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Integral of d{x}:
Integral of x*2^(-x)
Integral of ex^2
Integral of (1/x^2)dx
Integral of x*cos(x/2)
Derivative of:
sin^8(x/4)
Identical expressions
sin^ eight (x/ four)
sinus of to the power of 8(x divide by 4)
sinus of to the power of eight (x divide by four)
sin8(x/4)
sin8x/4
sin⁸(x/4)
sin^8x/4
sin^8(x divide by 4)
sin^8(x/4)dx

Integral of sin^8(x/4) dx

The solution

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 2*p          
  /           
 |            
 |     8/x\   
 |  sin |-| dx
 |      \4/   
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/             
0

$$\int\limits_{0}^{2 p} \sin^{8}{\left(\frac{x}{4} \right)}\, dx$$

Integral(sin(x/4)^8, (x, 0, 2*p))

Detail solution

Rewrite the integrand:

$\sin^{8}{\left(\frac{x}{4} \right)} = \left(\frac{1}{2} - \frac{\cos{\left(\frac{x}{2} \right)}}{2}\right)^{4}$
There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(\frac{1}{2} - \frac{\cos{\left(\frac{x}{2} \right)}}{2}\right)^{4} = \frac{\cos^{4}{\left(\frac{x}{2} \right)}}{16} - \frac{\cos^{3}{\left(\frac{x}{2} \right)}}{4} + \frac{3 \cos^{2}{\left(\frac{x}{2} \right)}}{8} - \frac{\cos{\left(\frac{x}{2} \right)}}{4} + \frac{1}{16}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\cos^{4}{\left(\frac{x}{2} \right)}}{16}\, dx = \frac{\int \cos^{4}{\left(\frac{x}{2} \right)}\, dx}{16}$
    1. Rewrite the integrand:
      
      $\cos^{4}{\left(\frac{x}{2} \right)} = \left(\frac{\cos{\left(x \right)}}{2} + \frac{1}{2}\right)^{2}$
    2. Rewrite the integrand:
      
      $\left(\frac{\cos{\left(x \right)}}{2} + \frac{1}{2}\right)^{2} = \frac{\cos^{2}{\left(x \right)}}{4} + \frac{\cos{\left(x \right)}}{2} + \frac{1}{4}$
    3. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{2}{\left(x \right)}}{4}\, dx = \frac{\int \cos^{2}{\left(x \right)}\, dx}{4}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(x \right)} = \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}$
      
      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 $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du$
      
      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: $\frac{\sin{\left(2 x \right)}}{4}$
      
      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: $\frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(x \right)}}{2}\, dx = \frac{\int \cos{\left(x \right)}\, dx}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$
      
      So, the result is: $\frac{\sin{\left(x \right)}}{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{4}\, dx = \frac{x}{4}$
      
      The result is: $\frac{3 x}{8} + \frac{\sin{\left(x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{16}$
    So, the result is: $\frac{3 x}{128} + \frac{\sin{\left(x \right)}}{32} + \frac{\sin{\left(2 x \right)}}{256}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos^{3}{\left(\frac{x}{2} \right)}}{4}\right)\, dx = - \frac{\int \cos^{3}{\left(\frac{x}{2} \right)}\, dx}{4}$
    1. Rewrite the integrand:
      
      $\cos^{3}{\left(\frac{x}{2} \right)} = \left(1 - \sin^{2}{\left(\frac{x}{2} \right)}\right) \cos{\left(\frac{x}{2} \right)}$
    2. There are multiple ways to do this integral.
      
      Method #1
      
      Let $u = \sin{\left(\frac{x}{2} \right)}$ .
      
      Then let $du = \frac{\cos{\left(\frac{x}{2} \right)} dx}{2}$ and substitute $du$ :
      
      $\int \left(2 - 2 u^{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 2\, du = 2 u$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 u^{2}\right)\, du = - 2 \int u^{2}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      So, the result is: $- \frac{2 u^{3}}{3}$
      
      The result is: $- \frac{2 u^{3}}{3} + 2 u$
      
      Now substitute $u$ back in:
      
      $- \frac{2 \sin^{3}{\left(\frac{x}{2} \right)}}{3} + 2 \sin{\left(\frac{x}{2} \right)}$
      
      Method #2
      
      Rewrite the integrand:
      
      $\left(1 - \sin^{2}{\left(\frac{x}{2} \right)}\right) \cos{\left(\frac{x}{2} \right)} = - \sin^{2}{\left(\frac{x}{2} \right)} \cos{\left(\frac{x}{2} \right)} + \cos{\left(\frac{x}{2} \right)}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin^{2}{\left(\frac{x}{2} \right)} \cos{\left(\frac{x}{2} \right)}\right)\, dx = - \int \sin^{2}{\left(\frac{x}{2} \right)} \cos{\left(\frac{x}{2} \right)}\, dx$
      
      Let $u = \sin{\left(\frac{x}{2} \right)}$ .
      
      Then let $du = \frac{\cos{\left(\frac{x}{2} \right)} dx}{2}$ and substitute $2 du$ :
      
      $\int 2 u^{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{2}\, du = 2 \int u^{2}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      So, the result is: $\frac{2 u^{3}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{2 \sin^{3}{\left(\frac{x}{2} \right)}}{3}$
      
      So, the result is: $- \frac{2 \sin^{3}{\left(\frac{x}{2} \right)}}{3}$
      
      Let $u = \frac{x}{2}$ .
      
      Then let $du = \frac{dx}{2}$ and substitute $2 du$ :
      
      $\int 2 \cos{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = 2 \int \cos{\left(u \right)}\, du$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $2 \sin{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $2 \sin{\left(\frac{x}{2} \right)}$
      
      The result is: $- \frac{2 \sin^{3}{\left(\frac{x}{2} \right)}}{3} + 2 \sin{\left(\frac{x}{2} \right)}$
      
      Method #3
      
      Rewrite the integrand:
      
      $\left(1 - \sin^{2}{\left(\frac{x}{2} \right)}\right) \cos{\left(\frac{x}{2} \right)} = - \sin^{2}{\left(\frac{x}{2} \right)} \cos{\left(\frac{x}{2} \right)} + \cos{\left(\frac{x}{2} \right)}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin^{2}{\left(\frac{x}{2} \right)} \cos{\left(\frac{x}{2} \right)}\right)\, dx = - \int \sin^{2}{\left(\frac{x}{2} \right)} \cos{\left(\frac{x}{2} \right)}\, dx$
      
      Let $u = \sin{\left(\frac{x}{2} \right)}$ .
      
      Then let $du = \frac{\cos{\left(\frac{x}{2} \right)} dx}{2}$ and substitute $2 du$ :
      
      $\int 2 u^{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{2}\, du = 2 \int u^{2}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      So, the result is: $\frac{2 u^{3}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{2 \sin^{3}{\left(\frac{x}{2} \right)}}{3}$
      
      So, the result is: $- \frac{2 \sin^{3}{\left(\frac{x}{2} \right)}}{3}$
      
      Let $u = \frac{x}{2}$ .
      
      Then let $du = \frac{dx}{2}$ and substitute $2 du$ :
      
      $\int 2 \cos{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = 2 \int \cos{\left(u \right)}\, du$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $2 \sin{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $2 \sin{\left(\frac{x}{2} \right)}$
      
      The result is: $- \frac{2 \sin^{3}{\left(\frac{x}{2} \right)}}{3} + 2 \sin{\left(\frac{x}{2} \right)}$
    So, the result is: $\frac{\sin^{3}{\left(\frac{x}{2} \right)}}{6} - \frac{\sin{\left(\frac{x}{2} \right)}}{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{3 \cos^{2}{\left(\frac{x}{2} \right)}}{8}\, dx = \frac{3 \int \cos^{2}{\left(\frac{x}{2} \right)}\, dx}{8}$
    1. Rewrite the integrand:
      
      $\cos^{2}{\left(\frac{x}{2} \right)} = \frac{\cos{\left(x \right)}}{2} + \frac{1}{2}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(x \right)}}{2}\, dx = \frac{\int \cos{\left(x \right)}\, dx}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$
      
      So, the result is: $\frac{\sin{\left(x \right)}}{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{x}{2} + \frac{\sin{\left(x \right)}}{2}$
    So, the result is: $\frac{3 x}{16} + \frac{3 \sin{\left(x \right)}}{16}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos{\left(\frac{x}{2} \right)}}{4}\right)\, dx = - \frac{\int \cos{\left(\frac{x}{2} \right)}\, dx}{4}$
    1. Let $u = \frac{x}{2}$ .
      
      Then let $du = \frac{dx}{2}$ and substitute $2 du$ :
      
      $\int 2 \cos{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = 2 \int \cos{\left(u \right)}\, du$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $2 \sin{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $2 \sin{\left(\frac{x}{2} \right)}$
    So, the result is: $- \frac{\sin{\left(\frac{x}{2} \right)}}{2}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \frac{1}{16}\, dx = \frac{x}{16}$
  The result is: $\frac{35 x}{128} + \frac{\sin^{3}{\left(\frac{x}{2} \right)}}{6} - \sin{\left(\frac{x}{2} \right)} + \frac{7 \sin{\left(x \right)}}{32} + \frac{\sin{\left(2 x \right)}}{256}$
Method #2
1. Rewrite the integrand:
  
  $\left(\frac{1}{2} - \frac{\cos{\left(\frac{x}{2} \right)}}{2}\right)^{4} = \frac{\cos^{4}{\left(\frac{x}{2} \right)}}{16} - \frac{\cos^{3}{\left(\frac{x}{2} \right)}}{4} + \frac{3 \cos^{2}{\left(\frac{x}{2} \right)}}{8} - \frac{\cos{\left(\frac{x}{2} \right)}}{4} + \frac{1}{16}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\cos^{4}{\left(\frac{x}{2} \right)}}{16}\, dx = \frac{\int \cos^{4}{\left(\frac{x}{2} \right)}\, dx}{16}$
    1. Rewrite the integrand:
      
      $\cos^{4}{\left(\frac{x}{2} \right)} = \left(\frac{\cos{\left(x \right)}}{2} + \frac{1}{2}\right)^{2}$
    2. Rewrite the integrand:
      
      $\left(\frac{\cos{\left(x \right)}}{2} + \frac{1}{2}\right)^{2} = \frac{\cos^{2}{\left(x \right)}}{4} + \frac{\cos{\left(x \right)}}{2} + \frac{1}{4}$
    3. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{2}{\left(x \right)}}{4}\, dx = \frac{\int \cos^{2}{\left(x \right)}\, dx}{4}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(x \right)} = \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}$
      
      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 $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du$
      
      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: $\frac{\sin{\left(2 x \right)}}{4}$
      
      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: $\frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(x \right)}}{2}\, dx = \frac{\int \cos{\left(x \right)}\, dx}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$
      
      So, the result is: $\frac{\sin{\left(x \right)}}{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{4}\, dx = \frac{x}{4}$
      
      The result is: $\frac{3 x}{8} + \frac{\sin{\left(x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{16}$
    So, the result is: $\frac{3 x}{128} + \frac{\sin{\left(x \right)}}{32} + \frac{\sin{\left(2 x \right)}}{256}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos^{3}{\left(\frac{x}{2} \right)}}{4}\right)\, dx = - \frac{\int \cos^{3}{\left(\frac{x}{2} \right)}\, dx}{4}$
    1. Rewrite the integrand:
      
      $\cos^{3}{\left(\frac{x}{2} \right)} = \left(1 - \sin^{2}{\left(\frac{x}{2} \right)}\right) \cos{\left(\frac{x}{2} \right)}$
    2. Let $u = \sin{\left(\frac{x}{2} \right)}$ .
      
      Then let $du = \frac{\cos{\left(\frac{x}{2} \right)} dx}{2}$ and substitute $du$ :
      
      $\int \left(2 - 2 u^{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 2\, du = 2 u$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 u^{2}\right)\, du = - 2 \int u^{2}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      So, the result is: $- \frac{2 u^{3}}{3}$
      
      The result is: $- \frac{2 u^{3}}{3} + 2 u$
      
      Now substitute $u$ back in:
      
      $- \frac{2 \sin^{3}{\left(\frac{x}{2} \right)}}{3} + 2 \sin{\left(\frac{x}{2} \right)}$
    So, the result is: $\frac{\sin^{3}{\left(\frac{x}{2} \right)}}{6} - \frac{\sin{\left(\frac{x}{2} \right)}}{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{3 \cos^{2}{\left(\frac{x}{2} \right)}}{8}\, dx = \frac{3 \int \cos^{2}{\left(\frac{x}{2} \right)}\, dx}{8}$
    1. Rewrite the integrand:
      
      $\cos^{2}{\left(\frac{x}{2} \right)} = \frac{\cos{\left(x \right)}}{2} + \frac{1}{2}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(x \right)}}{2}\, dx = \frac{\int \cos{\left(x \right)}\, dx}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$
      
      So, the result is: $\frac{\sin{\left(x \right)}}{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{x}{2} + \frac{\sin{\left(x \right)}}{2}$
    So, the result is: $\frac{3 x}{16} + \frac{3 \sin{\left(x \right)}}{16}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos{\left(\frac{x}{2} \right)}}{4}\right)\, dx = - \frac{\int \cos{\left(\frac{x}{2} \right)}\, dx}{4}$
    1. Let $u = \frac{x}{2}$ .
      
      Then let $du = \frac{dx}{2}$ and substitute $2 du$ :
      
      $\int 2 \cos{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = 2 \int \cos{\left(u \right)}\, du$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $2 \sin{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $2 \sin{\left(\frac{x}{2} \right)}$
    So, the result is: $- \frac{\sin{\left(\frac{x}{2} \right)}}{2}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \frac{1}{16}\, dx = \frac{x}{16}$
  The result is: $\frac{35 x}{128} + \frac{\sin^{3}{\left(\frac{x}{2} \right)}}{6} - \sin{\left(\frac{x}{2} \right)} + \frac{7 \sin{\left(x \right)}}{32} + \frac{\sin{\left(2 x \right)}}{256}$
Now simplify:

$\frac{35 x}{128} - \frac{7 \sin{\left(\frac{x}{2} \right)}}{8} + \frac{7 \sin{\left(x \right)}}{32} - \frac{\sin{\left(\frac{3 x}{2} \right)}}{24} + \frac{\sin{\left(2 x \right)}}{256}$
Add the constant of integration:

$\frac{35 x}{128} - \frac{7 \sin{\left(\frac{x}{2} \right)}}{8} + \frac{7 \sin{\left(x \right)}}{32} - \frac{\sin{\left(\frac{3 x}{2} \right)}}{24} + \frac{\sin{\left(2 x \right)}}{256}+ \mathrm{constant}$

The answer is:

$\frac{35 x}{128} - \frac{7 \sin{\left(\frac{x}{2} \right)}}{8} + \frac{7 \sin{\left(x \right)}}{32} - \frac{\sin{\left(\frac{3 x}{2} \right)}}{24} + \frac{\sin{\left(2 x \right)}}{256}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                             3/x\                             
 |                           sin |-|                             
 |    8/x\             /x\       \2/   sin(2*x)   7*sin(x)   35*x
 | sin |-| dx = C - sin|-| + ------- + -------- + -------- + ----
 |     \4/             \2/      6        256         32      128 
 |                                                               
/

$$\int \sin^{8}{\left(\frac{x}{4} \right)}\, dx = C + \frac{35 x}{128} + \frac{\sin^{3}{\left(\frac{x}{2} \right)}}{6} - \sin{\left(\frac{x}{2} \right)} + \frac{7 \sin{\left(x \right)}}{32} + \frac{\sin{\left(2 x \right)}}{256}$$

Simplify

The answer [src]

             /p\    /p\         3/p\    /p\        5/p\    /p\      7/p\    /p\
       35*cos|-|*sin|-|   35*sin |-|*cos|-|   7*sin |-|*cos|-|   sin |-|*cos|-|
35*p         \2/    \2/          \2/    \2/         \2/    \2/       \2/    \2/
---- - ---------------- - ----------------- - ---------------- - --------------
 64           32                  48                 12                2

$$\frac{35 p}{64} - \frac{\sin^{7}{\left(\frac{p}{2} \right)} \cos{\left(\frac{p}{2} \right)}}{2} - \frac{7 \sin^{5}{\left(\frac{p}{2} \right)} \cos{\left(\frac{p}{2} \right)}}{12} - \frac{35 \sin^{3}{\left(\frac{p}{2} \right)} \cos{\left(\frac{p}{2} \right)}}{48} - \frac{35 \sin{\left(\frac{p}{2} \right)} \cos{\left(\frac{p}{2} \right)}}{32}$$

             /p\    /p\         3/p\    /p\        5/p\    /p\      7/p\    /p\
       35*cos|-|*sin|-|   35*sin |-|*cos|-|   7*sin |-|*cos|-|   sin |-|*cos|-|
35*p         \2/    \2/          \2/    \2/         \2/    \2/       \2/    \2/
---- - ---------------- - ----------------- - ---------------- - --------------
 64           32                  48                 12                2

35*p/64 - 35*cos(p/2)*sin(p/2)/32 - 35*sin(p/2)^3*cos(p/2)/48 - 7*sin(p/2)^5*cos(p/2)/12 - sin(p/2)^7*cos(p/2)/2

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

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Identical expressions

Integral of sin^8(x/4) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #3

Method #2