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Solving integrals/ sin^5x-cosx

Integral of sin^5x-cosx dx

The solution

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  1                      
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 |  \sin (x) - cos(x)/ dx
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0

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

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

Detail solution

Integrate term-by-term:
1. Rewrite the integrand:
  
  $\sin^{5}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right)^{2} \sin{\left(x \right)}$
2. 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 $- du$ :
      
      $\int u^{4}\, du$
      
      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:
      
      $\int \left(- 2 \sin{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 2 \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
      
      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 $- du$ :
      
      $\int u^{4}\, du$
      
      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:
      
      $\int \left(- 2 \sin{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 2 \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
      
      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: $- \frac{\cos^{5}{\left(x \right)}}{5} + \frac{2 \cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \cos{\left(x \right)}\right)\, dx = - \int \cos{\left(x \right)}\, dx$
  1. The integral of cosine is sine:
    
    $\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$
  So, the result is: $- \sin{\left(x \right)}$
The result is: $- \sin{\left(x \right)} - \frac{\cos^{5}{\left(x \right)}}{5} + \frac{2 \cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

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

$$-\sin x-{{\cos ^5x}\over{5}}+{{2\,\cos ^3x}\over{3}}-\cos x$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

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

$$-{{15\,\sin 1+3\,\cos ^51-10\,\cos ^31+15\,\cos 1-8}\over{15}}$$

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

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

Simplify

Numerical answer [src]

-0.752496588356321

-0.752496588356321

The graph

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

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How to use it?

Identical expressions

Similar expressions

Integral of sin^5x-cosx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2