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Identical expressions
x^ two *cos4(x)
x squared multiply by co sinus of e of 4(x)
x to the power of two multiply by co sinus of e of 4(x)
x2*cos4(x)
x2*cos4x
x²*cos4(x)
x to the power of 2*cos4(x)
x^2cos4(x)
x2cos4(x)
x2cos4x
x^2cos4x
x^2*cos4(x)dx

Solving integrals/ x^2*cos4(x)

Integral of x^2*cos4(x) dx

The solution

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 2*pi             
   /              
  |               
  |   2    4      
  |  x *cos (x) dx
  |               
 /                
 pi

$$\int\limits_{\pi}^{2 \pi} x^{2} \cos^{4}{\left(x \right)}\, dx$$

Integral(x^2*cos(x)^4, (x, pi, 2*pi))

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 $\operatorname{dv}{\left(x \right)} = \cos^{4}{\left(x \right)}$ .

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

To find $v{\left(x \right)}$ :
1. Rewrite the integrand:
  
  $\cos^{4}{\left(x \right)} = \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{2}$
2. There are multiple ways to do this integral.
  Method #1
  1. Rewrite the integrand:
    
    $\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{2} = \frac{\cos^{2}{\left(2 x \right)}}{4} + \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{4}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{2}{\left(2 x \right)}}{4}\, dx = \frac{\int \cos^{2}{\left(2 x \right)}\, dx}{4}$
      
      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: $\frac{x}{8} + \frac{\sin{\left(4 x \right)}}{32}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 x \right)}}{2}\, dx = \frac{\int \cos{\left(2 x \right)}\, dx}{2}$
      
      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: $\frac{3 x}{8} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{32}$
  Method #2
  1. Rewrite the integrand:
    
    $\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)^{2} = \frac{\cos^{2}{\left(2 x \right)}}{4} + \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{4}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{2}{\left(2 x \right)}}{4}\, dx = \frac{\int \cos^{2}{\left(2 x \right)}\, dx}{4}$
      
      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: $\frac{x}{8} + \frac{\sin{\left(4 x \right)}}{32}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 x \right)}}{2}\, dx = \frac{\int \cos{\left(2 x \right)}\, dx}{2}$
      
      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: $\frac{3 x}{8} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{32}$
Now evaluate the sub-integral.
Use integration by parts:

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

Let $u{\left(x \right)} = \frac{x}{16}$ and let $\operatorname{dv}{\left(x \right)} = 12 x + 8 \sin{\left(2 x \right)} + \sin{\left(4 x \right)}$ .

Then $\operatorname{du}{\left(x \right)} = \frac{1}{16}$ .

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:
    
    $\int 12 x\, dx = 12 \int x\, dx$
    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: $6 x^{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 8 \sin{\left(2 x \right)}\, dx = 8 \int \sin{\left(2 x \right)}\, dx$
    1. There are multiple ways to do this integral.
      Method #1
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{2}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(2 x \right)}}{2}$
      Method #2
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 \sin{\left(x \right)} \cos{\left(x \right)}\, dx = 2 \int \sin{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      There are multiple ways to do this integral.
      
      Method #1
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u\right)\, du = - \int u\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      So, the result is: $- \frac{u^{2}}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{2}{\left(x \right)}}{2}$
      
      Method #2
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{2}{\left(x \right)}}{2}$
      
      So, the result is: $- \cos^{2}{\left(x \right)}$
    So, the result is: $- 4 \cos{\left(2 x \right)}$
  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:
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      1. The integral of sine is negative cosine:
        
        $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(4 x \right)}}{4}$
  The result is: $6 x^{2} - 4 \cos{\left(2 x \right)} - \frac{\cos{\left(4 x \right)}}{4}$
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:
  
  $\int \frac{3 x^{2}}{8}\, dx = \frac{3 \int x^{2}\, dx}{8}$
  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: $\frac{x^{3}}{8}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{\cos{\left(2 x \right)}}{4}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{4}$
  1. Let $u = 2 x$ .
    
    Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{\cos{\left(u \right)}}{4}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      1. The integral of cosine is sine:
        
        $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
    Now substitute $u$ back in:
    
    $\frac{\sin{\left(2 x \right)}}{2}$
  So, the result is: $- \frac{\sin{\left(2 x \right)}}{8}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{\cos{\left(4 x \right)}}{64}\right)\, dx = - \frac{\int \cos{\left(4 x \right)}\, dx}{64}$
  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: $- \frac{\sin{\left(4 x \right)}}{256}$
The result is: $\frac{x^{3}}{8} - \frac{\sin{\left(2 x \right)}}{8} - \frac{\sin{\left(4 x \right)}}{256}$
Now simplify:

$\frac{x^{3}}{8} + \frac{x^{2} \sin{\left(2 x \right)}}{4} + \frac{x^{2} \sin{\left(4 x \right)}}{32} + \frac{x \cos{\left(2 x \right)}}{4} + \frac{x \cos{\left(4 x \right)}}{64} - \frac{\sin{\left(2 x \right)}}{8} - \frac{\sin{\left(4 x \right)}}{256}$
Add the constant of integration:

$\frac{x^{3}}{8} + \frac{x^{2} \sin{\left(2 x \right)}}{4} + \frac{x^{2} \sin{\left(4 x \right)}}{32} + \frac{x \cos{\left(2 x \right)}}{4} + \frac{x \cos{\left(4 x \right)}}{64} - \frac{\sin{\left(2 x \right)}}{8} - \frac{\sin{\left(4 x \right)}}{256}+ \mathrm{constant}$

The answer is:

$\frac{x^{3}}{8} + \frac{x^{2} \sin{\left(2 x \right)}}{4} + \frac{x^{2} \sin{\left(4 x \right)}}{32} + \frac{x \cos{\left(2 x \right)}}{4} + \frac{x \cos{\left(4 x \right)}}{64} - \frac{\sin{\left(2 x \right)}}{8} - \frac{\sin{\left(4 x \right)}}{256}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                  /                 2   cos(4*x)\
 |                                            3                                    x*|-4*cos(2*x) + 6*x  - --------|
 |  2    4             sin(2*x)   sin(4*x)   x     2 /sin(2*x)   sin(4*x)   3*x\     \                        4    /
 | x *cos (x) dx = C - -------- - -------- + -- + x *|-------- + -------- + ---| - ---------------------------------
 |                        8         256      8       \   4          32       8 /                   16               
/

$${{\left(8\,x^2-1\right)\,\sin \left(4\,x\right)+4\,x\,\cos \left(4 \,x\right)+\left(64\,x^2-32\right)\,\sin \left(2\,x\right)+64\,x\, \cos \left(2\,x\right)+32\,x^3}\over{256}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

    3        
7*pi    17*pi
----- + -----
  8       64

$${{\left(32\,\pi^2-1\right)\,\sin \left(8\,\pi\right)+8\,\pi\,\cos \left(8\,\pi\right)+\left(256\,\pi^2-32\right)\,\sin \left(4\,\pi \right)+128\,\pi\,\cos \left(4\,\pi\right)+256\,\pi^3}\over{256}}-{{ \left(8\,\pi^2-1\right)\,\sin \left(4\,\pi\right)+4\,\pi\,\cos \left(4\,\pi\right)+\left(64\,\pi^2-32\right)\,\sin \left(2\,\pi \right)+64\,\pi\,\cos \left(2\,\pi\right)+32\,\pi^3}\over{256}}$$

    3        
7*pi    17*pi
----- + -----
  8       64

$$\frac{17 \pi}{64} + \frac{7 \pi^{3}}{8}$$

Упростить

Numerical answer [src]

27.9649776438721

27.9649776438721

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of x^2*cos4(x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #1

Method #2

Method #1

Method #2