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Identical expressions
3sin(3x)+cos(2x+ four)
3 sinus of (3x) plus co sinus of e of (2x plus 4)
3 sinus of (3x) plus co sinus of e of (2x plus four)
3sin3x+cos2x+4
3sin(3x)+cos(2x+4)dx
Similar expressions
3sin(3x)-cos(2x+4)
3sin(3x)+cos(2x-4)

Integral of 3sin(3x)+cos(2x+4) dx

The solution

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  1                               
  /                               
 |                                
 |  (3*sin(3*x) + cos(2*x + 4)) dx
 |                                
/                                 
0

$$\int\limits_{0}^{1} \left(3 \sin{\left(3 x \right)} + \cos{\left(2 x + 4 \right)}\right)\, dx$$

Integral(3*sin(3*x) + cos(2*x + 4), (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:
  
  $\int 3 \sin{\left(3 x \right)}\, dx = 3 \int \sin{\left(3 x \right)}\, dx$
  1. Let $u = 3 x$ .
    
    Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
    
    $\int \frac{\sin{\left(u \right)}}{3}\, du$
    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:
    
    $- \frac{\cos{\left(3 x \right)}}{3}$
  So, the result is: $- \cos{\left(3 x \right)}$
1. Let $u = 2 x + 4$ .
  
  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:
    
    $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
    1. The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
    So, the result is: $\frac{\sin{\left(u \right)}}{2}$
  Now substitute $u$ back in:
  
  $\frac{\sin{\left(2 x + 4 \right)}}{2}$
The result is: $\frac{\sin{\left(2 x + 4 \right)}}{2} - \cos{\left(3 x \right)}$
Now simplify:

$\frac{\sin{\left(2 x + 4 \right)}}{2} - \cos{\left(3 x \right)}$
Add the constant of integration:

$\frac{\sin{\left(2 x + 4 \right)}}{2} - \cos{\left(3 x \right)}+ \mathrm{constant}$

The answer is:

$\frac{\sin{\left(2 x + 4 \right)}}{2} - \cos{\left(3 x \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                            
 |                                      sin(2*x + 4)           
 | (3*sin(3*x) + cos(2*x + 4)) dx = C + ------------ - cos(3*x)
 |                                           2                 
/

$$\int \left(3 \sin{\left(3 x \right)} + \cos{\left(2 x + 4 \right)}\right)\, dx = C + \frac{\sin{\left(2 x + 4 \right)}}{2} - \cos{\left(3 x \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

    sin(6)            sin(4)
1 + ------ - cos(3) - ------
      2                 2

$$\frac{\sin{\left(6 \right)}}{2} - \frac{\sin{\left(4 \right)}}{2} - \cos{\left(3 \right)} + 1$$

    sin(6)            sin(4)
1 + ------ - cos(3) - ------
      2                 2

$$\frac{\sin{\left(6 \right)}}{2} - \frac{\sin{\left(4 \right)}}{2} - \cos{\left(3 \right)} + 1$$

1 + sin(6)/2 - cos(3) - sin(4)/2

Numerical answer [src]

2.22868599515495

2.22868599515495

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

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Integral of 3sin(3x)+cos(2x+4) dx

Limits of integration:

The graph:

Piecewise:

The solution