Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of e^(2*x+1)
Integral of dx/(9*x^2-1)
Integral of -cos(2x)
Integral of dx/(5-2*x)
Identical expressions
sin^5x/cosx
sinus of to the power of 5x divide by co sinus of e of x
sin5x/cosx
sin⁵x/cosx
sin^5x divide by cosx
sin^5x/cosxdx

Solving integrals/ sin^5x/cosx

Integral of sin^5x/cosx dx

The solution

You have entered [src]

  1           
  /           
 |            
 |     5      
 |  sin (x)   
 |  ------- dx
 |   cos(x)   
 |            
/             
0

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

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

Detail solution

Rewrite the integrand:

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

$- \frac{\log{\left(\cos^{2}{\left(x \right)} \right)}}{2} - \frac{\cos^{4}{\left(x \right)}}{4} + \cos^{2}{\left(x \right)}+ \mathrm{constant}$

The answer is:

$- \frac{\log{\left(\cos^{2}{\left(x \right)} \right)}}{2} - \frac{\cos^{4}{\left(x \right)}}{4} + \cos^{2}{\left(x \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                 
 |                                                  
 |    5                          /   2   \      4   
 | sin (x)             2      log\cos (x)/   cos (x)
 | ------- dx = C + cos (x) - ------------ - -------
 |  cos(x)                         2            4   
 |                                                  
/

$$\int \frac{\sin^{5}{\left(x \right)}}{\cos{\left(x \right)}}\, dx = C - \frac{\log{\left(\cos^{2}{\left(x \right)} \right)}}{2} - \frac{\cos^{4}{\left(x \right)}}{4} + \cos^{2}{\left(x \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                                 4   
  3      2                    cos (1)
- - + cos (1) - log(cos(1)) - -------
  4                              4

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

                                 4   
  3      2                    cos (1)
- - + cos (1) - log(cos(1)) - -------
  4                              4

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

-3/4 + cos(1)^2 - log(cos(1)) - cos(1)^4/4

Numerical answer [src]

0.136247769832824

0.136247769832824

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of sin^5x/cosx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3