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Identical expressions
(8x+ one)sin3pix/ four
(8x plus 1) sinus of 3 Pi x divide by 4
(8x plus one) sinus of 3 Pi x divide by four
8x+1sin3pix/4
(8x+1)sin3pix divide by 4
(8x+1)sin3pix/4dx
Similar expressions
(8x-1)sin3pix/4

Integral of (8x+1)sin3pix/4 dx

The solution

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  0                         
  /                         
 |                          
 |  (8*x + 1)*sin(3*pi*x)   
 |  --------------------- dx
 |            4             
 |                          
/                           
-1

$$\int\limits_{-1}^{0} \frac{\left(8 x + 1\right) \sin{\left(3 \pi x \right)}}{4}\, dx$$

Integral(((8*x + 1)*sin((3*pi)*x))/4, (x, -1, 0))

Detail solution

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

$\int \frac{\left(8 x + 1\right) \sin{\left(3 \pi x \right)}}{4}\, dx = \frac{\int \left(8 x + 1\right) \sin{\left(3 \pi x \right)}\, dx}{4}$
1. There are multiple ways to do this integral.
  Method #1
  1. Rewrite the integrand:
    
    $\left(8 x + 1\right) \sin{\left(3 \pi x \right)} = 8 x \sin{\left(3 \pi x \right)} + \sin{\left(3 \pi x \right)}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 8 x \sin{\left(3 \pi x \right)}\, dx = 8 \int x \sin{\left(3 \pi x \right)}\, dx$
      
      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 $\operatorname{dv}{\left(x \right)} = \sin{\left(3 \pi x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 3 \pi x$ .
      
      Then let $du = 3 \pi dx$ and substitute $\frac{du}{3 \pi}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3 \pi}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3 \pi}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3 \pi}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(3 \pi x \right)}}{3 \pi}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(3 \pi x \right)}}{3 \pi}\right)\, dx = - \frac{\int \cos{\left(3 \pi x \right)}\, dx}{3 \pi}$
      
      Let $u = 3 \pi x$ .
      
      Then let $du = 3 \pi dx$ and substitute $\frac{du}{3 \pi}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3 \pi}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3 \pi}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{3 \pi}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(3 \pi x \right)}}{3 \pi}$
      
      So, the result is: $- \frac{\sin{\left(3 \pi x \right)}}{9 \pi^{2}}$
      
      So, the result is: $- \frac{8 x \cos{\left(3 \pi x \right)}}{3 \pi} + \frac{8 \sin{\left(3 \pi x \right)}}{9 \pi^{2}}$
    1. Let $u = 3 \pi x$ .
      
      Then let $du = 3 \pi dx$ and substitute $\frac{du}{3 \pi}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3 \pi}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3 \pi}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3 \pi}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(3 \pi x \right)}}{3 \pi}$
    The result is: $- \frac{8 x \cos{\left(3 \pi x \right)}}{3 \pi} + \frac{8 \sin{\left(3 \pi x \right)}}{9 \pi^{2}} - \frac{\cos{\left(3 \pi x \right)}}{3 \pi}$
  Method #2
  1. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(x \right)} = 8 x + 1$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(3 \pi x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 8$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 3 \pi x$ .
      
      Then let $du = 3 \pi dx$ and substitute $\frac{du}{3 \pi}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3 \pi}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3 \pi}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3 \pi}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(3 \pi x \right)}}{3 \pi}$
    Now evaluate the sub-integral.
  2. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{8 \cos{\left(3 \pi x \right)}}{3 \pi}\right)\, dx = - \frac{8 \int \cos{\left(3 \pi x \right)}\, dx}{3 \pi}$
    1. Let $u = 3 \pi x$ .
      
      Then let $du = 3 \pi dx$ and substitute $\frac{du}{3 \pi}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3 \pi}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3 \pi}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{3 \pi}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(3 \pi x \right)}}{3 \pi}$
    So, the result is: $- \frac{8 \sin{\left(3 \pi x \right)}}{9 \pi^{2}}$
  Method #3
  1. Rewrite the integrand:
    
    $\left(8 x + 1\right) \sin{\left(3 \pi x \right)} = 8 x \sin{\left(3 \pi x \right)} + \sin{\left(3 \pi x \right)}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 8 x \sin{\left(3 \pi x \right)}\, dx = 8 \int x \sin{\left(3 \pi x \right)}\, dx$
      
      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 $\operatorname{dv}{\left(x \right)} = \sin{\left(3 \pi x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 3 \pi x$ .
      
      Then let $du = 3 \pi dx$ and substitute $\frac{du}{3 \pi}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3 \pi}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3 \pi}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3 \pi}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(3 \pi x \right)}}{3 \pi}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(3 \pi x \right)}}{3 \pi}\right)\, dx = - \frac{\int \cos{\left(3 \pi x \right)}\, dx}{3 \pi}$
      
      Let $u = 3 \pi x$ .
      
      Then let $du = 3 \pi dx$ and substitute $\frac{du}{3 \pi}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3 \pi}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3 \pi}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{3 \pi}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(3 \pi x \right)}}{3 \pi}$
      
      So, the result is: $- \frac{\sin{\left(3 \pi x \right)}}{9 \pi^{2}}$
      
      So, the result is: $- \frac{8 x \cos{\left(3 \pi x \right)}}{3 \pi} + \frac{8 \sin{\left(3 \pi x \right)}}{9 \pi^{2}}$
    1. Let $u = 3 \pi x$ .
      
      Then let $du = 3 \pi dx$ and substitute $\frac{du}{3 \pi}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3 \pi}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3 \pi}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3 \pi}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(3 \pi x \right)}}{3 \pi}$
    The result is: $- \frac{8 x \cos{\left(3 \pi x \right)}}{3 \pi} + \frac{8 \sin{\left(3 \pi x \right)}}{9 \pi^{2}} - \frac{\cos{\left(3 \pi x \right)}}{3 \pi}$
So, the result is: $- \frac{2 x \cos{\left(3 \pi x \right)}}{3 \pi} + \frac{2 \sin{\left(3 \pi x \right)}}{9 \pi^{2}} - \frac{\cos{\left(3 \pi x \right)}}{12 \pi}$
Now simplify:

$\frac{- 3 \pi \left(8 x + 1\right) \cos{\left(3 \pi x \right)} + 8 \sin{\left(3 \pi x \right)}}{36 \pi^{2}}$
Add the constant of integration:

$\frac{- 3 \pi \left(8 x + 1\right) \cos{\left(3 \pi x \right)} + 8 \sin{\left(3 \pi x \right)}}{36 \pi^{2}}+ \mathrm{constant}$

The answer is:

$\frac{- 3 \pi \left(8 x + 1\right) \cos{\left(3 \pi x \right)} + 8 \sin{\left(3 \pi x \right)}}{36 \pi^{2}}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                            
 |                                                                             
 | (8*x + 1)*sin(3*pi*x)          cos(3*pi*x)   2*sin(3*pi*x)   2*x*cos(3*pi*x)
 | --------------------- dx = C - ----------- + ------------- - ---------------
 |           4                       12*pi              2             3*pi     
 |                                                  9*pi                       
/

$$\int \frac{\left(8 x + 1\right) \sin{\left(3 \pi x \right)}}{4}\, dx = C - \frac{2 x \cos{\left(3 \pi x \right)}}{3 \pi} + \frac{2 \sin{\left(3 \pi x \right)}}{9 \pi^{2}} - \frac{\cos{\left(3 \pi x \right)}}{12 \pi}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

 1  
----
2*pi

$$\frac{1}{2 \pi}$$

 1  
----
2*pi

$$\frac{1}{2 \pi}$$

1/(2*pi)

Numerical answer [src]

0.159154943091895

0.159154943091895

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

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Identical expressions

Similar expressions

Integral of (8x+1)sin3pix/4 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3