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Identical expressions
(-4x+ one)tsin((pix)/(two . five))
( minus 4x plus 1)t sinus of (( Pi x) divide by (2.5))
( minus 4x plus one)t sinus of (( Pi x) divide by (two . five))
-4x+1tsinpix/2.5
(-4x+1)tsin((pix) divide by (2.5))
(-4x+1)tsin((pix)/(2.5))dx
Similar expressions
(4x+1)tsin((pix)/(2.5))
(-4x-1)tsin((pix)/(2.5))

Solving integrals/ (-4x+1)tsin((pix)/(2.5))

Integral of (-4x+1)tsin((pix)/(2.5)) dx

The solution

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 5/2                         
  /                          
 |                           
 |                  /pi*x\   
 |  (-4*x + 1)*t*sin|----| dx
 |                  \5/2 /   
 |                           
/                            
0

$$\int\limits_{0}^{\frac{5}{2}} t \left(1 - 4 x\right) \sin{\left(\frac{\pi x}{\frac{5}{2}} \right)}\, dx$$

Integral(((-4*x + 1)*t)*sin((pi*x)/(5/2)), (x, 0, 5/2))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $t \left(1 - 4 x\right) \sin{\left(\frac{\pi x}{\frac{5}{2}} \right)} = - 4 t x \sin{\left(\frac{2 \pi x}{5} \right)} + t \sin{\left(\frac{2 \pi x}{5} \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 4 t x \sin{\left(\frac{2 \pi x}{5} \right)}\right)\, dx = - 4 t \int x \sin{\left(\frac{2 \pi x}{5} \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \sin{\left(\frac{2 \pi x}{5} \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = \frac{2 \pi x}{5}$ .
      
      Then let $du = \frac{2 \pi dx}{5}$ and substitute $\frac{5 du}{2 \pi}$ :
      
      $\int \frac{5 \sin{\left(u \right)}}{2 \pi}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{5 \int \sin{\left(u \right)}\, du}{2 \pi}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{5 \cos{\left(u \right)}}{2 \pi}$
      
      Now substitute $u$ back in:
      
      $- \frac{5 \cos{\left(\frac{2 \pi x}{5} \right)}}{2 \pi}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{5 \cos{\left(\frac{2 \pi x}{5} \right)}}{2 \pi}\right)\, dx = - \frac{5 \int \cos{\left(\frac{2 \pi x}{5} \right)}\, dx}{2 \pi}$
      
      Let $u = \frac{2 \pi x}{5}$ .
      
      Then let $du = \frac{2 \pi dx}{5}$ and substitute $\frac{5 du}{2 \pi}$ :
      
      $\int \frac{5 \cos{\left(u \right)}}{2 \pi}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{5 \int \cos{\left(u \right)}\, du}{2 \pi}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{5 \sin{\left(u \right)}}{2 \pi}$
      
      Now substitute $u$ back in:
      
      $\frac{5 \sin{\left(\frac{2 \pi x}{5} \right)}}{2 \pi}$
      
      So, the result is: $- \frac{25 \sin{\left(\frac{2 \pi x}{5} \right)}}{4 \pi^{2}}$
    So, the result is: $- 4 t \left(- \frac{5 x \cos{\left(\frac{2 \pi x}{5} \right)}}{2 \pi} + \frac{25 \sin{\left(\frac{2 \pi x}{5} \right)}}{4 \pi^{2}}\right)$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int t \sin{\left(\frac{2 \pi x}{5} \right)}\, dx = t \int \sin{\left(\frac{2 \pi x}{5} \right)}\, dx$
    1. Let $u = \frac{2 \pi x}{5}$ .
      
      Then let $du = \frac{2 \pi dx}{5}$ and substitute $\frac{5 du}{2 \pi}$ :
      
      $\int \frac{5 \sin{\left(u \right)}}{2 \pi}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{5 \int \sin{\left(u \right)}\, du}{2 \pi}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{5 \cos{\left(u \right)}}{2 \pi}$
      
      Now substitute $u$ back in:
      
      $- \frac{5 \cos{\left(\frac{2 \pi x}{5} \right)}}{2 \pi}$
    So, the result is: $- \frac{5 t \cos{\left(\frac{2 \pi x}{5} \right)}}{2 \pi}$
  The result is: $- 4 t \left(- \frac{5 x \cos{\left(\frac{2 \pi x}{5} \right)}}{2 \pi} + \frac{25 \sin{\left(\frac{2 \pi x}{5} \right)}}{4 \pi^{2}}\right) - \frac{5 t \cos{\left(\frac{2 \pi x}{5} \right)}}{2 \pi}$
Method #2
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = t \left(1 - 4 x\right)$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(\frac{2 \pi x}{5} \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = - 4 t$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = \frac{2 \pi x}{5}$ .
    
    Then let $du = \frac{2 \pi dx}{5}$ and substitute $\frac{5 du}{2 \pi}$ :
    
    $\int \frac{5 \sin{\left(u \right)}}{2 \pi}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{5 \int \sin{\left(u \right)}\, du}{2 \pi}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{5 \cos{\left(u \right)}}{2 \pi}$
    Now substitute $u$ back in:
    
    $- \frac{5 \cos{\left(\frac{2 \pi x}{5} \right)}}{2 \pi}$
  Now evaluate the sub-integral.
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{10 t \cos{\left(\frac{2 \pi x}{5} \right)}}{\pi}\, dx = \frac{10 t \int \cos{\left(\frac{2 \pi x}{5} \right)}\, dx}{\pi}$
  1. Let $u = \frac{2 \pi x}{5}$ .
    
    Then let $du = \frac{2 \pi dx}{5}$ and substitute $\frac{5 du}{2 \pi}$ :
    
    $\int \frac{5 \cos{\left(u \right)}}{2 \pi}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{5 \int \cos{\left(u \right)}\, du}{2 \pi}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{5 \sin{\left(u \right)}}{2 \pi}$
    Now substitute $u$ back in:
    
    $\frac{5 \sin{\left(\frac{2 \pi x}{5} \right)}}{2 \pi}$
  So, the result is: $\frac{25 t \sin{\left(\frac{2 \pi x}{5} \right)}}{\pi^{2}}$
Method #3
1. Rewrite the integrand:
  
  $t \left(1 - 4 x\right) \sin{\left(\frac{\pi x}{\frac{5}{2}} \right)} = - 4 t x \sin{\left(\frac{2 \pi x}{5} \right)} + t \sin{\left(\frac{2 \pi x}{5} \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 4 t x \sin{\left(\frac{2 \pi x}{5} \right)}\right)\, dx = - 4 t \int x \sin{\left(\frac{2 \pi x}{5} \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \sin{\left(\frac{2 \pi x}{5} \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = \frac{2 \pi x}{5}$ .
      
      Then let $du = \frac{2 \pi dx}{5}$ and substitute $\frac{5 du}{2 \pi}$ :
      
      $\int \frac{5 \sin{\left(u \right)}}{2 \pi}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{5 \int \sin{\left(u \right)}\, du}{2 \pi}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{5 \cos{\left(u \right)}}{2 \pi}$
      
      Now substitute $u$ back in:
      
      $- \frac{5 \cos{\left(\frac{2 \pi x}{5} \right)}}{2 \pi}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{5 \cos{\left(\frac{2 \pi x}{5} \right)}}{2 \pi}\right)\, dx = - \frac{5 \int \cos{\left(\frac{2 \pi x}{5} \right)}\, dx}{2 \pi}$
      
      Let $u = \frac{2 \pi x}{5}$ .
      
      Then let $du = \frac{2 \pi dx}{5}$ and substitute $\frac{5 du}{2 \pi}$ :
      
      $\int \frac{5 \cos{\left(u \right)}}{2 \pi}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{5 \int \cos{\left(u \right)}\, du}{2 \pi}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{5 \sin{\left(u \right)}}{2 \pi}$
      
      Now substitute $u$ back in:
      
      $\frac{5 \sin{\left(\frac{2 \pi x}{5} \right)}}{2 \pi}$
      
      So, the result is: $- \frac{25 \sin{\left(\frac{2 \pi x}{5} \right)}}{4 \pi^{2}}$
    So, the result is: $- 4 t \left(- \frac{5 x \cos{\left(\frac{2 \pi x}{5} \right)}}{2 \pi} + \frac{25 \sin{\left(\frac{2 \pi x}{5} \right)}}{4 \pi^{2}}\right)$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int t \sin{\left(\frac{2 \pi x}{5} \right)}\, dx = t \int \sin{\left(\frac{2 \pi x}{5} \right)}\, dx$
    1. Let $u = \frac{2 \pi x}{5}$ .
      
      Then let $du = \frac{2 \pi dx}{5}$ and substitute $\frac{5 du}{2 \pi}$ :
      
      $\int \frac{5 \sin{\left(u \right)}}{2 \pi}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{5 \int \sin{\left(u \right)}\, du}{2 \pi}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{5 \cos{\left(u \right)}}{2 \pi}$
      
      Now substitute $u$ back in:
      
      $- \frac{5 \cos{\left(\frac{2 \pi x}{5} \right)}}{2 \pi}$
    So, the result is: $- \frac{5 t \cos{\left(\frac{2 \pi x}{5} \right)}}{2 \pi}$
  The result is: $- 4 t \left(- \frac{5 x \cos{\left(\frac{2 \pi x}{5} \right)}}{2 \pi} + \frac{25 \sin{\left(\frac{2 \pi x}{5} \right)}}{4 \pi^{2}}\right) - \frac{5 t \cos{\left(\frac{2 \pi x}{5} \right)}}{2 \pi}$
Now simplify:

$\frac{5 t \left(\pi \left(4 x - 1\right) \cos{\left(\frac{2 \pi x}{5} \right)} - 10 \sin{\left(\frac{2 \pi x}{5} \right)}\right)}{2 \pi^{2}}$
Add the constant of integration:

$\frac{5 t \left(\pi \left(4 x - 1\right) \cos{\left(\frac{2 \pi x}{5} \right)} - 10 \sin{\left(\frac{2 \pi x}{5} \right)}\right)}{2 \pi^{2}}+ \mathrm{constant}$

The answer is:

$\frac{5 t \left(\pi \left(4 x - 1\right) \cos{\left(\frac{2 \pi x}{5} \right)} - 10 \sin{\left(\frac{2 \pi x}{5} \right)}\right)}{2 \pi^{2}}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                    /      /2*pi*x\          /2*pi*x\\          /2*pi*x\
 |                                     |25*sin|------|   5*x*cos|------||   5*t*cos|------|
 |                 /pi*x\              |      \  5   /          \  5   /|          \  5   /
 | (-4*x + 1)*t*sin|----| dx = C - 4*t*|-------------- - ---------------| - ---------------
 |                 \5/2 /              |        2              2*pi     |         2*pi     
 |                                     \    4*pi                        /                  
/

$$\int t \left(1 - 4 x\right) \sin{\left(\frac{\pi x}{\frac{5}{2}} \right)}\, dx = C - 4 t \left(- \frac{5 x \cos{\left(\frac{2 \pi x}{5} \right)}}{2 \pi} + \frac{25 \sin{\left(\frac{2 \pi x}{5} \right)}}{4 \pi^{2}}\right) - \frac{5 t \cos{\left(\frac{2 \pi x}{5} \right)}}{2 \pi}$$

Simplify

The answer [src]

-20*t
-----
  pi

$$- \frac{20 t}{\pi}$$

-20*t
-----
  pi

$$- \frac{20 t}{\pi}$$

-20*t/pi

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

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Identical expressions

Similar expressions

Integral of (-4x+1)tsin((pix)/(2.5)) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3