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Identical expressions
(two *x- three)*sin(five *x)
(2 multiply by x minus 3) multiply by sinus of (5 multiply by x)
(two multiply by x minus three) multiply by sinus of (five multiply by x)
(2x-3)sin(5x)
2x-3sin5x
(2*x-3)*sin(5*x)dx
Similar expressions
(2*x+3)*sin(5*x)
(x^2+2x-3)*sin*5*xdx

Integral of (2x-3)sin(5*x) dx

The solution

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 |  (2*x - 3)*sin(5*x) dx
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0

$$\int\limits_{0}^{1} \left(2 x - 3\right) \sin{\left(5 x \right)}\, dx$$

Integral((2*x - 3)*sin(5*x), (x, 0, 1))

Detail solution

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

$- \frac{2 x \cos{\left(5 x \right)}}{5} + \frac{2 \sin{\left(5 x \right)}}{25} + \frac{3 \cos{\left(5 x \right)}}{5}+ \mathrm{constant}$

The answer is:

$- \frac{2 x \cos{\left(5 x \right)}}{5} + \frac{2 \sin{\left(5 x \right)}}{25} + \frac{3 \cos{\left(5 x \right)}}{5}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                  
 |                             2*sin(5*x)   3*cos(5*x)   2*x*cos(5*x)
 | (2*x - 3)*sin(5*x) dx = C + ---------- + ---------- - ------------
 |                                 25           5             5      
/

$$\int \left(2 x - 3\right) \sin{\left(5 x \right)}\, dx = C - \frac{2 x \cos{\left(5 x \right)}}{5} + \frac{2 \sin{\left(5 x \right)}}{25} + \frac{3 \cos{\left(5 x \right)}}{5}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

  3   cos(5)   2*sin(5)
- - + ------ + --------
  5     5         25

$$- \frac{3}{5} + \frac{2 \sin{\left(5 \right)}}{25} + \frac{\cos{\left(5 \right)}}{5}$$

  3   cos(5)   2*sin(5)
- - + ------ + --------
  5     5         25

$$- \frac{3}{5} + \frac{2 \sin{\left(5 \right)}}{25} + \frac{\cos{\left(5 \right)}}{5}$$

-3/5 + cos(5)/5 + 2*sin(5)/25

Numerical answer [src]

-0.619981504880406

-0.619981504880406

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (2x-3)sin(5*x) 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 (2*x-3)*sin(5*x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Integral of (2x-3)sin(5*x) dx