Integral of x^2*sin(2x) dx

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 |  x *sin(2*x) dx
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0

\int\limits_{0}^{1} x^{2} \sin{\left(2 x \right)}\, dx

Integral(x^2*sin(2*x), (x, 0, 1))

Detail solution

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 $u{\left(x \right)} = x^{2}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(2 x \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = 2 x$ .
  
  To find $v{\left(x \right)}$ :
  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:
    
    $\int 2 \sin{\left(x \right)} \cos{\left(x \right)}\, dx = 2 \int \sin{\left(x \right)} \cos{\left(x \right)}\, dx$
    
    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:
    
    $- \frac{\cos^{2}{\left(x \right)}}{2}$
    
    So, the result is: $- \cos^{2}{\left(x \right)}$
  Now evaluate the sub-integral.
2. 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 $\operatorname{dv}{\left(x \right)} = \cos{\left(2 x \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = -1$ .
  
  To find $v{\left(x \right)}$ :
  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:
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      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:
    
    $\frac{\sin{\left(2 x \right)}}{2}$
  Now evaluate the sub-integral.
3. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{\sin{\left(2 x \right)}}{2}\right)\, dx = - \frac{\int \sin{\left(2 x \right)}\, dx}{2}$
  1. Let $u = 2 x$ .
    
    Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{\sin{\left(u \right)}}{4}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{2}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
      
      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:
    
    $- \frac{\cos{\left(2 x \right)}}{2}$
  So, the result is: $\frac{\cos{\left(2 x \right)}}{4}$
Method #2
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 2 x^{2} \sin{\left(x \right)} \cos{\left(x \right)}\, dx = 2 \int x^{2} \sin{\left(x \right)} \cos{\left(x \right)}\, dx$
  1. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(x \right)} = x^{2}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(x \right)} \cos{\left(x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 2 x$ .
    
    To find $v{\left(x \right)}$ :
    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:
      
      $- \frac{\cos^{2}{\left(x \right)}}{2}$
    Now evaluate the sub-integral.
  2. 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 $\operatorname{dv}{\left(x \right)} = \cos^{2}{\left(x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = -1$ .
    
    To find $v{\left(x \right)}$ :
    1. Rewrite the integrand:
      
      $\cos^{2}{\left(x \right)} = \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 x \right)}}{2}\, dx = \frac{\int \cos{\left(2 x \right)}\, dx}{2}$
      
      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: $\frac{\sin{\left(2 x \right)}}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4}$
    Now evaluate the sub-integral.
  3. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{x}{2}\right)\, dx = - \frac{\int x\, dx}{2}$
      
      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: $- \frac{x^{2}}{4}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\sin{\left(2 x \right)}}{4}\right)\, dx = - \frac{\int \sin{\left(2 x \right)}\, dx}{4}$
      
      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:
      
      $- \frac{\cos{\left(2 x \right)}}{2}$
      
      So, the result is: $\frac{\cos{\left(2 x \right)}}{8}$
    The result is: $- \frac{x^{2}}{4} + \frac{\cos{\left(2 x \right)}}{8}$
  So, the result is: $- x^{2} \cos^{2}{\left(x \right)} - \frac{x^{2}}{2} + 2 x \left(\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4}\right) + \frac{\cos{\left(2 x \right)}}{4}$
Add the constant of integration:

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

The answer is:

- \frac{x^{2} \cos{\left(2 x \right)}}{2} + \frac{x \sin{\left(2 x \right)}}{2} + \frac{\cos{\left(2 x \right)}}{4}+ \mathrm{constant}

The answer (Indefinite) [src]

  /                                                        
 |                                               2         
 |  2                   cos(2*x)   x*sin(2*x)   x *cos(2*x)
 | x *sin(2*x) dx = C + -------- + ---------- - -----------
 |                         4           2             2     
/

{{4\,x\,\sin \left(2\,x\right)+\left(2-4\,x^2\right)\,\cos \left(2 \,x\right)}\over{8}}

Simplify

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Plot the graph f(x)

The answer [src]

  1   sin(2)   cos(2)
- - + ------ - ------
  4     2        4

{{2\,\sin 2-\cos 2}\over{4}}-{{1}\over{4}}

=

  1   sin(2)   cos(2)
- - + ------ - ------
  4     2        4

- \frac{1}{4} - \frac{\cos{\left(2 \right)}}{4} + \frac{\sin{\left(2 \right)}}{2}

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Numerical answer [src]

0.308685422549626

0.308685422549626

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Identical expressions

Integral of x^2*sin(2x) dx

Limits of integration:

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The solution

Method #1

Method #1

Method #2

Method #2