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Integral of d{x}:
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Identical expressions
(-5x- two)cos(4x+ thirteen)
( minus 5x minus 2) co sinus of e of (4x plus 13)
( minus 5x minus two) co sinus of e of (4x plus thirteen)
-5x-2cos4x+13
(-5x-2)cos(4x+13)dx
Similar expressions
(-5x-2)cos(4x-13)
(5x-2)cos(4x+13)
(-5x+2)cos(4x+13)

Integral of (-5x-2)cos(4x+13) dx

The solution

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 |  (-5*x - 2)*cos(4*x + 13) dx
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0

$$\int\limits_{0}^{1} \left(- 5 x - 2\right) \cos{\left(4 x + 13 \right)}\, dx$$

Integral((-5*x - 2)*cos(4*x + 13), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(- 5 x - 2\right) \cos{\left(4 x + 13 \right)} = - 5 x \cos{\left(4 x + 13 \right)} - 2 \cos{\left(4 x + 13 \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 5 x \cos{\left(4 x + 13 \right)}\right)\, dx = - 5 \int x \cos{\left(4 x + 13 \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \cos{\left(4 x + 13 \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 4 x + 13$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      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:
      
      $\frac{\sin{\left(4 x + 13 \right)}}{4}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(4 x + 13 \right)}}{4}\, dx = \frac{\int \sin{\left(4 x + 13 \right)}\, dx}{4}$
      
      Let $u = 4 x + 13$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      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:
      
      $- \frac{\cos{\left(4 x + 13 \right)}}{4}$
      
      So, the result is: $- \frac{\cos{\left(4 x + 13 \right)}}{16}$
    So, the result is: $- \frac{5 x \sin{\left(4 x + 13 \right)}}{4} - \frac{5 \cos{\left(4 x + 13 \right)}}{16}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 \cos{\left(4 x + 13 \right)}\right)\, dx = - 2 \int \cos{\left(4 x + 13 \right)}\, dx$
    1. Let $u = 4 x + 13$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      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:
      
      $\frac{\sin{\left(4 x + 13 \right)}}{4}$
    So, the result is: $- \frac{\sin{\left(4 x + 13 \right)}}{2}$
  The result is: $- \frac{5 x \sin{\left(4 x + 13 \right)}}{4} - \frac{\sin{\left(4 x + 13 \right)}}{2} - \frac{5 \cos{\left(4 x + 13 \right)}}{16}$
Method #2
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = - 5 x - 2$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(4 x + 13 \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = -5$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 4 x + 13$ .
    
    Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
    
    $\int \frac{\cos{\left(u \right)}}{4}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      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:
    
    $\frac{\sin{\left(4 x + 13 \right)}}{4}$
  Now evaluate the sub-integral.
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{5 \sin{\left(4 x + 13 \right)}}{4}\right)\, dx = - \frac{5 \int \sin{\left(4 x + 13 \right)}\, dx}{4}$
  1. Let $u = 4 x + 13$ .
    
    Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
    
    $\int \frac{\sin{\left(u \right)}}{4}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      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:
    
    $- \frac{\cos{\left(4 x + 13 \right)}}{4}$
  So, the result is: $\frac{5 \cos{\left(4 x + 13 \right)}}{16}$
Method #3
1. Rewrite the integrand:
  
  $\left(- 5 x - 2\right) \cos{\left(4 x + 13 \right)} = - 5 x \cos{\left(4 x + 13 \right)} - 2 \cos{\left(4 x + 13 \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 5 x \cos{\left(4 x + 13 \right)}\right)\, dx = - 5 \int x \cos{\left(4 x + 13 \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \cos{\left(4 x + 13 \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 4 x + 13$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      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:
      
      $\frac{\sin{\left(4 x + 13 \right)}}{4}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(4 x + 13 \right)}}{4}\, dx = \frac{\int \sin{\left(4 x + 13 \right)}\, dx}{4}$
      
      Let $u = 4 x + 13$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      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:
      
      $- \frac{\cos{\left(4 x + 13 \right)}}{4}$
      
      So, the result is: $- \frac{\cos{\left(4 x + 13 \right)}}{16}$
    So, the result is: $- \frac{5 x \sin{\left(4 x + 13 \right)}}{4} - \frac{5 \cos{\left(4 x + 13 \right)}}{16}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 \cos{\left(4 x + 13 \right)}\right)\, dx = - 2 \int \cos{\left(4 x + 13 \right)}\, dx$
    1. Let $u = 4 x + 13$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      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:
      
      $\frac{\sin{\left(4 x + 13 \right)}}{4}$
    So, the result is: $- \frac{\sin{\left(4 x + 13 \right)}}{2}$
  The result is: $- \frac{5 x \sin{\left(4 x + 13 \right)}}{4} - \frac{\sin{\left(4 x + 13 \right)}}{2} - \frac{5 \cos{\left(4 x + 13 \right)}}{16}$
Add the constant of integration:

$- \frac{5 x \sin{\left(4 x + 13 \right)}}{4} - \frac{\sin{\left(4 x + 13 \right)}}{2} - \frac{5 \cos{\left(4 x + 13 \right)}}{16}+ \mathrm{constant}$

The answer is:

$- \frac{5 x \sin{\left(4 x + 13 \right)}}{4} - \frac{\sin{\left(4 x + 13 \right)}}{2} - \frac{5 \cos{\left(4 x + 13 \right)}}{16}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                     
 |                                   5*cos(13 + 4*x)   sin(13 + 4*x)   5*x*sin(13 + 4*x)
 | (-5*x - 2)*cos(4*x + 13) dx = C - --------------- - ------------- - -----------------
 |                                          16               2                 4        
/

$$\int \left(- 5 x - 2\right) \cos{\left(4 x + 13 \right)}\, dx = C - \frac{5 x \sin{\left(4 x + 13 \right)}}{4} - \frac{\sin{\left(4 x + 13 \right)}}{2} - \frac{5 \cos{\left(4 x + 13 \right)}}{16}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

sin(13)   7*sin(17)   5*cos(17)   5*cos(13)
------- - --------- - --------- + ---------
   2          4           16          16

$$- \frac{5 \cos{\left(17 \right)}}{16} + \frac{\sin{\left(13 \right)}}{2} + \frac{5 \cos{\left(13 \right)}}{16} - \frac{7 \sin{\left(17 \right)}}{4}$$

sin(13)   7*sin(17)   5*cos(17)   5*cos(13)
------- - --------- - --------- + ---------
   2          4           16          16

$$- \frac{5 \cos{\left(17 \right)}}{16} + \frac{\sin{\left(13 \right)}}{2} + \frac{5 \cos{\left(13 \right)}}{16} - \frac{7 \sin{\left(17 \right)}}{4}$$

sin(13)/2 - 7*sin(17)/4 - 5*cos(17)/16 + 5*cos(13)/16

Numerical answer [src]

2.26209479154686

2.26209479154686

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

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Identical expressions

Similar expressions

Integral of (-5x-2)cos(4x+13) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3