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Solving integrals/ 1/sin(x)

Integral of 1/sin(x) dx

Limits of integration:

from to
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Piecewise:

The solution

You have entered [src] 
  1            
  /            
 |             
 |      1      
 |  1*------ dx
 |    sin(x)   
 |             
/              
0
$$\int\limits_{0}^{1} 1 \cdot \frac{1}{\sin{\left(x \right)}}\, dx$$ 
Integral(1/sin(x), (x, 0, 1))
Detail solution
We have the integral:
  /             
 |              
 |       1      
 | 1*1*------ dx
 |     sin(x)   
 |              
/
The integrand
    1   
1*------
  sin(x)
Multiply numerator and denominator by
sin(x)
we get
    1      1*sin(1*x)
1*------ = ----------
  sin(x)      2      
           sin (1*x)
Because
sin(a)^2 + cos(a)^2 = 1
then
   2             2   
sin (x) = 1 - cos (x)
transform the denominator
1*sin(1*x)     1*sin(1*x) 
---------- = -------------
   2                2     
sin (1*x)    1 - cos (1*x)
do replacement
u = cos(x)
then the integral
  /                  
 |                   
 |   1*sin(1*x)      
 | ------------- dx  
 |        2         =
 | 1 - cos (1*x)     
 |                   
/                    
  
  /                  
 |                   
 |   1*sin(1*x)      
 | ------------- dx  
 |        2         =
 | 1 - cos (1*x)     
 |                   
/
Because du = -dx*sin(x)
  /                
 |                 
 |   1*sin(1*x)    
 | ------------- du
 |        2        
 | 1 - cos (1*x)   
 |                 
/
Rewrite the integrand
  1*sin(1*x)    1*-1 /  1       1  \
------------- = ----*|----- + -----|
       2         2   \1 - u   1 + u/
1 - cos (1*x)
then
                          /             /          
                         |             |           
                         |   1         |   1       
                         | ----- du    | ----- du  
  /                      | 1 + u       | 1 - u     
 |                       |             |           
 |   1*sin(1*x)         /             /           =
 | ------------- du = - ----------- - -----------  
 |        2                  2             2       
 | 1 - cos (1*x)                                   
 |                                                 
/                                                  
  
= u*sin(x)/(1 - cos(x)^2)
do backward replacement
u = cos(x)
The answer
  /                                                       
 |                                                        
 |       1         log(-1 + cos(x))   log(1 + cos(x))     
 | 1*1*------ dx = ---------------- - --------------- + C0
 |     sin(x)             2                  2            
 |                                                        
/
where C0 is constant, independent of x
The answer (Indefinite) [src] 
  /                                                    
 |                                                     
 |     1             log(-1 + cos(x))   log(1 + cos(x))
 | 1*------ dx = C + ---------------- - ---------------
 |   sin(x)                 2                  2       
 |                                                     
/
$${{\log \left(\cos x-1\right)}\over{2}}-{{\log \left(\cos x+1\right) }\over{2}}$$
Simplify
The answer [src] 
     pi*I
oo + ----
      2
$${\it \%a}$$ 
=
= 
     pi*I
oo + ----
      2
$$\infty + \frac{i \pi}{2}$$
Упростить
Numerical answer [src] 
44.1790108686112
44.1790108686112

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

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