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Integral of d{x}:
Integral of x*exp(-4*x)
Integral of 1/(x^2-2*x)
Integral of x(1+x)^1/2
Integral of tg^5x/cos^2x
Identical expressions
tg^5x/cos^2x
tg to the power of 5x divide by co sinus of e of squared x
tg5x/cos2x
tg⁵x/cos²x
tg to the power of 5x/cos to the power of 2x
tg^5x divide by cos^2x
tg^5x/cos^2xdx

Solving integrals/ tg^5x/cos^2x

Integral of tg^5x/cos^2x dx

The solution

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  1           
  /           
 |            
 |     5      
 |  tan (x)   
 |  ------- dx
 |     2      
 |  cos (x)   
 |            
/             
0

$$\int\limits_{0}^{1} \frac{\tan^{5}{\left(x \right)}}{\cos^{2}{\left(x \right)}}\, dx$$

Integral(tan(x)^5/cos(x)^2, (x, 0, 1))

Detail solution

Rewrite the integrand:

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

$\frac{\tan^{6}{\left(x \right)}}{6}$
Add the constant of integration:

$\frac{\tan^{6}{\left(x \right)}}{6}+ \mathrm{constant}$

The answer is:

$\frac{\tan^{6}{\left(x \right)}}{6}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                
 |                                3
 |    5             /        2   \ 
 | tan (x)          \-1 + sec (x)/ 
 | ------- dx = C + ---------------
 |    2                    6       
 | cos (x)                         
 |                                 
/

$$\int \frac{\tan^{5}{\left(x \right)}}{\cos^{2}{\left(x \right)}}\, dx = C + \frac{\left(\sec^{2}{\left(x \right)} - 1\right)^{3}}{6}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

               2           4   
  1   1 - 3*cos (1) + 3*cos (1)
- - + -------------------------
  6                6           
              6*cos (1)

$$- \frac{1}{6} + \frac{- 3 \cos^{2}{\left(1 \right)} + 3 \cos^{4}{\left(1 \right)} + 1}{6 \cos^{6}{\left(1 \right)}}$$

               2           4   
  1   1 - 3*cos (1) + 3*cos (1)
- - + -------------------------
  6                6           
              6*cos (1)

$$- \frac{1}{6} + \frac{- 3 \cos^{2}{\left(1 \right)} + 3 \cos^{4}{\left(1 \right)} + 1}{6 \cos^{6}{\left(1 \right)}}$$

-1/6 + (1 - 3*cos(1)^2 + 3*cos(1)^4)/(6*cos(1)^6)

Numerical answer [src]

2.37827842589157

2.37827842589157

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^5x/cos^2x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3