left(4 x \right)} - \frac{\cos{\left(8 x \right)}}{8}$
Now evaluate the sub-integral.
Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{3 x^{2}}{128}\, dx = \frac{3 \int x^{2}\, dx}{128}$
  1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
    
    $\int x^{2}\, dx = \frac{x^{3}}{3}$
  So, the result is: $\frac{x^{3}}{128}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{\cos{\left(4 x \right)}}{256}\, dx = \frac{\int \cos{\left(4 x \right)}\, dx}{256}$
  1. Let $u = 4 x$ .
    
    Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
    
    $\int \frac{\cos{\left(u \right)}}{16}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      1. 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: $\frac{\sin{\left(4 x \right)}}{1024}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{\cos{\left(8 x \right)}}{4096}\right)\, dx = - \frac{\int \cos{\left(8 x \right)}\, dx}{4096}$
  1. Let $u = 8 x$ .
    
    Then let $du = 8 dx$ and substitute $\frac{du}{8}$ :
    
    $\int \frac{\cos{\left(u \right)}}{64}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{8}\, du = \frac{\int \cos{\left(u \right)}\, du}{8}$
      1. 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: $- \frac{\sin{\left(8 x \right)}}{32768}$
The result is: $\frac{x^{3}}{128} + \frac{\sin{\left(4 x \right)}}{1024} - \frac{\sin{\left(8 x \right)}}{32768}$
Now simplify:

$\frac{x^{3}}{128} - \frac{x^{2} \sin{\left(4 x \right)}}{128} + \frac{x^{2} \sin{\left(8 x \right)}}{1024} - \frac{x \cos{\left(4 x \right)}}{256} + \frac{x \cos{\left(8 x \right)}}{4096} + \frac{\sin{\left(4 x \right)}}{1024} - \frac{\sin{\left(8 x \right)}}{32768}$
Add the constant of integration:

$\frac{x^{3}}{128} - \frac{x^{2} \sin{\left(4 x \right)}}{128} + \frac{x^{2} \sin{\left(8 x \right)}}{1024} - \frac{x \cos{\left(4 x \right)}}{256} + \frac{x \cos{\left(8 x \right)}}{4096} + \frac{\sin{\left(4 x \right)}}{1024} - \frac{\sin{\left(8 x \right)}}{32768}+ \mathrm{constant}$

The answer is:

$\frac{x^{3}}{128} - \frac{x^{2} \sin{\left(4 x \right)}}{128} + \frac{x^{2} \sin{\left(8 x \right)}}{1024} - \frac{x \cos{\left(4 x \right)}}{256} + \frac{x \cos{\left(8 x \right)}}{4096} + \frac{\sin{\left(4 x \right)}}{1024} - \frac{\sin{\left(8 x \right)}}{32768}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                              /                 2   cos(8*x)\
 |                                           3                                                 x*|2*cos(4*x) + 12*x  - --------|
 |    4         4               sin(8*x)    x    sin(4*x)    2 /  sin(4*x)   sin(8*x)   3*x\     \                        8    /
 | sin (x)*x*cos (x)*x dx = C - -------- + --- + -------- + x *|- -------- + -------- + ---| - ---------------------------------
 |                               32768     128     1024        \    128        1024     128/                  512               
/

$${{\left(32\,x^2-1\right)\,\sin \left(8\,x\right)+8\,x\,\cos \left(8 \,x\right)+\left(32-256\,x^2\right)\,\sin \left(4\,x\right)-128\,x\, \cos \left(4\,x\right)+256\,x^3}\over{32768}}$$

Simplify

The answer [src]

      8            8             5       3            7                   2       6            6       2            7                    4       4             3       5   
17*cos (1)   17*sin (1)   329*cos (1)*sin (1)   81*cos (1)*sin(1)   41*cos (1)*sin (1)   41*cos (1)*sin (1)   81*sin (1)*cos(1)   211*cos (1)*sin (1)   329*cos (1)*sin (1)
---------- + ---------- - ------------------- - ----------------- + ------------------ + ------------------ + ----------------- + ------------------- + -------------------
   4096         4096              4096                 4096                1024                 1024                 4096                 2048                  4096

$${{31\,\sin 8+8\,\cos 8-224\,\sin 4-128\,\cos 4+256}\over{32768}}$$

      8            8             5       3            7                   2       6            6       2            7                    4       4             3       5   
17*cos (1)   17*sin (1)   329*cos (1)*sin (1)   81*cos (1)*sin(1)   41*cos (1)*sin (1)   41*cos (1)*sin (1)   81*sin (1)*cos(1)   211*cos (1)*sin (1)   329*cos (1)*sin (1)
---------- + ---------- - ------------------- - ----------------- + ------------------ + ------------------ + ----------------- + ------------------- + -------------------
   4096         4096              4096                 4096                1024                 1024                 4096                 2048                  4096

$$- \frac{329 \sin^{3}{\left(1 \right)} \cos^{5}{\left(1 \right)}}{4096} - \frac{81 \sin{\left(1 \right)} \cos^{7}{\left(1 \right)}}{4096} + \frac{17 \cos^{8}{\left(1 \right)}}{4096} + \frac{41 \sin^{2}{\left(1 \right)} \cos^{6}{\left(1 \right)}}{1024} + \frac{17 \sin^{8}{\left(1 \right)}}{4096} + \frac{81 \sin^{7}{\left(1 \right)} \cos{\left(1 \right)}}{4096} + \frac{41 \sin^{6}{\left(1 \right)} \cos^{2}{\left(1 \right)}}{1024} + \frac{211 \sin^{4}{\left(1 \right)} \cos^{4}{\left(1 \right)}}{2048} + \frac{329 \sin^{5}{\left(1 \right)} \cos^{3}{\left(1 \right)}}{4096}$$

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Numerical answer [src]

0.0164397048277092

0.0164397048277092

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

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Identical expressions

Integral of (sin^4)x×(cos^4)x dx

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The solution

Method #1

Method #1

Method #2

Method #2