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Identical expressions
(8x- one)sin(5x+ three)
(8x minus 1) sinus of (5x plus 3)
(8x minus one) sinus of (5x plus three)
8x-1sin5x+3
(8x-1)sin(5x+3)dx
Similar expressions
(8x+1)sin(5x+3)
(8x-1)sin(5x-3)

Integral of (8x-1)sin(5x+3) dx

The solution

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 |  (8*x - 1)*sin(5*x + 3) dx
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0

$$\int\limits_{0}^{1} \left(8 x - 1\right) \sin{\left(5 x + 3 \right)}\, dx$$

Integral((8*x - 1)*sin(5*x + 3), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(8 x - 1\right) \sin{\left(5 x + 3 \right)} = 8 x \sin{\left(5 x + 3 \right)} - \sin{\left(5 x + 3 \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 8 x \sin{\left(5 x + 3 \right)}\, dx = 8 \int x \sin{\left(5 x + 3 \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \sin{\left(5 x + 3 \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 5 x + 3$ .
      
      Then let $du = 5 dx$ and substitute $\frac{du}{5}$ :
      
      $\int \frac{\sin{\left(u \right)}}{5}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{5}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{5}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(5 x + 3 \right)}}{5}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(5 x + 3 \right)}}{5}\right)\, dx = - \frac{\int \cos{\left(5 x + 3 \right)}\, dx}{5}$
      
      Let $u = 5 x + 3$ .
      
      Then let $du = 5 dx$ and substitute $\frac{du}{5}$ :
      
      $\int \frac{\cos{\left(u \right)}}{5}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{5}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{5}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(5 x + 3 \right)}}{5}$
      
      So, the result is: $- \frac{\sin{\left(5 x + 3 \right)}}{25}$
    So, the result is: $- \frac{8 x \cos{\left(5 x + 3 \right)}}{5} + \frac{8 \sin{\left(5 x + 3 \right)}}{25}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \sin{\left(5 x + 3 \right)}\right)\, dx = - \int \sin{\left(5 x + 3 \right)}\, dx$
    1. Let $u = 5 x + 3$ .
      
      Then let $du = 5 dx$ and substitute $\frac{du}{5}$ :
      
      $\int \frac{\sin{\left(u \right)}}{5}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{5}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{5}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(5 x + 3 \right)}}{5}$
    So, the result is: $\frac{\cos{\left(5 x + 3 \right)}}{5}$
  The result is: $- \frac{8 x \cos{\left(5 x + 3 \right)}}{5} + \frac{8 \sin{\left(5 x + 3 \right)}}{25} + \frac{\cos{\left(5 x + 3 \right)}}{5}$
Method #2
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = 8 x - 1$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(5 x + 3 \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = 8$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 5 x + 3$ .
    
    Then let $du = 5 dx$ and substitute $\frac{du}{5}$ :
    
    $\int \frac{\sin{\left(u \right)}}{5}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{5}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{5}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(5 x + 3 \right)}}{5}$
  Now evaluate the sub-integral.
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{8 \cos{\left(5 x + 3 \right)}}{5}\right)\, dx = - \frac{8 \int \cos{\left(5 x + 3 \right)}\, dx}{5}$
  1. Let $u = 5 x + 3$ .
    
    Then let $du = 5 dx$ and substitute $\frac{du}{5}$ :
    
    $\int \frac{\cos{\left(u \right)}}{5}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{5}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{5}$
    Now substitute $u$ back in:
    
    $\frac{\sin{\left(5 x + 3 \right)}}{5}$
  So, the result is: $- \frac{8 \sin{\left(5 x + 3 \right)}}{25}$
Method #3
1. Rewrite the integrand:
  
  $\left(8 x - 1\right) \sin{\left(5 x + 3 \right)} = 8 x \sin{\left(5 x + 3 \right)} - \sin{\left(5 x + 3 \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 8 x \sin{\left(5 x + 3 \right)}\, dx = 8 \int x \sin{\left(5 x + 3 \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \sin{\left(5 x + 3 \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 5 x + 3$ .
      
      Then let $du = 5 dx$ and substitute $\frac{du}{5}$ :
      
      $\int \frac{\sin{\left(u \right)}}{5}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{5}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{5}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(5 x + 3 \right)}}{5}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(5 x + 3 \right)}}{5}\right)\, dx = - \frac{\int \cos{\left(5 x + 3 \right)}\, dx}{5}$
      
      Let $u = 5 x + 3$ .
      
      Then let $du = 5 dx$ and substitute $\frac{du}{5}$ :
      
      $\int \frac{\cos{\left(u \right)}}{5}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{5}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{5}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(5 x + 3 \right)}}{5}$
      
      So, the result is: $- \frac{\sin{\left(5 x + 3 \right)}}{25}$
    So, the result is: $- \frac{8 x \cos{\left(5 x + 3 \right)}}{5} + \frac{8 \sin{\left(5 x + 3 \right)}}{25}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \sin{\left(5 x + 3 \right)}\right)\, dx = - \int \sin{\left(5 x + 3 \right)}\, dx$
    1. Let $u = 5 x + 3$ .
      
      Then let $du = 5 dx$ and substitute $\frac{du}{5}$ :
      
      $\int \frac{\sin{\left(u \right)}}{5}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{5}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{5}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(5 x + 3 \right)}}{5}$
    So, the result is: $\frac{\cos{\left(5 x + 3 \right)}}{5}$
  The result is: $- \frac{8 x \cos{\left(5 x + 3 \right)}}{5} + \frac{8 \sin{\left(5 x + 3 \right)}}{25} + \frac{\cos{\left(5 x + 3 \right)}}{5}$
Add the constant of integration:

$- \frac{8 x \cos{\left(5 x + 3 \right)}}{5} + \frac{8 \sin{\left(5 x + 3 \right)}}{25} + \frac{\cos{\left(5 x + 3 \right)}}{5}+ \mathrm{constant}$

The answer is:

$- \frac{8 x \cos{\left(5 x + 3 \right)}}{5} + \frac{8 \sin{\left(5 x + 3 \right)}}{25} + \frac{\cos{\left(5 x + 3 \right)}}{5}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                
 |                                 cos(3 + 5*x)   8*sin(3 + 5*x)   8*x*cos(3 + 5*x)
 | (8*x - 1)*sin(5*x + 3) dx = C + ------------ + -------------- - ----------------
 |                                      5               25                5        
/

$$\int \left(8 x - 1\right) \sin{\left(5 x + 3 \right)}\, dx = C - \frac{8 x \cos{\left(5 x + 3 \right)}}{5} + \frac{8 \sin{\left(5 x + 3 \right)}}{25} + \frac{\cos{\left(5 x + 3 \right)}}{5}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

  8*sin(3)   7*cos(8)   cos(3)   8*sin(8)
- -------- - -------- - ------ + --------
     25         5         5         25

$$- \frac{8 \sin{\left(3 \right)}}{25} - \frac{\cos{\left(3 \right)}}{5} - \frac{7 \cos{\left(8 \right)}}{5} + \frac{8 \sin{\left(8 \right)}}{25}$$

  8*sin(3)   7*cos(8)   cos(3)   8*sin(8)
- -------- - -------- - ------ + --------
     25         5         5         25

$$- \frac{8 \sin{\left(3 \right)}}{25} - \frac{\cos{\left(3 \right)}}{5} - \frac{7 \cos{\left(8 \right)}}{5} + \frac{8 \sin{\left(8 \right)}}{25}$$

-8*sin(3)/25 - 7*cos(8)/5 - cos(3)/5 + 8*sin(8)/25

Numerical answer [src]

0.673134782992473

0.673134782992473

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

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Identical expressions

Similar expressions

Integral of (8x-1)sin(5x+3) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3