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sin(x)(1-cos(x))^4dx
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Integral of sin(x)(1+cos(x))^4dx dx

The solution

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  1                        
  /                        
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 |                     4   
 |  sin(x)*(1 + cos(x))  dx
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/                          
0

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

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

Detail solution

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

$- \frac{\left(\cos{\left(x \right)} + 1\right)^{5}}{5}+ \mathrm{constant}$

The answer is:

$- \frac{\left(\cos{\left(x \right)} + 1\right)^{5}}{5}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                           
 |                                           5
 |                    4          (1 + cos(x)) 
 | sin(x)*(1 + cos(x))  dx = C - -------------
 |                                     5      
/

$$\int \left(\cos{\left(x \right)} + 1\right)^{4} \sin{\left(x \right)}\, dx = C - \frac{\left(\cos{\left(x \right)} + 1\right)^{5}}{5}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                                                   5   
31      4                    2           3      cos (1)
-- - cos (1) - cos(1) - 2*cos (1) - 2*cos (1) - -------
5                                                  5

$$- 2 \cos^{2}{\left(1 \right)} - \cos{\left(1 \right)} - 2 \cos^{3}{\left(1 \right)} - \cos^{4}{\left(1 \right)} - \frac{\cos^{5}{\left(1 \right)}}{5} + \frac{31}{5}$$

                                                   5   
31      4                    2           3      cos (1)
-- - cos (1) - cos(1) - 2*cos (1) - 2*cos (1) - -------
5                                                  5

$$- 2 \cos^{2}{\left(1 \right)} - \cos{\left(1 \right)} - 2 \cos^{3}{\left(1 \right)} - \cos^{4}{\left(1 \right)} - \frac{\cos^{5}{\left(1 \right)}}{5} + \frac{31}{5}$$

31/5 - cos(1)^4 - cos(1) - 2*cos(1)^2 - 2*cos(1)^3 - cos(1)^5/5

Numerical answer [src]

4.66595715654426

4.66595715654426

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

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Identical expressions

Similar expressions

Integral of sin(x)(1+cos(x))^4dx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3