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Identical expressions
(one +sin4x)/(sin two x)^2
(1 plus sinus of 4x) divide by ( sinus of 2x) squared
(one plus sinus of 4x) divide by ( sinus of two x) squared
(1+sin4x)/(sin2x)2
1+sin4x/sin2x2
(1+sin4x)/(sin2x)²
(1+sin4x)/(sin2x) to the power of 2
1+sin4x/sin2x^2
(1+sin4x) divide by (sin2x)^2
(1+sin4x)/(sin2x)^2dx
Similar expressions
(1-sin4x)/(sin2x)^2

Integral of (1+sin4x)/(sin2x)^2 dx

The solution

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  1                
  /                
 |                 
 |  1 + sin(4*x)   
 |  ------------ dx
 |      2          
 |   sin (2*x)     
 |                 
/                  
0

$$\int\limits_{0}^{1} \frac{\sin{\left(4 x \right)} + 1}{\sin^{2}{\left(2 x \right)}}\, dx$$

Integral((1 + sin(4*x))/sin(2*x)^2, (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{\sin{\left(4 x \right)} + 1}{\sin^{2}{\left(2 x \right)}} = \frac{\sin{\left(4 x \right)}}{\sin^{2}{\left(2 x \right)}} + \frac{1}{\sin^{2}{\left(2 x \right)}}$
2. Integrate term-by-term:
  1. Rewrite the integrand:
    
    $\frac{\sin{\left(4 x \right)}}{\sin^{2}{\left(2 x \right)}} = - \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{1}{\sin{\left(x \right)} \cos{\left(x \right)}}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}}\right)\, dx = - 2 \int \frac{\sin{\left(x \right)}}{\cos{\left(x \right)}}\, dx$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int \left(- \frac{1}{u}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u}\, du = - \int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $- \log{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- \log{\left(\cos{\left(x \right)} \right)}$
      
      So, the result is: $2 \log{\left(\cos{\left(x \right)} \right)}$
    1. Don't know the steps in finding this integral.
      
      But the integral is
      
      $- \frac{\log{\left(\sin^{2}{\left(x \right)} - 1 \right)}}{2} + \log{\left(\sin{\left(x \right)} \right)}$
    The result is: $- \frac{\log{\left(\sin^{2}{\left(x \right)} - 1 \right)}}{2} + \log{\left(\sin{\left(x \right)} \right)} + 2 \log{\left(\cos{\left(x \right)} \right)}$
  1. Don't know the steps in finding this integral.
    
    But the integral is
    
    $- \frac{\cos{\left(2 x \right)}}{2 \sin{\left(2 x \right)}}$
  The result is: $- \frac{\log{\left(\sin^{2}{\left(x \right)} - 1 \right)}}{2} + \log{\left(\sin{\left(x \right)} \right)} + 2 \log{\left(\cos{\left(x \right)} \right)} - \frac{\cos{\left(2 x \right)}}{2 \sin{\left(2 x \right)}}$
Method #2
1. Rewrite the integrand:
  
  $\frac{\sin{\left(4 x \right)} + 1}{\sin^{2}{\left(2 x \right)}} = - \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{1}{\sin{\left(x \right)} \cos{\left(x \right)}} + \frac{1}{4 \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{2 \sin{\left(x \right)}}{\cos{\left(x \right)}}\right)\, dx = - 2 \int \frac{\sin{\left(x \right)}}{\cos{\left(x \right)}}\, dx$
    1. Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int \left(- \frac{1}{u}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u}\, du = - \int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $- \log{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- \log{\left(\cos{\left(x \right)} \right)}$
    So, the result is: $2 \log{\left(\cos{\left(x \right)} \right)}$
  1. Don't know the steps in finding this integral.
    
    But the integral is
    
    $- \frac{\log{\left(\sin^{2}{\left(x \right)} - 1 \right)}}{2} + \log{\left(\sin{\left(x \right)} \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{4 \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}}\, dx = \frac{\int \frac{1}{\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}}\, dx}{4}$
    1. Don't know the steps in finding this integral.
      
      But the integral is
      
      $- \frac{2 \cos{\left(2 x \right)}}{\sin{\left(2 x \right)}}$
    So, the result is: $- \frac{\cos{\left(2 x \right)}}{2 \sin{\left(2 x \right)}}$
  The result is: $- \frac{\log{\left(\sin^{2}{\left(x \right)} - 1 \right)}}{2} + \log{\left(\sin{\left(x \right)} \right)} + 2 \log{\left(\cos{\left(x \right)} \right)} - \frac{\cos{\left(2 x \right)}}{2 \sin{\left(2 x \right)}}$
Now simplify:

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

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

The answer is:

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

The answer (Indefinite) [src]

  /                                                                                  
 |                                          /        2   \                           
 | 1 + sin(4*x)                          log\-1 + sin (x)/    cos(2*x)               
 | ------------ dx = C + 2*log(cos(x)) - ----------------- - ---------- + log(sin(x))
 |     2                                         2           2*sin(2*x)              
 |  sin (2*x)                                                                        
 |                                                                                   
/

$$\int \frac{\sin{\left(4 x \right)} + 1}{\sin^{2}{\left(2 x \right)}}\, dx = C - \frac{\log{\left(\sin^{2}{\left(x \right)} - 1 \right)}}{2} + \log{\left(\sin{\left(x \right)} \right)} + 2 \log{\left(\cos{\left(x \right)} \right)} - \frac{\cos{\left(2 x \right)}}{2 \sin{\left(2 x \right)}}$$

Simplify

The answer [src]

     pi*I
oo - ----
      2

$$\infty - \frac{i \pi}{2}$$

     pi*I
oo - ----
      2

$$\infty - \frac{i \pi}{2}$$

oo - pi*i/2

Numerical answer [src]

3.44830919487149e+18

3.44830919487149e+18

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

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Identical expressions

Similar expressions

Integral of (1+sin4x)/(sin2x)^2 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2