Integral of sin⁵xcos⁴x dx

The solution

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  1                   
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 |     5       4      
 |  sin (x)*cos (x) dx
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0

$$\int\limits_{0}^{1} \sin^{5}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$$

Integral(sin(x)^5*cos(x)^4, (x, 0, 1))

Detail solution

Rewrite the integrand:

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

$- \frac{\left(35 \sin^{4}{\left(x \right)} + 20 \sin^{2}{\left(x \right)} + 8\right) \cos^{5}{\left(x \right)}}{315}$
Add the constant of integration:

$- \frac{\left(35 \sin^{4}{\left(x \right)} + 20 \sin^{2}{\left(x \right)} + 8\right) \cos^{5}{\left(x \right)}}{315}+ \mathrm{constant}$

The answer is:

$- \frac{\left(35 \sin^{4}{\left(x \right)} + 20 \sin^{2}{\left(x \right)} + 8\right) \cos^{5}{\left(x \right)}}{315}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                      
 |                             5         9           7   
 |    5       4             cos (x)   cos (x)   2*cos (x)
 | sin (x)*cos (x) dx = C - ------- - ------- + ---------
 |                             5         9          7    
/

$$-{{35\,\cos ^9x-90\,\cos ^7x+63\,\cos ^5x}\over{315}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

         5         9           7   
 8    cos (1)   cos (1)   2*cos (1)
--- - ------- - ------- + ---------
315      5         9          7

$${{8}\over{315}}-{{35\,\cos ^91-90\,\cos ^71+63\,\cos ^51}\over{315 }}$$

         5         9           7   
 8    cos (1)   cos (1)   2*cos (1)
--- - ------- - ------- + ---------
315      5         9          7

$$- \frac{\cos^{5}{\left(1 \right)}}{5} - \frac{\cos^{9}{\left(1 \right)}}{9} + \frac{2 \cos^{7}{\left(1 \right)}}{7} + \frac{8}{315}$$

Simplify

Numerical answer [src]

0.0195923055782434

0.0195923055782434

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of sin⁵xcos⁴x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3