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Solving integrals/ 1/pi((pi*x-1)sin(2x))

Integral of 1/pi((pi*x-1)sin(2x)) dx

The solution

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 pi                       
  /                       
 |                        
 |  (pi*x - 1)*sin(2*x)   
 |  ------------------- dx
 |           pi           
 |                        
/                         
-pi

$$\int\limits_{- \pi}^{\pi} \frac{\left(\pi x - 1\right) \sin{\left(2 x \right)}}{\pi}\, dx$$

Integral(((pi*x - 1)*sin(2*x))/pi, (x, -pi, pi))

Detail solution

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

$\int \frac{\left(\pi x - 1\right) \sin{\left(2 x \right)}}{\pi}\, dx = \frac{\int \left(\pi x - 1\right) \sin{\left(2 x \right)}\, dx}{\pi}$
1. There are multiple ways to do this integral.
  Method #1
  1. Rewrite the integrand:
    
    $\left(\pi x - 1\right) \sin{\left(2 x \right)} = \pi x \sin{\left(2 x \right)} - \sin{\left(2 x \right)}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \pi x \sin{\left(2 x \right)}\, dx = \pi \int x \sin{\left(2 x \right)}\, dx$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(2 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      There are multiple ways to do this integral.
      
      Method #1
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\sin{\left(u \right)}}{2}\, du$
      
      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
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 \sin{\left(x \right)} \cos{\left(x \right)}\, dx = 2 \int \sin{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      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:
      
      $- \frac{\cos^{2}{\left(x \right)}}{2}$
      
      So, the result is: $- \cos^{2}{\left(x \right)}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(2 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{2}$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      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:
      
      $\frac{\sin{\left(2 x \right)}}{2}$
      
      So, the result is: $- \frac{\sin{\left(2 x \right)}}{4}$
      
      So, the result is: $\pi \left(- \frac{x \cos{\left(2 x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{4}\right)$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin{\left(2 x \right)}\right)\, dx = - \int \sin{\left(2 x \right)}\, dx$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\sin{\left(u \right)}}{2}\, du$
      
      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:
      
      $- \frac{\cos{\left(2 x \right)}}{2}$
      
      So, the result is: $\frac{\cos{\left(2 x \right)}}{2}$
    The result is: $\pi \left(- \frac{x \cos{\left(2 x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{4}\right) + \frac{\cos{\left(2 x \right)}}{2}$
  Method #2
  1. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(x \right)} = \pi x - 1$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(2 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = \pi$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\sin{\left(u \right)}}{2}\, du$
      
      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:
      
      $- \frac{\cos{\left(2 x \right)}}{2}$
    Now evaluate the sub-integral.
  2. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\pi \cos{\left(2 x \right)}}{2}\right)\, dx = - \frac{\pi \int \cos{\left(2 x \right)}\, dx}{2}$
    1. Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      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:
      
      $\frac{\sin{\left(2 x \right)}}{2}$
    So, the result is: $- \frac{\pi \sin{\left(2 x \right)}}{4}$
  Method #3
  1. Rewrite the integrand:
    
    $\left(\pi x - 1\right) \sin{\left(2 x \right)} = \pi x \sin{\left(2 x \right)} - \sin{\left(2 x \right)}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \pi x \sin{\left(2 x \right)}\, dx = \pi \int x \sin{\left(2 x \right)}\, dx$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(2 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\sin{\left(u \right)}}{2}\, du$
      
      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:
      
      $- \frac{\cos{\left(2 x \right)}}{2}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(2 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{2}$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      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:
      
      $\frac{\sin{\left(2 x \right)}}{2}$
      
      So, the result is: $- \frac{\sin{\left(2 x \right)}}{4}$
      
      So, the result is: $\pi \left(- \frac{x \cos{\left(2 x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{4}\right)$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin{\left(2 x \right)}\right)\, dx = - \int \sin{\left(2 x \right)}\, dx$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\sin{\left(u \right)}}{2}\, du$
      
      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:
      
      $- \frac{\cos{\left(2 x \right)}}{2}$
      
      So, the result is: $\frac{\cos{\left(2 x \right)}}{2}$
    The result is: $\pi \left(- \frac{x \cos{\left(2 x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{4}\right) + \frac{\cos{\left(2 x \right)}}{2}$
  Method #4
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 \left(\pi x - 1\right) \sin{\left(x \right)} \cos{\left(x \right)}\, dx = 2 \int \left(\pi x - 1\right) \sin{\left(x \right)} \cos{\left(x \right)}\, dx$
    1. Rewrite the integrand:
      
      $\left(\pi x - 1\right) \sin{\left(x \right)} \cos{\left(x \right)} = \pi x \sin{\left(x \right)} \cos{\left(x \right)} - \sin{\left(x \right)} \cos{\left(x \right)}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \pi x \sin{\left(x \right)} \cos{\left(x \right)}\, dx = \pi \int x \sin{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(x \right)} \cos{\left(x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(2 x \right)}}{2}\, dx = \frac{\int \sin{\left(2 x \right)}\, dx}{2}$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\sin{\left(u \right)}}{2}\, du$
      
      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:
      
      $- \frac{\cos{\left(2 x \right)}}{2}$
      
      So, the result is: $- \frac{\cos{\left(2 x \right)}}{4}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(2 x \right)}}{4}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{4}$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      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:
      
      $\frac{\sin{\left(2 x \right)}}{2}$
      
      So, the result is: $- \frac{\sin{\left(2 x \right)}}{8}$
      
      So, the result is: $\pi \left(- \frac{x \cos{\left(2 x \right)}}{4} + \frac{\sin{\left(2 x \right)}}{8}\right)$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos{\left(x \right)}\, dx$
      
      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:
      
      $- \frac{\cos^{2}{\left(x \right)}}{2}$
      
      So, the result is: $\frac{\cos^{2}{\left(x \right)}}{2}$
      
      The result is: $\pi \left(- \frac{x \cos{\left(2 x \right)}}{4} + \frac{\sin{\left(2 x \right)}}{8}\right) + \frac{\cos^{2}{\left(x \right)}}{2}$
    So, the result is: $2 \pi \left(- \frac{x \cos{\left(2 x \right)}}{4} + \frac{\sin{\left(2 x \right)}}{8}\right) + \cos^{2}{\left(x \right)}$
So, the result is: $\frac{\pi \left(- \frac{x \cos{\left(2 x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{4}\right) + \frac{\cos{\left(2 x \right)}}{2}}{\pi}$
Now simplify:

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

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

The answer is:

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

The answer (Indefinite) [src]

  /                             cos(2*x)      /sin(2*x)   x*cos(2*x)\
 |                              -------- + pi*|-------- - ----------|
 | (pi*x - 1)*sin(2*x)             2          \   4           2     /
 | ------------------- dx = C + -------------------------------------
 |          pi                                    pi                 
 |                                                                   
/

$$\int \frac{\left(\pi x - 1\right) \sin{\left(2 x \right)}}{\pi}\, dx = C + \frac{\pi \left(- \frac{x \cos{\left(2 x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{4}\right) + \frac{\cos{\left(2 x \right)}}{2}}{\pi}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

      2         2
1   pi    1   pi 
- - ---   - + ---
2    2    2    2 
------- - -------
   pi        pi

$$- \frac{\frac{1}{2} + \frac{\pi^{2}}{2}}{\pi} + \frac{\frac{1}{2} - \frac{\pi^{2}}{2}}{\pi}$$

      2         2
1   pi    1   pi 
- - ---   - + ---
2    2    2    2 
------- - -------
   pi        pi

$$- \frac{\frac{1}{2} + \frac{\pi^{2}}{2}}{\pi} + \frac{\frac{1}{2} - \frac{\pi^{2}}{2}}{\pi}$$

(1/2 - pi^2/2)/pi - (1/2 + pi^2/2)/pi

Numerical answer [src]

-3.14159265358979

-3.14159265358979

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

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How to use it?

Identical expressions

Similar expressions

Integral of 1/pi((pi*x-1)sin(2x)) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2

Method #3

Method #4