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Similar expressions
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Solving integrals/ (2tan(x))÷(1-(tan(x))^(2))

Integral of (2tan(x))÷(1-(tan(x))^(2)) dx

The solution

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  1               
  /               
 |                
 |    2*tan(x)    
 |  ----------- dx
 |         2      
 |  1 - tan (x)   
 |                
/                 
0

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

Integral(2*tan(x)/(1 - tan(x)^2), (x, 0, 1))

Detail solution

The integral of a constant times a function is the constant times the integral of the function:

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

$- \frac{\log{\left(\tan^{2}{\left(x \right)} - 1 \right)}}{2} + \frac{\log{\left(\frac{1}{\cos^{2}{\left(x \right)}} \right)}}{2}$
Add the constant of integration:

$- \frac{\log{\left(\tan^{2}{\left(x \right)} - 1 \right)}}{2} + \frac{\log{\left(\frac{1}{\cos^{2}{\left(x \right)}} \right)}}{2}+ \mathrm{constant}$

The answer is:

$- \frac{\log{\left(\tan^{2}{\left(x \right)} - 1 \right)}}{2} + \frac{\log{\left(\frac{1}{\cos^{2}{\left(x \right)}} \right)}}{2}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                         
 |                         /       2   \      /        2   \
 |   2*tan(x)           log\1 + tan (x)/   log\-1 + tan (x)/
 | ----------- dx = C + ---------------- - -----------------
 |        2                    2                   2        
 | 1 - tan (x)                                              
 |                                                          
/

$$-{{\log \left(2\,\sin ^2x-1\right)}\over{2}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

nan

$$-{{\log \left(2\,\sin ^21-1\right)}\over{2}}$$

nan

$$\text{NaN}$$

Упростить

Numerical answer [src]

-0.96117546625349

-0.96117546625349

The graph

Integral of (2tan(x))÷(1-(tan(x))^(2)) dx

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

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Identical expressions

Similar expressions

Integral of (2tan(x))÷(1-(tan(x))^(2)) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2