Integral of -sin2x dx

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 2*pi            
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  |  -sin(2*x) dx
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 0

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

Integral(-sin(2*x), (x, 0, 2*pi))

Detail solution

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

$\int \left(- \sin{\left(2 x \right)}\right)\, dx = - \int \sin{\left(2 x \right)}\, dx$
1. There are multiple ways to do this integral.
  Method #1
  1. Let $u = 2 x$ .
    
    Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{\sin{\left(u \right)}}{2}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{2}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(2 x \right)}}{2}$
  Method #2
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 \sin{\left(x \right)} \cos{\left(x \right)}\, dx = 2 \int \sin{\left(x \right)} \cos{\left(x \right)}\, dx$
    1. There are multiple ways to do this integral.
      
      Method #1
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int \left(- u\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u\, du = - \int u\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      So, the result is: $- \frac{u^{2}}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{2}{\left(x \right)}}{2}$
      
      Method #2
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{2}{\left(x \right)}}{2}$
    So, the result is: $- \cos^{2}{\left(x \right)}$
So, the result is: $\frac{\cos{\left(2 x \right)}}{2}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

$$0$$

=

$$0$$

Numerical answer [src]

5.4435142372695e-22

5.4435142372695e-22

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Integral of -sin2x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #1

Method #2