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Identical expressions
(eight *x^ two + sixteen *x+ seventeen)cos(4x)
(8 multiply by x squared plus 16 multiply by x plus 17) co sinus of e of (4x)
(eight multiply by x to the power of two plus sixteen multiply by x plus seventeen) co sinus of e of (4x)
(8*x2+16*x+17)cos(4x)
8*x2+16*x+17cos4x
(8*x²+16*x+17)cos(4x)
(8*x to the power of 2+16*x+17)cos(4x)
(8x^2+16x+17)cos(4x)
(8x2+16x+17)cos(4x)
8x2+16x+17cos4x
8x^2+16x+17cos4x
(8*x^2+16*x+17)cos(4x)dx
Similar expressions
(8*x^2-16*x+17)cos(4x)
(8*x^2+16*x-17)cos(4x)

Solving integrals/ (8*x^2+16*x+17)cos(4x)

Integral of (8x^2+16x+17)cos(4x) dx

The solution

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0

$$\int\limits_{0}^{1} \left(8 x^{2} + 16 x + 17\right) \cos{\left(4 x \right)}\, dx$$

Integral((8*x^2 + 16*x + 17)*cos(4*x), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(8 x^{2} + 16 x + 17\right) \cos{\left(4 x \right)} = 8 x^{2} \cos{\left(4 x \right)} + 16 x \cos{\left(4 x \right)} + 17 \cos{\left(4 x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 8 x^{2} \cos{\left(4 x \right)}\, dx = 8 \int x^{2} \cos{\left(4 x \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \cos{\left(4 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 2 x$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      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 $u{\left(x \right)} = \frac{x}{2}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(4 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = \frac{1}{2}$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(4 x \right)}}{4}$
      
      Now evaluate the sub-integral.
    3. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(4 x \right)}}{8}\right)\, dx = - \frac{\int \cos{\left(4 x \right)}\, dx}{8}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      So, the result is: $- \frac{\sin{\left(4 x \right)}}{32}$
    So, the result is: $2 x^{2} \sin{\left(4 x \right)} + x \cos{\left(4 x \right)} - \frac{\sin{\left(4 x \right)}}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 16 x \cos{\left(4 x \right)}\, dx = 16 \int x \cos{\left(4 x \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \cos{\left(4 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(4 x \right)}}{4}\, dx = \frac{\int \sin{\left(4 x \right)}\, dx}{4}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(4 x \right)}}{4}$
      
      So, the result is: $- \frac{\cos{\left(4 x \right)}}{16}$
    So, the result is: $4 x \sin{\left(4 x \right)} + \cos{\left(4 x \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 17 \cos{\left(4 x \right)}\, dx = 17 \int \cos{\left(4 x \right)}\, dx$
    1. Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
    So, the result is: $\frac{17 \sin{\left(4 x \right)}}{4}$
  The result is: $2 x^{2} \sin{\left(4 x \right)} + 4 x \sin{\left(4 x \right)} + x \cos{\left(4 x \right)} + 4 \sin{\left(4 x \right)} + \cos{\left(4 x \right)}$
Method #2
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = 8 x^{2} + 16 x + 17$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(4 x \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = 16 x + 16$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 4 x$ .
    
    Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
    
    $\int \frac{\cos{\left(u \right)}}{16}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
    Now substitute $u$ back in:
    
    $\frac{\sin{\left(4 x \right)}}{4}$
  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 $u{\left(x \right)} = 4 x + 4$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(4 x \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = 4$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 4 x$ .
    
    Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
    
    $\int \frac{\sin{\left(u \right)}}{16}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(4 x \right)}}{4}$
  Now evaluate the sub-integral.
3. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \cos{\left(4 x \right)}\right)\, dx = - \int \cos{\left(4 x \right)}\, dx$
  1. Let $u = 4 x$ .
    
    Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
    
    $\int \frac{\cos{\left(u \right)}}{16}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
    Now substitute $u$ back in:
    
    $\frac{\sin{\left(4 x \right)}}{4}$
  So, the result is: $- \frac{\sin{\left(4 x \right)}}{4}$
Method #3
1. Rewrite the integrand:
  
  $\left(8 x^{2} + 16 x + 17\right) \cos{\left(4 x \right)} = 8 x^{2} \cos{\left(4 x \right)} + 16 x \cos{\left(4 x \right)} + 17 \cos{\left(4 x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 8 x^{2} \cos{\left(4 x \right)}\, dx = 8 \int x^{2} \cos{\left(4 x \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \cos{\left(4 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 2 x$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      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 $u{\left(x \right)} = \frac{x}{2}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(4 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = \frac{1}{2}$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(4 x \right)}}{4}$
      
      Now evaluate the sub-integral.
    3. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(4 x \right)}}{8}\right)\, dx = - \frac{\int \cos{\left(4 x \right)}\, dx}{8}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      So, the result is: $- \frac{\sin{\left(4 x \right)}}{32}$
    So, the result is: $2 x^{2} \sin{\left(4 x \right)} + x \cos{\left(4 x \right)} - \frac{\sin{\left(4 x \right)}}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 16 x \cos{\left(4 x \right)}\, dx = 16 \int x \cos{\left(4 x \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \cos{\left(4 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(4 x \right)}}{4}\, dx = \frac{\int \sin{\left(4 x \right)}\, dx}{4}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(4 x \right)}}{4}$
      
      So, the result is: $- \frac{\cos{\left(4 x \right)}}{16}$
    So, the result is: $4 x \sin{\left(4 x \right)} + \cos{\left(4 x \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 17 \cos{\left(4 x \right)}\, dx = 17 \int \cos{\left(4 x \right)}\, dx$
    1. Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
    So, the result is: $\frac{17 \sin{\left(4 x \right)}}{4}$
  The result is: $2 x^{2} \sin{\left(4 x \right)} + 4 x \sin{\left(4 x \right)} + x \cos{\left(4 x \right)} + 4 \sin{\left(4 x \right)} + \cos{\left(4 x \right)}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

  /                                                                                                      
 |                                                                                                       
 | /   2            \                                                2                                   
 | \8*x  + 16*x + 17/*cos(4*x) dx = C + 4*sin(4*x) + x*cos(4*x) + 2*x *sin(4*x) + 4*x*sin(4*x) + cos(4*x)
 |                                                                                                       
/

$${{{{\left(16\,x^2-2\right)\,\sin \left(4\,x\right)+8\,x\,\cos \left(4\,x\right)}\over{2}}+4\,\left(4\,x\,\sin \left(4\,x\right)+ \cos \left(4\,x\right)\right)+17\,\sin \left(4\,x\right)}\over{4}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

-1 + 2*cos(4) + 10*sin(4)

$$10\,\sin 4+2\,\cos 4-1$$

-1 + 2*cos(4) + 10*sin(4)

$$10 \sin{\left(4 \right)} + 2 \cos{\left(4 \right)} - 1$$

Simplify

Numerical answer [src]

-9.87531219480651

-9.87531219480651

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 (8x^2+16x+17)cos(4x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (8*x^2+16*x+17)cos(4x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Integral of (8x^2+16x+17)cos(4x) dx