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Integral of d{x}:
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Integral of 1/(1+x^2)^1/2
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Identical expressions
tan^ two (x)sec^ four (x)
tangent of squared (x)sec to the power of 4(x)
tangent of to the power of two (x)sec to the power of four (x)
tan2(x)sec4(x)
tan2xsec4x
tan²(x)sec⁴(x)
tan to the power of 2(x)sec to the power of 4(x)
tan^2xsec^4x
tan^2(x)sec^4(x)dx

Solving integrals/ tan^2(x)sec^4(x)

Integral of tan^2(x)sec^4(x) dx

The solution

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  1                   
  /                   
 |                    
 |     2       4      
 |  tan (x)*sec (x) dx
 |                    
/                     
0

$$\int\limits_{0}^{1} \tan^{2}{\left(x \right)} \sec^{4}{\left(x \right)}\, dx$$

Integral(tan(x)^2*sec(x)^4, (x, 0, 1))

Detail solution

Rewrite the integrand:

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

$\frac{\tan^{5}{\left(x \right)}}{5} + \frac{\tan^{3}{\left(x \right)}}{3}+ \mathrm{constant}$

The answer is:

$\frac{\tan^{5}{\left(x \right)}}{5} + \frac{\tan^{3}{\left(x \right)}}{3}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                          
 |                             3         5   
 |    2       4             tan (x)   tan (x)
 | tan (x)*sec (x) dx = C + ------- + -------
 |                             3         5   
/

$${{3\,\tan ^5x+5\,\tan ^3x}\over{15}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

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

$${{3\,\tan ^51+5\,\tan ^31}\over{15}}$$

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

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

Упростить

Numerical answer [src]

3.09166393502534

3.09166393502534

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 tan^2(x)sec^4(x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3