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Integral of 2sin(4x) dx

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  4              
  /              
 |               
 |  2*sin(4*x) dx
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0

\int\limits_{0}^{4} 2 \sin{\left(4 x \right)}\, dx

Integral(2*sin(4*x), (x, 0, 4))

Detail solution

The integral of a constant times a function is the constant times the integral of the function:

$\int 2 \sin{\left(4 x \right)}\, dx = 2 \int \sin{\left(4 x \right)}\, dx$
1. Let $u = 4 x$ .
  
  Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
  
  $\int \frac{\sin{\left(u \right)}}{4}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
    1. The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
    So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
  Now substitute $u$ back in:
  
  $- \frac{\cos{\left(4 x \right)}}{4}$
So, the result is: $- \frac{\cos{\left(4 x \right)}}{2}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

1   cos(16)
- - -------
2      2

\frac{1}{2} - \frac{\cos{\left(16 \right)}}{2}

1   cos(16)
- - -------
2      2

\frac{1}{2} - \frac{\cos{\left(16 \right)}}{2}

1/2 - cos(16)/2

Numerical answer [src]

0.978829740161692

0.978829740161692

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

Integral of 2*sin(4*x) dx