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Integral of d{x}:
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Integral of sin^4(Pix/l)
Integral of sin^2(2x)cos^2(2x)
Identical expressions
sin^ four (Pix/l)
sinus of to the power of 4(Pix divide by l)
sinus of to the power of four (Pix divide by l)
sin4(Pix/l)
sin4Pix/l
sin⁴(Pix/l)
sin^4Pix/l
sin^4(Pix divide by l)
sin^4(Pix/l)dx

Integral of sin^4(Pix/l) dx

The solution

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  l              
  /              
 |               
 |     4/pi*x\   
 |  sin |----| dx
 |      \ l  /   
 |               
/                
0

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

Integral(sin((pi*x)/l)^4, (x, 0, l))

Detail solution

Rewrite the integrand:

$\sin^{4}{\left(\frac{\pi x}{l} \right)} = \left(\frac{1}{2} - \frac{\cos{\left(\frac{2 \pi x}{l} \right)}}{2}\right)^{2}$
There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(\frac{1}{2} - \frac{\cos{\left(\frac{2 \pi x}{l} \right)}}{2}\right)^{2} = \frac{\cos^{2}{\left(\frac{2 \pi x}{l} \right)}}{4} - \frac{\cos{\left(\frac{2 \pi x}{l} \right)}}{2} + \frac{1}{4}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\cos^{2}{\left(\frac{2 \pi x}{l} \right)}}{4}\, dx = \frac{\int \cos^{2}{\left(\frac{2 \pi x}{l} \right)}\, dx}{4}$
    1. Rewrite the integrand:
      
      $\cos^{2}{\left(\frac{2 \pi x}{l} \right)} = \frac{\cos{\left(\frac{4 \pi x}{l} \right)}}{2} + \frac{1}{2}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(\frac{4 \pi x}{l} \right)}}{2}\, dx = \frac{\int \cos{\left(\frac{4 \pi x}{l} \right)}\, dx}{2}$
      
      Let $u = \frac{4 \pi x}{l}$ .
      
      Then let $du = \frac{4 \pi dx}{l}$ and substitute $\frac{du l}{4 \pi}$ :
      
      $\int \frac{l \cos{\left(u \right)}}{4 \pi}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{l \int \cos{\left(u \right)}\, du}{4 \pi}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{l \sin{\left(u \right)}}{4 \pi}$
      
      Now substitute $u$ back in:
      
      $\frac{l \sin{\left(\frac{4 \pi x}{l} \right)}}{4 \pi}$
      
      So, the result is: $\frac{l \sin{\left(\frac{4 \pi x}{l} \right)}}{8 \pi}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{l \sin{\left(\frac{4 \pi x}{l} \right)}}{8 \pi} + \frac{x}{2}$
    So, the result is: $\frac{l \sin{\left(\frac{4 \pi x}{l} \right)}}{32 \pi} + \frac{x}{8}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos{\left(\frac{2 \pi x}{l} \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(\frac{2 \pi x}{l} \right)}\, dx}{2}$
    1. Let $u = \frac{2 \pi x}{l}$ .
      
      Then let $du = \frac{2 \pi dx}{l}$ and substitute $\frac{du l}{2 \pi}$ :
      
      $\int \frac{l \cos{\left(u \right)}}{2 \pi}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{l \int \cos{\left(u \right)}\, du}{2 \pi}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{l \sin{\left(u \right)}}{2 \pi}$
      
      Now substitute $u$ back in:
      
      $\frac{l \sin{\left(\frac{2 \pi x}{l} \right)}}{2 \pi}$
    So, the result is: $- \frac{l \sin{\left(\frac{2 \pi x}{l} \right)}}{4 \pi}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \frac{1}{4}\, dx = \frac{x}{4}$
  The result is: $- \frac{l \sin{\left(\frac{2 \pi x}{l} \right)}}{4 \pi} + \frac{l \sin{\left(\frac{4 \pi x}{l} \right)}}{32 \pi} + \frac{3 x}{8}$
Method #2
1. Rewrite the integrand:
  
  $\left(\frac{1}{2} - \frac{\cos{\left(\frac{2 \pi x}{l} \right)}}{2}\right)^{2} = \frac{\cos^{2}{\left(\frac{2 \pi x}{l} \right)}}{4} - \frac{\cos{\left(\frac{2 \pi x}{l} \right)}}{2} + \frac{1}{4}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\cos^{2}{\left(\frac{2 \pi x}{l} \right)}}{4}\, dx = \frac{\int \cos^{2}{\left(\frac{2 \pi x}{l} \right)}\, dx}{4}$
    1. Rewrite the integrand:
      
      $\cos^{2}{\left(\frac{2 \pi x}{l} \right)} = \frac{\cos{\left(\frac{4 \pi x}{l} \right)}}{2} + \frac{1}{2}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(\frac{4 \pi x}{l} \right)}}{2}\, dx = \frac{\int \cos{\left(\frac{4 \pi x}{l} \right)}\, dx}{2}$
      
      Let $u = \frac{4 \pi x}{l}$ .
      
      Then let $du = \frac{4 \pi dx}{l}$ and substitute $\frac{du l}{4 \pi}$ :
      
      $\int \frac{l \cos{\left(u \right)}}{4 \pi}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{l \int \cos{\left(u \right)}\, du}{4 \pi}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{l \sin{\left(u \right)}}{4 \pi}$
      
      Now substitute $u$ back in:
      
      $\frac{l \sin{\left(\frac{4 \pi x}{l} \right)}}{4 \pi}$
      
      So, the result is: $\frac{l \sin{\left(\frac{4 \pi x}{l} \right)}}{8 \pi}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{l \sin{\left(\frac{4 \pi x}{l} \right)}}{8 \pi} + \frac{x}{2}$
    So, the result is: $\frac{l \sin{\left(\frac{4 \pi x}{l} \right)}}{32 \pi} + \frac{x}{8}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos{\left(\frac{2 \pi x}{l} \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(\frac{2 \pi x}{l} \right)}\, dx}{2}$
    1. Let $u = \frac{2 \pi x}{l}$ .
      
      Then let $du = \frac{2 \pi dx}{l}$ and substitute $\frac{du l}{2 \pi}$ :
      
      $\int \frac{l \cos{\left(u \right)}}{2 \pi}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{l \int \cos{\left(u \right)}\, du}{2 \pi}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{l \sin{\left(u \right)}}{2 \pi}$
      
      Now substitute $u$ back in:
      
      $\frac{l \sin{\left(\frac{2 \pi x}{l} \right)}}{2 \pi}$
    So, the result is: $- \frac{l \sin{\left(\frac{2 \pi x}{l} \right)}}{4 \pi}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \frac{1}{4}\, dx = \frac{x}{4}$
  The result is: $- \frac{l \sin{\left(\frac{2 \pi x}{l} \right)}}{4 \pi} + \frac{l \sin{\left(\frac{4 \pi x}{l} \right)}}{32 \pi} + \frac{3 x}{8}$
Now simplify:

$\frac{- 8 l \sin{\left(\frac{2 \pi x}{l} \right)} + l \sin{\left(\frac{4 \pi x}{l} \right)} + 12 \pi x}{32 \pi}$
Add the constant of integration:

$\frac{- 8 l \sin{\left(\frac{2 \pi x}{l} \right)} + l \sin{\left(\frac{4 \pi x}{l} \right)} + 12 \pi x}{32 \pi}+ \mathrm{constant}$

The answer is:

\frac{- 8 l \sin{\left(\frac{2 \pi x}{l} \right)} + l \sin{\left(\frac{4 \pi x}{l} \right)} + 12 \pi x}{32 \pi}+ \mathrm{constant}

The answer (Indefinite) [src]

  /                               /2*pi*x\        /4*pi*x\
 |                           l*sin|------|   l*sin|------|
 |    4/pi*x\          3*x        \  l   /        \  l   /
 | sin |----| dx = C + --- - ------------- + -------------
 |     \ l  /           8         4*pi           32*pi    
 |                                                        
/

\int \sin^{4}{\left(\frac{\pi x}{l} \right)}\, dx = C - \frac{l \sin{\left(\frac{2 \pi x}{l} \right)}}{4 \pi} + \frac{l \sin{\left(\frac{4 \pi x}{l} \right)}}{32 \pi} + \frac{3 x}{8}

Simplify

The answer [src]

/3*l                                  
|---  for And(l > -oo, l < oo, l != 0)
< 8                                   
|                                     
\ 0              otherwise

\begin{cases} \frac{3 l}{8} & \text{for}\: l > -\infty \wedge l < \infty \wedge l \neq 0 \\0 & \text{otherwise} \end{cases}

/3*l                                  
|---  for And(l > -oo, l < oo, l != 0)
< 8                                   
|                                     
\ 0              otherwise

\begin{cases} \frac{3 l}{8} & \text{for}\: l > -\infty \wedge l < \infty \wedge l \neq 0 \\0 & \text{otherwise} \end{cases}

Piecewise((3*l/8, (l > -oo)∧(l < oo)∧(Ne(l, 0))), (0, True))

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

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Identical expressions

Integral of sin^4(Pix/l) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2