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Identical expressions
x^3sin2xdx
x cubed sinus of 2xdx
x3sin2xdx
x³sin2xdx
x to the power of 3sin2xdx

Solving integrals/ x^3sin2xdx

Integral of x^3sin2xdx dx

The solution

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  1               
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 |   3            
 |  x *sin(2*x) dx
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0

$$\int\limits_{0}^{1} x^{3} \sin{\left(2 x \right)}\, dx$$

Integral(x^3*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 $u{\left(x \right)} = x^{3}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(2 x \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = 3 x^{2}$ .
  
  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 $\frac{du}{2}$ :
    
    $\int \frac{\sin{\left(u \right)}}{2}\, du$
    
    The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
    
    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 $- du$ :
    
    $\int \left(- u\right)\, du$
    
    The integral of a constant times a function is the constant times the integral of the function:
    
    $\int u\, du = - \int u\, du$
    
    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 $u{\left(x \right)} = - \frac{3 x^{2}}{2}$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(2 x \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = - 3 x$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 2 x$ .
    
    Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{\cos{\left(u \right)}}{2}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      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. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = - \frac{3 x}{2}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(2 x \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = - \frac{3}{2}$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 2 x$ .
    
    Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{\sin{\left(u \right)}}{2}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
      
      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:
    
    $- \frac{\cos{\left(2 x \right)}}{2}$
  Now evaluate the sub-integral.
4. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{3 \cos{\left(2 x \right)}}{4}\, dx = \frac{3 \int \cos{\left(2 x \right)}\, dx}{4}$
  1. Let $u = 2 x$ .
    
    Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{\cos{\left(u \right)}}{2}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
    Now substitute $u$ back in:
    
    $\frac{\sin{\left(2 x \right)}}{2}$
  So, the result is: $\frac{3 \sin{\left(2 x \right)}}{8}$
Method #2
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 2 x^{3} \sin{\left(x \right)} \cos{\left(x \right)}\, dx = 2 \int x^{3} \sin{\left(x \right)} \cos{\left(x \right)}\, dx$
  1. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(x \right)} = x^{3}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(x \right)} \cos{\left(x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 3 x^{2}$ .
    
    To find $v{\left(x \right)}$ :
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(2 x \right)}}{2}\, dx = \frac{\int \sin{\left(2 x \right)}\, dx}{2}$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\sin{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
      
      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: $- \frac{\cos{\left(2 x \right)}}{4}$
    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 $u{\left(x \right)} = - \frac{3 x^{2}}{4}$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(2 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = - \frac{3 x}{2}$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      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. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(x \right)} = - \frac{3 x}{4}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(2 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = - \frac{3}{4}$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\sin{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
      
      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:
      
      $- \frac{\cos{\left(2 x \right)}}{2}$
    Now evaluate the sub-integral.
  4. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{3 \cos{\left(2 x \right)}}{8}\, dx = \frac{3 \int \cos{\left(2 x \right)}\, dx}{8}$
    1. Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(2 x \right)}}{2}$
    So, the result is: $\frac{3 \sin{\left(2 x \right)}}{16}$
  So, the result is: $- \frac{x^{3} \cos{\left(2 x \right)}}{2} + \frac{3 x^{2} \sin{\left(2 x \right)}}{4} + \frac{3 x \cos{\left(2 x \right)}}{4} - \frac{3 \sin{\left(2 x \right)}}{8}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

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

$$\int x^{3} \sin{\left(2 x \right)}\, dx = C - \frac{x^{3} \cos{\left(2 x \right)}}{2} + \frac{3 x^{2} \sin{\left(2 x \right)}}{4} + \frac{3 x \cos{\left(2 x \right)}}{4} - \frac{3 \sin{\left(2 x \right)}}{8}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

cos(2)   3*sin(2)
------ + --------
  4         8

$$\frac{\cos{\left(2 \right)}}{4} + \frac{3 \sin{\left(2 \right)}}{8}$$

cos(2)   3*sin(2)
------ + --------
  4         8

$$\frac{\cos{\left(2 \right)}}{4} + \frac{3 \sin{\left(2 \right)}}{8}$$

cos(2)/4 + 3*sin(2)/8

Numerical answer [src]

0.236949825922845

0.236949825922845

The graph

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

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How to use it?

Identical expressions

Integral of x^3sin2xdx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2