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Integral of d{x}:
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Identical expressions
(x-pi/ four)*sin(two *x)
(x minus Pi divide by 4) multiply by sinus of (2 multiply by x)
(x minus Pi divide by four) multiply by sinus of (two multiply by x)
(x-pi/4)sin(2x)
x-pi/4sin2x
(x-pi divide by 4)*sin(2*x)
(x-pi/4)*sin(2*x)dx
Similar expressions
(x+pi/4)*sin(2*x)

Integral of (x-pi/4)sin(2x) dx

The solution

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 pi                     
 --                     
 4                      
  /                     
 |                      
 |  /    pi\            
 |  |x - --|*sin(2*x) dx
 |  \    4 /            
 |                      
/                       
0

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

Integral((x - pi/4)*sin(2*x), (x, 0, pi/4))

Detail solution

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

$- \frac{x \cos{\left(2 x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\pi \cos{\left(2 x \right)}}{8}+ \mathrm{constant}$

The answer is:

$- \frac{x \cos{\left(2 x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\pi \cos{\left(2 x \right)}}{8}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                              
 |                                                               
 | /    pi\                   sin(2*x)   x*cos(2*x)   pi*cos(2*x)
 | |x - --|*sin(2*x) dx = C + -------- - ---------- + -----------
 | \    4 /                      4           2             8     
 |                                                               
/

$$\int \left(x - \frac{\pi}{4}\right) \sin{\left(2 x \right)}\, dx = C - \frac{x \cos{\left(2 x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\pi \cos{\left(2 x \right)}}{8}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

1   pi
- - --
4   8

$$\frac{1}{4} - \frac{\pi}{8}$$

1   pi
- - --
4   8

$$\frac{1}{4} - \frac{\pi}{8}$$

1/4 - pi/8

Numerical answer [src]

-0.142699081698724

-0.142699081698724

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (x-pi/4)sin(2x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2

Method #3

Method #4

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (x-pi/4)*sin(2*x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2

Method #3

Method #4

Integral of (x-pi/4)sin(2x) dx