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Integral of sin2x/(1-sin^2x) dx

The solution

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  p               
  -               
  4               
  /               
 |                
 |    sin(2*x)    
 |  ----------- dx
 |         2      
 |  1 - sin (x)   
 |                
/                 
p                 
-                 
6

$$\int\limits_{\frac{p}{6}}^{\frac{p}{4}} \frac{\sin{\left(2 x \right)}}{1 - \sin^{2}{\left(x \right)}}\, dx$$

Integral(sin(2*x)/(1 - sin(x)^2), (x, p/6, p/4))

Detail solution

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

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

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

The answer is:

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

The answer (Indefinite) [src]

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

$$\int \frac{\sin{\left(2 x \right)}}{1 - \sin^{2}{\left(x \right)}}\, dx = C - \log{\left(\sin^{2}{\left(x \right)} - 1 \right)}$$

Simplify

The answer [src]

     /        2/p\\      /        2/p\\
- log|-1 + sin |-|| + log|-1 + sin |-||
     \         \4//      \         \6//

$$\log{\left(\sin^{2}{\left(\frac{p}{6} \right)} - 1 \right)} - \log{\left(\sin^{2}{\left(\frac{p}{4} \right)} - 1 \right)}$$

     /        2/p\\      /        2/p\\
- log|-1 + sin |-|| + log|-1 + sin |-||
     \         \4//      \         \6//

$$\log{\left(\sin^{2}{\left(\frac{p}{6} \right)} - 1 \right)} - \log{\left(\sin^{2}{\left(\frac{p}{4} \right)} - 1 \right)}$$

-log(-1 + sin(p/4)^2) + log(-1 + sin(p/6)^2)

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

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Identical expressions

Similar expressions

Integral of sin2x/(1-sin^2x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3