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Identical expressions
(6x- five)sin(-2x- five)
(6x minus 5) sinus of ( minus 2x minus 5)
(6x minus five) sinus of ( minus 2x minus five)
6x-5sin-2x-5
(6x-5)sin(-2x-5)dx
Similar expressions
(6x-5)sin(-2x+5)
(6x+5)sin(-2x-5)
(6x-5)sin(2x-5)

Integral of (6x-5)sin(-2x-5) dx

The solution

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  1                           
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 |  (6*x - 5)*sin(-2*x - 5) dx
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0

$$\int\limits_{0}^{1} \left(6 x - 5\right) \sin{\left(- 2 x - 5 \right)}\, dx$$

Integral((6*x - 5)*sin(-2*x - 5), (x, 0, 1))

Detail solution

Rewrite the integrand:

$\left(6 x - 5\right) \sin{\left(- 2 x - 5 \right)} = - 6 x \sin{\left(2 x + 5 \right)} + 5 \sin{\left(2 x + 5 \right)}$
Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 6 x \sin{\left(2 x + 5 \right)}\right)\, dx = - 6 \int x \sin{\left(2 x + 5 \right)}\, dx$
  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 $\operatorname{dv}{\left(x \right)} = \sin{\left(2 x + 5 \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 1$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 2 x + 5$ .
      
      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:
      
      $- \frac{\cos{\left(2 x + 5 \right)}}{2}$
    Now evaluate the sub-integral.
  2. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos{\left(2 x + 5 \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(2 x + 5 \right)}\, dx}{2}$
    1. Let $u = 2 x + 5$ .
      
      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:
        
        $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
        
        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:
      
      $\frac{\sin{\left(2 x + 5 \right)}}{2}$
    So, the result is: $- \frac{\sin{\left(2 x + 5 \right)}}{4}$
  So, the result is: $3 x \cos{\left(2 x + 5 \right)} - \frac{3 \sin{\left(2 x + 5 \right)}}{2}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 5 \sin{\left(2 x + 5 \right)}\, dx = 5 \int \sin{\left(2 x + 5 \right)}\, dx$
  1. Let $u = 2 x + 5$ .
    
    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:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
      1. The integral of sine is negative cosine:
        
        $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      So, the result is: $- \frac{\cos{\left(u \right)}}{2}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(2 x + 5 \right)}}{2}$
  So, the result is: $- \frac{5 \cos{\left(2 x + 5 \right)}}{2}$
The result is: $3 x \cos{\left(2 x + 5 \right)} - \frac{3 \sin{\left(2 x + 5 \right)}}{2} - \frac{5 \cos{\left(2 x + 5 \right)}}{2}$
Add the constant of integration:

$3 x \cos{\left(2 x + 5 \right)} - \frac{3 \sin{\left(2 x + 5 \right)}}{2} - \frac{5 \cos{\left(2 x + 5 \right)}}{2}+ \mathrm{constant}$

The answer is:

$3 x \cos{\left(2 x + 5 \right)} - \frac{3 \sin{\left(2 x + 5 \right)}}{2} - \frac{5 \cos{\left(2 x + 5 \right)}}{2}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                   
 |                                  5*cos(5 + 2*x)   3*sin(5 + 2*x)                   
 | (6*x - 5)*sin(-2*x - 5) dx = C - -------------- - -------------- + 3*x*cos(5 + 2*x)
 |                                        2                2                          
/

$$\int \left(6 x - 5\right) \sin{\left(- 2 x - 5 \right)}\, dx = C + 3 x \cos{\left(2 x + 5 \right)} - \frac{3 \sin{\left(2 x + 5 \right)}}{2} - \frac{5 \cos{\left(2 x + 5 \right)}}{2}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

cos(7)   3*sin(7)   3*sin(5)   5*cos(5)
------ - -------- + -------- + --------
  2         2          2          2

$$\frac{3 \sin{\left(5 \right)}}{2} - \frac{3 \sin{\left(7 \right)}}{2} + \frac{\cos{\left(7 \right)}}{2} + \frac{5 \cos{\left(5 \right)}}{2}$$

cos(7)   3*sin(7)   3*sin(5)   5*cos(5)
------ - -------- + -------- + --------
  2         2          2          2

$$\frac{3 \sin{\left(5 \right)}}{2} - \frac{3 \sin{\left(7 \right)}}{2} + \frac{\cos{\left(7 \right)}}{2} + \frac{5 \cos{\left(5 \right)}}{2}$$

cos(7)/2 - 3*sin(7)/2 + 3*sin(5)/2 + 5*cos(5)/2

Numerical answer [src]

-1.33775971924317

-1.33775971924317

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

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Integral of (6x-5)sin(-2x-5) dx

Limits of integration:

The graph:

Piecewise:

The solution