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Identical expressions
(- two x^2-12x+ six)cos6x
( minus 2x squared minus 12x plus 6) co sinus of e of 6x
( minus two x squared minus 12x plus six) co sinus of e of 6x
(-2x2-12x+6)cos6x
-2x2-12x+6cos6x
(-2x²-12x+6)cos6x
(-2x to the power of 2-12x+6)cos6x
-2x^2-12x+6cos6x
(-2x^2-12x+6)cos6xdx
Similar expressions
(-2x^2+12x+6)cos6x
(-2x^2-12x-6)cos6x
(2x^2-12x+6)cos6x

Solving integrals/ (-2x^2-12x+6)cos6x

Integral of (-2x^2-12x+6)cos6x dx

The solution

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  1                                
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 |  /     2           \            
 |  \- 2*x  - 12*x + 6/*cos(6*x) dx
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/                                  
0

$$\int\limits_{0}^{1} \left(- 2 x^{2} - 12 x + 6\right) \cos{\left(6 x \right)}\, dx$$

Integral((-2*x^2 - 12*x + 6)*cos(6*x), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(- 2 x^{2} - 12 x + 6\right) \cos{\left(6 x \right)} = - 2 x^{2} \cos{\left(6 x \right)} - 12 x \cos{\left(6 x \right)} + 6 \cos{\left(6 x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 x^{2} \cos{\left(6 x \right)}\right)\, dx = - 2 \int x^{2} \cos{\left(6 x \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \cos{\left(6 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 2 x$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 6 x$ .
      
      Then let $du = 6 dx$ and substitute $\frac{du}{6}$ :
      
      $\int \frac{\cos{\left(u \right)}}{36}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{6}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{6}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(6 x \right)}}{6}$
      
      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 $u{\left(x \right)} = \frac{x}{3}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(6 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = \frac{1}{3}$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 6 x$ .
      
      Then let $du = 6 dx$ and substitute $\frac{du}{6}$ :
      
      $\int \frac{\sin{\left(u \right)}}{36}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{6}\, du = \frac{\int \sin{\left(u \right)}\, du}{6}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{6}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(6 x \right)}}{6}$
      
      Now evaluate the sub-integral.
    3. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(6 x \right)}}{18}\right)\, dx = - \frac{\int \cos{\left(6 x \right)}\, dx}{18}$
      
      Let $u = 6 x$ .
      
      Then let $du = 6 dx$ and substitute $\frac{du}{6}$ :
      
      $\int \frac{\cos{\left(u \right)}}{36}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{6}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{6}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(6 x \right)}}{6}$
      
      So, the result is: $- \frac{\sin{\left(6 x \right)}}{108}$
    So, the result is: $- \frac{x^{2} \sin{\left(6 x \right)}}{3} - \frac{x \cos{\left(6 x \right)}}{9} + \frac{\sin{\left(6 x \right)}}{54}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 12 x \cos{\left(6 x \right)}\right)\, dx = - 12 \int x \cos{\left(6 x \right)}\, dx$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(6 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 6 x$ .
      
      Then let $du = 6 dx$ and substitute $\frac{du}{6}$ :
      
      $\int \frac{\cos{\left(u \right)}}{36}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{6}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{6}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(6 x \right)}}{6}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(6 x \right)}}{6}\, dx = \frac{\int \sin{\left(6 x \right)}\, dx}{6}$
      
      Let $u = 6 x$ .
      
      Then let $du = 6 dx$ and substitute $\frac{du}{6}$ :
      
      $\int \frac{\sin{\left(u \right)}}{36}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{6}\, du = \frac{\int \sin{\left(u \right)}\, du}{6}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{6}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(6 x \right)}}{6}$
      
      So, the result is: $- \frac{\cos{\left(6 x \right)}}{36}$
    So, the result is: $- 2 x \sin{\left(6 x \right)} - \frac{\cos{\left(6 x \right)}}{3}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 6 \cos{\left(6 x \right)}\, dx = 6 \int \cos{\left(6 x \right)}\, dx$
    1. Let $u = 6 x$ .
      
      Then let $du = 6 dx$ and substitute $\frac{du}{6}$ :
      
      $\int \frac{\cos{\left(u \right)}}{36}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{6}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{6}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(6 x \right)}}{6}$
    So, the result is: $\sin{\left(6 x \right)}$
  The result is: $- \frac{x^{2} \sin{\left(6 x \right)}}{3} - 2 x \sin{\left(6 x \right)} - \frac{x \cos{\left(6 x \right)}}{9} + \frac{55 \sin{\left(6 x \right)}}{54} - \frac{\cos{\left(6 x \right)}}{3}$
Method #2
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = - 2 x^{2} - 12 x + 6$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(6 x \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = - 4 x - 12$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 6 x$ .
    
    Then let $du = 6 dx$ and substitute $\frac{du}{6}$ :
    
    $\int \frac{\cos{\left(u \right)}}{36}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{6}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{6}$
    Now substitute $u$ back in:
    
    $\frac{\sin{\left(6 x \right)}}{6}$
  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 $u{\left(x \right)} = - \frac{2 x}{3} - 2$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(6 x \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = - \frac{2}{3}$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 6 x$ .
    
    Then let $du = 6 dx$ and substitute $\frac{du}{6}$ :
    
    $\int \frac{\sin{\left(u \right)}}{36}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{6}\, du = \frac{\int \sin{\left(u \right)}\, du}{6}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{6}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(6 x \right)}}{6}$
  Now evaluate the sub-integral.
3. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{\cos{\left(6 x \right)}}{9}\, dx = \frac{\int \cos{\left(6 x \right)}\, dx}{9}$
  1. Let $u = 6 x$ .
    
    Then let $du = 6 dx$ and substitute $\frac{du}{6}$ :
    
    $\int \frac{\cos{\left(u \right)}}{36}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{6}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{6}$
    Now substitute $u$ back in:
    
    $\frac{\sin{\left(6 x \right)}}{6}$
  So, the result is: $\frac{\sin{\left(6 x \right)}}{54}$
Method #3
1. Rewrite the integrand:
  
  $\left(- 2 x^{2} - 12 x + 6\right) \cos{\left(6 x \right)} = - 2 x^{2} \cos{\left(6 x \right)} - 12 x \cos{\left(6 x \right)} + 6 \cos{\left(6 x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 x^{2} \cos{\left(6 x \right)}\right)\, dx = - 2 \int x^{2} \cos{\left(6 x \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \cos{\left(6 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 2 x$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 6 x$ .
      
      Then let $du = 6 dx$ and substitute $\frac{du}{6}$ :
      
      $\int \frac{\cos{\left(u \right)}}{36}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{6}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{6}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(6 x \right)}}{6}$
      
      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 $u{\left(x \right)} = \frac{x}{3}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(6 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = \frac{1}{3}$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 6 x$ .
      
      Then let $du = 6 dx$ and substitute $\frac{du}{6}$ :
      
      $\int \frac{\sin{\left(u \right)}}{36}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{6}\, du = \frac{\int \sin{\left(u \right)}\, du}{6}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{6}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(6 x \right)}}{6}$
      
      Now evaluate the sub-integral.
    3. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(6 x \right)}}{18}\right)\, dx = - \frac{\int \cos{\left(6 x \right)}\, dx}{18}$
      
      Let $u = 6 x$ .
      
      Then let $du = 6 dx$ and substitute $\frac{du}{6}$ :
      
      $\int \frac{\cos{\left(u \right)}}{36}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{6}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{6}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(6 x \right)}}{6}$
      
      So, the result is: $- \frac{\sin{\left(6 x \right)}}{108}$
    So, the result is: $- \frac{x^{2} \sin{\left(6 x \right)}}{3} - \frac{x \cos{\left(6 x \right)}}{9} + \frac{\sin{\left(6 x \right)}}{54}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 12 x \cos{\left(6 x \right)}\right)\, dx = - 12 \int x \cos{\left(6 x \right)}\, dx$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(6 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 6 x$ .
      
      Then let $du = 6 dx$ and substitute $\frac{du}{6}$ :
      
      $\int \frac{\cos{\left(u \right)}}{36}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{6}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{6}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(6 x \right)}}{6}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(6 x \right)}}{6}\, dx = \frac{\int \sin{\left(6 x \right)}\, dx}{6}$
      
      Let $u = 6 x$ .
      
      Then let $du = 6 dx$ and substitute $\frac{du}{6}$ :
      
      $\int \frac{\sin{\left(u \right)}}{36}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{6}\, du = \frac{\int \sin{\left(u \right)}\, du}{6}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{6}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(6 x \right)}}{6}$
      
      So, the result is: $- \frac{\cos{\left(6 x \right)}}{36}$
    So, the result is: $- 2 x \sin{\left(6 x \right)} - \frac{\cos{\left(6 x \right)}}{3}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 6 \cos{\left(6 x \right)}\, dx = 6 \int \cos{\left(6 x \right)}\, dx$
    1. Let $u = 6 x$ .
      
      Then let $du = 6 dx$ and substitute $\frac{du}{6}$ :
      
      $\int \frac{\cos{\left(u \right)}}{36}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{6}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{6}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(6 x \right)}}{6}$
    So, the result is: $\sin{\left(6 x \right)}$
  The result is: $- \frac{x^{2} \sin{\left(6 x \right)}}{3} - 2 x \sin{\left(6 x \right)} - \frac{x \cos{\left(6 x \right)}}{9} + \frac{55 \sin{\left(6 x \right)}}{54} - \frac{\cos{\left(6 x \right)}}{3}$
Add the constant of integration:

$- \frac{x^{2} \sin{\left(6 x \right)}}{3} - 2 x \sin{\left(6 x \right)} - \frac{x \cos{\left(6 x \right)}}{9} + \frac{55 \sin{\left(6 x \right)}}{54} - \frac{\cos{\left(6 x \right)}}{3}+ \mathrm{constant}$

The answer is:

$- \frac{x^{2} \sin{\left(6 x \right)}}{3} - 2 x \sin{\left(6 x \right)} - \frac{x \cos{\left(6 x \right)}}{9} + \frac{55 \sin{\left(6 x \right)}}{54} - \frac{\cos{\left(6 x \right)}}{3}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                                      
 |                                                                                2                      
 | /     2           \                   cos(6*x)   55*sin(6*x)                  x *sin(6*x)   x*cos(6*x)
 | \- 2*x  - 12*x + 6/*cos(6*x) dx = C - -------- + ----------- - 2*x*sin(6*x) - ----------- - ----------
 |                                          3            54                           3            9     
/

$${{-{{\left(36\,x^2-2\right)\,\sin \left(6\,x\right)+12\,x\,\cos \left(6\,x\right)}\over{18}}-2\,\left(6\,x\,\sin \left(6\,x\right)+ \cos \left(6\,x\right)\right)+6\,\sin \left(6\,x\right)}\over{6}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

1   71*sin(6)   4*cos(6)
- - --------- - --------
3       54         9

$${{1}\over{3}}-{{71\,\sin 6+24\,\cos 6}\over{54}}$$

1   71*sin(6)   4*cos(6)
- - --------- - --------
3       54         9

$$- \frac{4 \cos{\left(6 \right)}}{9} + \frac{1}{3} - \frac{71 \sin{\left(6 \right)}}{54}$$

Упростить

Numerical answer [src]

0.273970620231758

0.273970620231758

The graph

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (-2x^2-12x+6)cos6x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3