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cos(19x- two)-sin(10x)
co sinus of e of (19x minus 2) minus sinus of (10x)
co sinus of e of (19x minus two) minus sinus of (10x)
cos19x-2-sin10x
cos(19x-2)-sin(10x)dx
Similar expressions
cos(19x-2)+sin(10x)
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Integral of cos(19x-2)-sin(10x) dx

The solution

You have entered [src]

  1                               
  /                               
 |                                
 |  (cos(19*x - 2) - sin(10*x)) dx
 |                                
/                                 
0

$$\int\limits_{0}^{1} \left(- \sin{\left(10 x \right)} + \cos{\left(19 x - 2 \right)}\right)\, dx$$

Integral(cos(19*x - 2) - sin(10*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:
  
  $\int \left(- \sin{\left(10 x \right)}\right)\, dx = - \int \sin{\left(10 x \right)}\, dx$
  1. Let $u = 10 x$ .
    
    Then let $du = 10 dx$ and substitute $\frac{du}{10}$ :
    
    $\int \frac{\sin{\left(u \right)}}{10}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{10}$
      1. The integral of sine is negative cosine:
        
        $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      So, the result is: $- \frac{\cos{\left(u \right)}}{10}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(10 x \right)}}{10}$
  So, the result is: $\frac{\cos{\left(10 x \right)}}{10}$
1. Let $u = 19 x - 2$ .
  
  Then let $du = 19 dx$ and substitute $\frac{du}{19}$ :
  
  $\int \frac{\cos{\left(u \right)}}{19}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{19}$
    1. The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
    So, the result is: $\frac{\sin{\left(u \right)}}{19}$
  Now substitute $u$ back in:
  
  $\frac{\sin{\left(19 x - 2 \right)}}{19}$
The result is: $\frac{\sin{\left(19 x - 2 \right)}}{19} + \frac{\cos{\left(10 x \right)}}{10}$
Now simplify:

$\frac{\sin{\left(19 x - 2 \right)}}{19} + \frac{\cos{\left(10 x \right)}}{10}$
Add the constant of integration:

$\frac{\sin{\left(19 x - 2 \right)}}{19} + \frac{\cos{\left(10 x \right)}}{10}+ \mathrm{constant}$

The answer is:

$\frac{\sin{\left(19 x - 2 \right)}}{19} + \frac{\cos{\left(10 x \right)}}{10}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                              
 |                                      cos(10*x)   sin(19*x - 2)
 | (cos(19*x - 2) - sin(10*x)) dx = C + --------- + -------------
 |                                          10            19     
/

$$\int \left(- \sin{\left(10 x \right)} + \cos{\left(19 x - 2 \right)}\right)\, dx = C + \frac{\sin{\left(19 x - 2 \right)}}{19} + \frac{\cos{\left(10 x \right)}}{10}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

  1    cos(10)   sin(2)   sin(17)
- -- + ------- + ------ + -------
  10      10       19        19

$$- \frac{1}{10} + \frac{\cos{\left(10 \right)}}{10} + \frac{\sin{\left(17 \right)}}{19} + \frac{\sin{\left(2 \right)}}{19}$$

  1    cos(10)   sin(2)   sin(17)
- -- + ------- + ------ + -------
  10      10       19        19

$$- \frac{1}{10} + \frac{\cos{\left(10 \right)}}{10} + \frac{\sin{\left(17 \right)}}{19} + \frac{\sin{\left(2 \right)}}{19}$$

-1/10 + cos(10)/10 + sin(2)/19 + sin(17)/19

Numerical answer [src]

-0.186649261594691

-0.186649261594691

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

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Integral of cos(19x-2)-sin(10x) dx

Limits of integration:

The graph:

Piecewise:

The solution