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sin(x/ two)^ four
sinus of (x divide by 2) to the power of 4
sinus of (x divide by two) to the power of four
sin(x/2)4
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sin(x/2)⁴
sinx/2^4
sin(x divide by 2)^4
sin(x/2)^4dx

Solving integrals/ sin(x/2)^4

Integral of sin(x/2)^4 dx

The solution

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

Integral(sin(x/2)^4, (x, 0, 1))

Detail solution

Rewrite the integrand:

$\sin^{4}{\left(\frac{x}{2} \right)} = \left(\frac{1}{2} - \frac{\cos{\left(x \right)}}{2}\right)^{2}$
There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(\frac{1}{2} - \frac{\cos{\left(x \right)}}{2}\right)^{2} = \frac{\cos^{2}{\left(x \right)}}{4} - \frac{\cos{\left(x \right)}}{2} + \frac{1}{4}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\cos^{2}{\left(x \right)}}{4}\, dx = \frac{\int \cos^{2}{\left(x \right)}\, dx}{4}$
    1. Rewrite the integrand:
      
      $\cos^{2}{\left(x \right)} = \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 x \right)}}{2}\, dx = \frac{\int \cos{\left(2 x \right)}\, dx}{2}$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      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: $\frac{\sin{\left(2 x \right)}}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4}$
    So, the result is: $\frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos{\left(x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(x \right)}\, dx}{2}$
    1. The integral of cosine is sine:
      
      $\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$
    So, the result is: $- \frac{\sin{\left(x \right)}}{2}$
  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: $\frac{3 x}{8} - \frac{\sin{\left(x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{16}$
Method #2
1. Rewrite the integrand:
  
  $\left(\frac{1}{2} - \frac{\cos{\left(x \right)}}{2}\right)^{2} = \frac{\cos^{2}{\left(x \right)}}{4} - \frac{\cos{\left(x \right)}}{2} + \frac{1}{4}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\cos^{2}{\left(x \right)}}{4}\, dx = \frac{\int \cos^{2}{\left(x \right)}\, dx}{4}$
    1. Rewrite the integrand:
      
      $\cos^{2}{\left(x \right)} = \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 x \right)}}{2}\, dx = \frac{\int \cos{\left(2 x \right)}\, dx}{2}$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      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: $\frac{\sin{\left(2 x \right)}}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4}$
    So, the result is: $\frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos{\left(x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(x \right)}\, dx}{2}$
    1. The integral of cosine is sine:
      
      $\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$
    So, the result is: $- \frac{\sin{\left(x \right)}}{2}$
  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: $\frac{3 x}{8} - \frac{\sin{\left(x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{16}$
Add the constant of integration:

$\frac{3 x}{8} - \frac{\sin{\left(x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{16}+ \mathrm{constant}$

The answer is:

$\frac{3 x}{8} - \frac{\sin{\left(x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{16}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                        
 |                                         
 |    4/x\          sin(x)   sin(2*x)   3*x
 | sin |-| dx = C - ------ + -------- + ---
 |     \2/            2         16       8 
 |                                         
/

$${{{{\sin \left(2\,x\right)}\over{2}}+x}\over{8}}-{{\sin x}\over{2}} +{{x}\over{4}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                             3              
3   3*cos(1/2)*sin(1/2)   sin (1/2)*cos(1/2)
- - ------------------- - ------------------
8            4                    2

$${{\sin 2-8\,\sin 1+6}\over{16}}$$

                             3              
3   3*cos(1/2)*sin(1/2)   sin (1/2)*cos(1/2)
- - ------------------- - ------------------
8            4                    2

$$- \frac{3 \sin{\left(\frac{1}{2} \right)} \cos{\left(\frac{1}{2} \right)}}{4} - \frac{\sin^{3}{\left(\frac{1}{2} \right)} \cos{\left(\frac{1}{2} \right)}}{2} + \frac{3}{8}$$

Упростить

Numerical answer [src]

0.0110955967726569

0.0110955967726569

The graph

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

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How to use it?

Identical expressions

Integral of sin(x/2)^4 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2