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Integral of d{x}:
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Integral of (1+x^2)^1/2
Identical expressions
cos(3x- two)-9sin(two -3x)
co sinus of e of (3x minus 2) minus 9 sinus of (2 minus 3x)
co sinus of e of (3x minus two) minus 9 sinus of (two minus 3x)
cos3x-2-9sin2-3x
cos(3x-2)-9sin(2-3x)dx
Similar expressions
cos(3x-2)+9sin(2-3x)
cos(3x-2)-9sin(2+3x)
cos(3x+2)-9sin(2-3x)

Integral of cos(3x-2)-9sin(2-3x) dx

The solution

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  1                                   
  /                                   
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 |  (cos(3*x - 2) - 9*sin(2 - 3*x)) dx
 |                                    
/                                     
0

$$\int\limits_{0}^{1} \left(- 9 \sin{\left(2 - 3 x \right)} + \cos{\left(3 x - 2 \right)}\right)\, dx$$

Integral(cos(3*x - 2) - 9*sin(2 - 3*x), (x, 0, 1))

Detail solution

Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 9 \sin{\left(2 - 3 x \right)}\right)\, dx = - 9 \int \sin{\left(2 - 3 x \right)}\, dx$
  1. Let $u = 2 - 3 x$ .
    
    Then let $du = - 3 dx$ and substitute $- \frac{du}{3}$ :
    
    $\int \left(- \frac{\sin{\left(u \right)}}{3}\right)\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = - \frac{\int \sin{\left(u \right)}\, du}{3}$
      1. The integral of sine is negative cosine:
        
        $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      So, the result is: $\frac{\cos{\left(u \right)}}{3}$
    Now substitute $u$ back in:
    
    $\frac{\cos{\left(3 x - 2 \right)}}{3}$
  So, the result is: $- 3 \cos{\left(3 x - 2 \right)}$
1. Let $u = 3 x - 2$ .
  
  Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
  
  $\int \frac{\cos{\left(u \right)}}{3}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
    1. The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
    So, the result is: $\frac{\sin{\left(u \right)}}{3}$
  Now substitute $u$ back in:
  
  $\frac{\sin{\left(3 x - 2 \right)}}{3}$
The result is: $\frac{\sin{\left(3 x - 2 \right)}}{3} - 3 \cos{\left(3 x - 2 \right)}$
Now simplify:

$\frac{\sin{\left(3 x - 2 \right)}}{3} - 3 \cos{\left(3 x - 2 \right)}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

  /                                                                       
 |                                                            sin(3*x - 2)
 | (cos(3*x - 2) - 9*sin(2 - 3*x)) dx = C - 3*cos(-2 + 3*x) + ------------
 |                                                                 3      
/

$$\int \left(- 9 \sin{\left(2 - 3 x \right)} + \cos{\left(3 x - 2 \right)}\right)\, dx = C + \frac{\sin{\left(3 x - 2 \right)}}{3} - 3 \cos{\left(3 x - 2 \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                       sin(1)   sin(2)
-3*cos(1) + 3*cos(2) + ------ + ------
                         3        3

$$- 3 \cos{\left(1 \right)} + 3 \cos{\left(2 \right)} + \frac{\sin{\left(1 \right)}}{3} + \frac{\sin{\left(2 \right)}}{3}$$

                       sin(1)   sin(2)
-3*cos(1) + 3*cos(2) + ------ + ------
                         3        3

$$- 3 \cos{\left(1 \right)} + 3 \cos{\left(2 \right)} + \frac{\sin{\left(1 \right)}}{3} + \frac{\sin{\left(2 \right)}}{3}$$

-3*cos(1) + 3*cos(2) + sin(1)/3 + sin(2)/3

Numerical answer [src]

-2.28575795670132

-2.28575795670132

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

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Integral of cos(3x-2)-9sin(2-3x) dx

Limits of integration:

The graph:

Piecewise:

The solution