Integral of sin2x-sinx dx

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 |  (sin(2*x) - sin(x)) dx
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$$\int\limits_{0}^{\frac{\pi}{3}} \left(- \sin{\left(x \right)} + \sin{\left(2 x \right)}\right)\, dx$$

Integral(sin(2*x) - sin(x), (x, 0, pi/3))

Detail solution

Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \sin{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)}\, dx$
  1. The integral of sine is negative cosine:
    
    $\int \sin{\left(x \right)}\, dx = - \cos{\left(x \right)}$
  So, the result is: $\cos{\left(x \right)}$
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: $- \cos^{2}{\left(x \right)}$
The result is: $\cos{\left(x \right)} - \frac{\cos{\left(2 x \right)}}{2}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

1/4

$$\frac{1}{4}$$

=

1/4

$$\frac{1}{4}$$

1/4

Numerical answer [src]

0.25

0.25

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Integral of sin2x-sinx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #1

Method #2