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Integral of d{x}:
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Integral of xe
Graphing y =:
sin(x)^9
Derivative of:
sin(x)^9
Identical expressions
sin(x)^ nine
sinus of (x) to the power of 9
sinus of (x) to the power of nine
sin(x)9
sinx9
sin(x)⁹
sinx^9
sin(x)^9dx
Similar expressions
sinx^9

Integral of sin(x)^9 dx

The solution

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  1           
  /           
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 |     9      
 |  sin (x) dx
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0

$$\int\limits_{0}^{1} \sin^{9}{\left(x \right)}\, dx$$

Integral(sin(x)^9, (x, 0, 1))

Detail solution

Rewrite the integrand:

$\sin^{9}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right)^{4} \sin{\left(x \right)}$
There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(1 - \cos^{2}{\left(x \right)}\right)^{4} \sin{\left(x \right)} = \sin{\left(x \right)} \cos^{8}{\left(x \right)} - 4 \sin{\left(x \right)} \cos^{6}{\left(x \right)} + 6 \sin{\left(x \right)} \cos^{4}{\left(x \right)} - 4 \sin{\left(x \right)} \cos^{2}{\left(x \right)} + \sin{\left(x \right)}$
2. Integrate term-by-term:
  1. Let $u = \cos{\left(x \right)}$ .
    
    Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
    
    $\int \left(- u^{8}\right)\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{8}\, du = - \int u^{8}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{8}\, du = \frac{u^{9}}{9}$
      
      So, the result is: $- \frac{u^{9}}{9}$
    Now substitute $u$ back in:
    
    $- \frac{\cos^{9}{\left(x \right)}}{9}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 4 \sin{\left(x \right)} \cos^{6}{\left(x \right)}\right)\, dx = - 4 \int \sin{\left(x \right)} \cos^{6}{\left(x \right)}\, dx$
    1. Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int \left(- u^{6}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{6}\, du = - \int u^{6}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      So, the result is: $- \frac{u^{7}}{7}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{7}{\left(x \right)}}{7}$
    So, the result is: $\frac{4 \cos^{7}{\left(x \right)}}{7}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 6 \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx = 6 \int \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$
    1. Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int \left(- u^{4}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{4}\, du = - \int u^{4}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
      
      So, the result is: $- \frac{u^{5}}{5}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{5}{\left(x \right)}}{5}$
    So, the result is: $- \frac{6 \cos^{5}{\left(x \right)}}{5}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 4 \sin{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 4 \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
    1. Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int \left(- u^{2}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{2}\, du = - \int u^{2}\, du$
      
      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: $- \frac{u^{3}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{3}{\left(x \right)}}{3}$
    So, the result is: $\frac{4 \cos^{3}{\left(x \right)}}{3}$
  1. The integral of sine is negative cosine:
    
    $\int \sin{\left(x \right)}\, dx = - \cos{\left(x \right)}$
  The result is: $- \frac{\cos^{9}{\left(x \right)}}{9} + \frac{4 \cos^{7}{\left(x \right)}}{7} - \frac{6 \cos^{5}{\left(x \right)}}{5} + \frac{4 \cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}$
Method #2
1. Rewrite the integrand:
  
  $\left(1 - \cos^{2}{\left(x \right)}\right)^{4} \sin{\left(x \right)} = \sin{\left(x \right)} \cos^{8}{\left(x \right)} - 4 \sin{\left(x \right)} \cos^{6}{\left(x \right)} + 6 \sin{\left(x \right)} \cos^{4}{\left(x \right)} - 4 \sin{\left(x \right)} \cos^{2}{\left(x \right)} + \sin{\left(x \right)}$
2. Integrate term-by-term:
  1. Let $u = \cos{\left(x \right)}$ .
    
    Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
    
    $\int \left(- u^{8}\right)\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{8}\, du = - \int u^{8}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{8}\, du = \frac{u^{9}}{9}$
      
      So, the result is: $- \frac{u^{9}}{9}$
    Now substitute $u$ back in:
    
    $- \frac{\cos^{9}{\left(x \right)}}{9}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 4 \sin{\left(x \right)} \cos^{6}{\left(x \right)}\right)\, dx = - 4 \int \sin{\left(x \right)} \cos^{6}{\left(x \right)}\, dx$
    1. Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int \left(- u^{6}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{6}\, du = - \int u^{6}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      So, the result is: $- \frac{u^{7}}{7}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{7}{\left(x \right)}}{7}$
    So, the result is: $\frac{4 \cos^{7}{\left(x \right)}}{7}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 6 \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx = 6 \int \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$
    1. Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int \left(- u^{4}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{4}\, du = - \int u^{4}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
      
      So, the result is: $- \frac{u^{5}}{5}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{5}{\left(x \right)}}{5}$
    So, the result is: $- \frac{6 \cos^{5}{\left(x \right)}}{5}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 4 \sin{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 4 \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
    1. Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int \left(- u^{2}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{2}\, du = - \int u^{2}\, du$
      
      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: $- \frac{u^{3}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{3}{\left(x \right)}}{3}$
    So, the result is: $\frac{4 \cos^{3}{\left(x \right)}}{3}$
  1. The integral of sine is negative cosine:
    
    $\int \sin{\left(x \right)}\, dx = - \cos{\left(x \right)}$
  The result is: $- \frac{\cos^{9}{\left(x \right)}}{9} + \frac{4 \cos^{7}{\left(x \right)}}{7} - \frac{6 \cos^{5}{\left(x \right)}}{5} + \frac{4 \cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}$
Now simplify:

$\frac{\left(- 378 \sin^{4}{\left(x \right)} - 35 \cos^{8}{\left(x \right)} + 180 \cos^{6}{\left(x \right)} - 336 \cos^{2}{\left(x \right)} + 63\right) \cos{\left(x \right)}}{315}$
Add the constant of integration:

$\frac{\left(- 378 \sin^{4}{\left(x \right)} - 35 \cos^{8}{\left(x \right)} + 180 \cos^{6}{\left(x \right)} - 336 \cos^{2}{\left(x \right)} + 63\right) \cos{\left(x \right)}}{315}+ \mathrm{constant}$

The answer is:

$\frac{\left(- 378 \sin^{4}{\left(x \right)} - 35 \cos^{8}{\left(x \right)} + 180 \cos^{6}{\left(x \right)} - 336 \cos^{2}{\left(x \right)} + 63\right) \cos{\left(x \right)}}{315}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                     
 |                                5         9           3           7   
 |    9                      6*cos (x)   cos (x)   4*cos (x)   4*cos (x)
 | sin (x) dx = C - cos(x) - --------- - ------- + --------- + ---------
 |                               5          9          3           7    
/

$$\int \sin^{9}{\left(x \right)}\, dx = C - \frac{\cos^{9}{\left(x \right)}}{9} + \frac{4 \cos^{7}{\left(x \right)}}{7} - \frac{6 \cos^{5}{\left(x \right)}}{5} + \frac{4 \cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                    5         9           3           7   
128            6*cos (1)   cos (1)   4*cos (1)   4*cos (1)
--- - cos(1) - --------- - ------- + --------- + ---------
315                5          9          3           7

$$- \cos{\left(1 \right)} - \frac{6 \cos^{5}{\left(1 \right)}}{5} - \frac{\cos^{9}{\left(1 \right)}}{9} + \frac{4 \cos^{7}{\left(1 \right)}}{7} + \frac{4 \cos^{3}{\left(1 \right)}}{3} + \frac{128}{315}$$

                    5         9           3           7   
128            6*cos (1)   cos (1)   4*cos (1)   4*cos (1)
--- - cos(1) - --------- - ------- + --------- + ---------
315                5          9          3           7

128/315 - cos(1) - 6*cos(1)^5/5 - cos(1)^9/9 + 4*cos(1)^3/3 + 4*cos(1)^7/7

Numerical answer [src]

0.028342532187773

0.028342532187773

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

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Identical expressions

Similar expressions

Integral of sin(x)^9 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2