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Integral of d{x}:
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Identical expressions
tg^ five (x)/cos^ five (x)
tg to the power of 5(x) divide by co sinus of e of to the power of 5(x)
tg to the power of five (x) divide by co sinus of e of to the power of five (x)
tg5(x)/cos5(x)
tg5x/cos5x
tg⁵(x)/cos⁵(x)
tg^5x/cos^5x
tg^5(x) divide by cos^5(x)
tg^5(x)/cos^5(x)dx

Solving integrals/ tg^5(x)/cos^5(x)

Integral of tg^5(x)/cos^5(x) dx

The solution

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

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

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

Detail solution

Rewrite the integrand:

$\tan^{5}{\left(x \right)} \sec^{5}{\left(x \right)} = \left(\sec^{2}{\left(x \right)} - 1\right)^{2} \tan{\left(x \right)} \sec^{5}{\left(x \right)}$
There are multiple ways to do this integral.
Method #1
1. Let $u = \sec{\left(x \right)}$ .
  
  Then let $du = \tan{\left(x \right)} \sec{\left(x \right)} dx$ and substitute $du$ :
  
  $\int \left(u^{8} - 2 u^{6} + u^{4}\right)\, du$
  1. Integrate term-by-term:
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{8}\, du = \frac{u^{9}}{9}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 u^{6}\right)\, du = - 2 \int u^{6}\, du$
      
      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 $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
    The result is: $\frac{u^{9}}{9} - \frac{2 u^{7}}{7} + \frac{u^{5}}{5}$
  Now substitute $u$ back in:
  
  $\frac{\sec^{9}{\left(x \right)}}{9} - \frac{2 \sec^{7}{\left(x \right)}}{7} + \frac{\sec^{5}{\left(x \right)}}{5}$
Method #2
1. Rewrite the integrand:
  
  $\left(\sec^{2}{\left(x \right)} - 1\right)^{2} \tan{\left(x \right)} \sec^{5}{\left(x \right)} = \tan{\left(x \right)} \sec^{9}{\left(x \right)} - 2 \tan{\left(x \right)} \sec^{7}{\left(x \right)} + \tan{\left(x \right)} \sec^{5}{\left(x \right)}$
2. Integrate term-by-term:
  1. Let $u = \sec^{9}{\left(x \right)}$ .
    
    Then let $du = 9 \tan{\left(x \right)} \sec^{9}{\left(x \right)} dx$ and substitute $\frac{du}{9}$ :
    
    $\int \frac{1}{81}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{9}\, du = \frac{\int 1\, du}{9}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{9}$
    Now substitute $u$ back in:
    
    $\frac{\sec^{9}{\left(x \right)}}{9}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 \tan{\left(x \right)} \sec^{7}{\left(x \right)}\right)\, dx = - 2 \int \tan{\left(x \right)} \sec^{7}{\left(x \right)}\, dx$
    1. Let $u = \sec^{7}{\left(x \right)}$ .
      
      Then let $du = 7 \tan{\left(x \right)} \sec^{7}{\left(x \right)} dx$ and substitute $\frac{du}{7}$ :
      
      $\int \frac{1}{49}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{7}\, du = \frac{\int 1\, du}{7}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{7}$
      
      Now substitute $u$ back in:
      
      $\frac{\sec^{7}{\left(x \right)}}{7}$
    So, the result is: $- \frac{2 \sec^{7}{\left(x \right)}}{7}$
  1. Let $u = \sec^{5}{\left(x \right)}$ .
    
    Then let $du = 5 \tan{\left(x \right)} \sec^{5}{\left(x \right)} dx$ and substitute $\frac{du}{5}$ :
    
    $\int \frac{1}{25}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{5}\, du = \frac{\int 1\, du}{5}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{5}$
    Now substitute $u$ back in:
    
    $\frac{\sec^{5}{\left(x \right)}}{5}$
  The result is: $\frac{\sec^{9}{\left(x \right)}}{9} - \frac{2 \sec^{7}{\left(x \right)}}{7} + \frac{\sec^{5}{\left(x \right)}}{5}$
Method #3
1. Rewrite the integrand:
  
  $\left(\sec^{2}{\left(x \right)} - 1\right)^{2} \tan{\left(x \right)} \sec^{5}{\left(x \right)} = \tan{\left(x \right)} \sec^{9}{\left(x \right)} - 2 \tan{\left(x \right)} \sec^{7}{\left(x \right)} + \tan{\left(x \right)} \sec^{5}{\left(x \right)}$
2. Integrate term-by-term:
  1. Let $u = \sec^{9}{\left(x \right)}$ .
    
    Then let $du = 9 \tan{\left(x \right)} \sec^{9}{\left(x \right)} dx$ and substitute $\frac{du}{9}$ :
    
    $\int \frac{1}{81}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{9}\, du = \frac{\int 1\, du}{9}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{9}$
    Now substitute $u$ back in:
    
    $\frac{\sec^{9}{\left(x \right)}}{9}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 \tan{\left(x \right)} \sec^{7}{\left(x \right)}\right)\, dx = - 2 \int \tan{\left(x \right)} \sec^{7}{\left(x \right)}\, dx$
    1. Let $u = \sec^{7}{\left(x \right)}$ .
      
      Then let $du = 7 \tan{\left(x \right)} \sec^{7}{\left(x \right)} dx$ and substitute $\frac{du}{7}$ :
      
      $\int \frac{1}{49}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{7}\, du = \frac{\int 1\, du}{7}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{7}$
      
      Now substitute $u$ back in:
      
      $\frac{\sec^{7}{\left(x \right)}}{7}$
    So, the result is: $- \frac{2 \sec^{7}{\left(x \right)}}{7}$
  1. Let $u = \sec^{5}{\left(x \right)}$ .
    
    Then let $du = 5 \tan{\left(x \right)} \sec^{5}{\left(x \right)} dx$ and substitute $\frac{du}{5}$ :
    
    $\int \frac{1}{25}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{5}\, du = \frac{\int 1\, du}{5}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{5}$
    Now substitute $u$ back in:
    
    $\frac{\sec^{5}{\left(x \right)}}{5}$
  The result is: $\frac{\sec^{9}{\left(x \right)}}{9} - \frac{2 \sec^{7}{\left(x \right)}}{7} + \frac{\sec^{5}{\left(x \right)}}{5}$
Now simplify:

$\frac{\left(35 \sec^{4}{\left(x \right)} - 90 \sec^{2}{\left(x \right)} + 63\right) \sec^{5}{\left(x \right)}}{315}$
Add the constant of integration:

$\frac{\left(35 \sec^{4}{\left(x \right)} - 90 \sec^{2}{\left(x \right)} + 63\right) \sec^{5}{\left(x \right)}}{315}+ \mathrm{constant}$

The answer is:

$\frac{\left(35 \sec^{4}{\left(x \right)} - 90 \sec^{2}{\left(x \right)} + 63\right) \sec^{5}{\left(x \right)}}{315}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

$${{63\,\cos ^4x-90\,\cos ^2x+35}\over{315\,\cos ^9x}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                    4            2   
   8    -35 - 63*cos (1) + 90*cos (1)
- --- - -----------------------------
  315                   9            
                 315*cos (1)

$${{1}\over{5\,\cos ^51}}-{{2}\over{7\,\cos ^71}}+{{1}\over{9\,\cos ^ 91}}-{{8}\over{315}}$$

                    4            2   
   8    -35 - 63*cos (1) + 90*cos (1)
- --- - -----------------------------
  315                   9            
                 315*cos (1)

$$- \frac{8}{315} - \frac{-35 - 63 \cos^{4}{\left(1 \right)} + 90 \cos^{2}{\left(1 \right)}}{315 \cos^{9}{\left(1 \right)}}$$

Упростить

Numerical answer [src]

11.3781444268107

11.3781444268107

The graph

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

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How to use it?

Identical expressions

Integral of tg^5(x)/cos^5(x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3