Integral of cos^9x dx

The solution

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  1           
  /           
 |            
 |     9      
 |  cos (x) dx
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/             
0

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

Simplify

Detail solution

Rewrite the integrand:

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

$\frac{\left(35 \sin^{8}{\left(x \right)} - 180 \sin^{6}{\left(x \right)} + 378 \sin^{4}{\left(x \right)} - 420 \sin^{2}{\left(x \right)} + 315\right) \sin{\left(x \right)}}{315}$
Add the constant of integration:

$\frac{\left(35 \sin^{8}{\left(x \right)} - 180 \sin^{6}{\left(x \right)} + 378 \sin^{4}{\left(x \right)} - 420 \sin^{2}{\left(x \right)} + 315\right) \sin{\left(x \right)}}{315}+ \mathrm{constant}$

The answer is:

$\frac{\left(35 \sin^{8}{\left(x \right)} - 180 \sin^{6}{\left(x \right)} + 378 \sin^{4}{\left(x \right)} - 420 \sin^{2}{\left(x \right)} + 315\right) \sin{\left(x \right)}}{315}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

$${{\sin ^9x}\over{9}}-{{4\,\sin ^7x}\over{7}}+{{6\,\sin ^5x}\over{5 }}-{{4\,\sin ^3x}\over{3}}+\sin x$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

       3           7         9           5            
  4*sin (1)   4*sin (1)   sin (1)   6*sin (1)         
- --------- - --------- + ------- + --------- + sin(1)
      3           7          9          5

$${{35\,\sin ^91-180\,\sin ^71+378\,\sin ^51-420\,\sin ^31+315\,\sin 1}\over{315}}$$

       3           7         9           5            
  4*sin (1)   4*sin (1)   sin (1)   6*sin (1)         
- --------- - --------- + ------- + --------- + sin(1)
      3           7          9          5

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

Simplify

Numerical answer [src]

0.406105224735144

0.406105224735144

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of cos^9x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2