Integral of sin⁵x dx

The solution

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  1           
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 |     5      
 |  sin (x) dx
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0

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

Integral(sin(x)^5, (x, 0, 1))

Detail solution

Rewrite the integrand:

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

$- \frac{\cos^{5}{\left(x \right)}}{5} + \frac{2 \cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}+ \mathrm{constant}$

The answer is:

$- \frac{\cos^{5}{\left(x \right)}}{5} + \frac{2 \cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                             
 |                              5           3   
 |    5                      cos (x)   2*cos (x)
 | sin (x) dx = C - cos(x) - ------- + ---------
 |                              5          3    
/

$$\int \sin^{5}{\left(x \right)}\, dx = C - \frac{\cos^{5}{\left(x \right)}}{5} + \frac{2 \cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                 5           3   
8             cos (1)   2*cos (1)
-- - cos(1) - ------- + ---------
15               5          3

$$- \cos{\left(1 \right)} - \frac{\cos^{5}{\left(1 \right)}}{5} + \frac{2 \cos^{3}{\left(1 \right)}}{3} + \frac{8}{15}$$

                 5           3   
8             cos (1)   2*cos (1)
-- - cos(1) - ------- + ---------
15               5          3

$$- \cos{\left(1 \right)} - \frac{\cos^{5}{\left(1 \right)}}{5} + \frac{2 \cos^{3}{\left(1 \right)}}{3} + \frac{8}{15}$$

Упростить

Numerical answer [src]

0.0889743964515759

0.0889743964515759

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of sin⁵x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2