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Integral of d{x}:
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Integral of e^5*x*dx
Integral of x^2cos2x
Integral of log(x+sqrt(1+x^2))
Identical expressions
x^2cos2x
x squared co sinus of e of 2x
x2cos2x
x²cos2x
x to the power of 2cos2x
x^2cos2xdx

Solving integrals/ x^2cos2x

Integral of x^2cos2x dx

The solution

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  1               
  /               
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 |   2            
 |  x *cos(2*x) dx
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/                 
0

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

Integral(x^2*cos(2*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 $\operatorname{dv}{\left(x \right)} = \cos{\left(2 x \right)}$ .

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

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:
    
    $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
    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:
  
  $\frac{\sin{\left(2 x \right)}}{2}$
Now evaluate the sub-integral.
Use integration by parts:

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

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

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

To find $v{\left(x \right)}$ :
1. There are multiple ways to do this integral.
  Method #1
  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:
    
    $- \frac{\cos{\left(2 x \right)}}{2}$
  Method #2
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 \sin{\left(x \right)} \cos{\left(x \right)}\, dx = 2 \int \sin{\left(x \right)} \cos{\left(x \right)}\, dx$
    1. 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 $- du$ :
      
      $\int \left(- u\right)\, du$
      
      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:
      
      $- \frac{\cos^{2}{\left(x \right)}}{2}$
      
      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: $- \cos^{2}{\left(x \right)}$
Now evaluate the sub-integral.
The integral of a constant times a function is the constant times the integral of the function:

$\int \left(- \frac{\cos{\left(2 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{2}$
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:
    
    $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
    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: $- \frac{\sin{\left(2 x \right)}}{4}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

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

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

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

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

cos(2)/2 + sin(2)/4

Numerical answer [src]

0.0192509384328492

0.0192509384328492

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^2cos2x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #1

Method #2