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(6x^ two)cos(3x)
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(6x2)cos(3x)
6x2cos3x
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6x^2cos3x
(6x^2)cos(3x)dx

Solving integrals/ (6x^2)cos(3x)

Integral of (6x^2)cos(3x) dx

The solution

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  1                 
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 |     2            
 |  6*x *cos(3*x) dx
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0

$$\int\limits_{0}^{1} 6 x^{2} \cos{\left(3 x \right)}\, dx$$

Integral(6*x^2*cos(3*x), (x, 0, 1))

Detail solution

The integral of a constant times a function is the constant times the integral of the function:

$\int 6 x^{2} \cos{\left(3 x \right)}\, dx = 6 \int x^{2} \cos{\left(3 x \right)}\, dx$
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 $\operatorname{dv}{\left(x \right)} = \cos{\left(3 x \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = 2 x$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 3 x$ .
    
    Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
    
    $\int \frac{\cos{\left(u \right)}}{9}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      1. The integral of cosine is sine:
        
        $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      So, the result is: $\frac{\sin{\left(u \right)}}{3}$
    Now substitute $u$ back in:
    
    $\frac{\sin{\left(3 x \right)}}{3}$
  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 $u{\left(x \right)} = \frac{2 x}{3}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = \frac{2}{3}$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 3 x$ .
    
    Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
    
    $\int \frac{\sin{\left(u \right)}}{9}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{3}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
      1. The integral of sine is negative cosine:
        
        $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      So, the result is: $- \frac{\cos{\left(u \right)}}{3}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(3 x \right)}}{3}$
  Now evaluate the sub-integral.
3. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{2 \cos{\left(3 x \right)}}{9}\right)\, dx = - \frac{2 \int \cos{\left(3 x \right)}\, dx}{9}$
  1. Let $u = 3 x$ .
    
    Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
    
    $\int \frac{\cos{\left(u \right)}}{9}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      1. The integral of cosine is sine:
        
        $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      So, the result is: $\frac{\sin{\left(u \right)}}{3}$
    Now substitute $u$ back in:
    
    $\frac{\sin{\left(3 x \right)}}{3}$
  So, the result is: $- \frac{2 \sin{\left(3 x \right)}}{27}$
So, the result is: $2 x^{2} \sin{\left(3 x \right)} + \frac{4 x \cos{\left(3 x \right)}}{3} - \frac{4 \sin{\left(3 x \right)}}{9}$
Add the constant of integration:

$2 x^{2} \sin{\left(3 x \right)} + \frac{4 x \cos{\left(3 x \right)}}{3} - \frac{4 \sin{\left(3 x \right)}}{9}+ \mathrm{constant}$

The answer is:

$2 x^{2} \sin{\left(3 x \right)} + \frac{4 x \cos{\left(3 x \right)}}{3} - \frac{4 \sin{\left(3 x \right)}}{9}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                
 |                                                                 
 |    2                   4*sin(3*x)      2            4*x*cos(3*x)
 | 6*x *cos(3*x) dx = C - ---------- + 2*x *sin(3*x) + ------------
 |                            9                             3      
/

$${{2\,\left(\left(9\,x^2-2\right)\,\sin \left(3\,x\right)+6\,x\, \cos \left(3\,x\right)\right)}\over{9}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

4*cos(3)   14*sin(3)
-------- + ---------
   3           9

$${{2\,\left(7\,\sin 3+6\,\cos 3\right)}\over{9}}$$

4*cos(3)   14*sin(3)
-------- + ---------
   3           9

$$\frac{4 \cos{\left(3 \right)}}{3} + \frac{14 \sin{\left(3 \right)}}{9}$$

Упростить

Numerical answer [src]

-1.10046998292969

-1.10046998292969

The graph

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

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Identical expressions

Integral of (6x^2)cos(3x) dx

Limits of integration:

The graph:

Piecewise:

The solution