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sinus of 2x plus one divide by sinus of 2 multiply by x divide by three
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sin2x+1/sin2*x/3dx
Similar expressions
sin2x-1/sin2*x/3

Integral of sin2x+1/sin2*x/3 dx

The solution

You have entered [src]

  1                           
  /                           
 |                            
 |  /               1     \   
 |  |sin(2*x) + ----------| dx
 |  \           sin(2*x)*3/   
 |                            
/                             
0

$$\int\limits_{0}^{1} \left(\sin{\left(2 x \right)} + \frac{1}{3 \sin{\left(2 x \right)}}\right)\, dx$$

Integral(sin(2*x) + 1/(sin(2*x)*3), (x, 0, 1))

Detail solution

Integrate term-by-term:
1. There are multiple ways to do this integral.
  Method #1
  1. Let $u = 2 x$ .
    
    Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{\sin{\left(u \right)}}{2}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{2}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(2 x \right)}}{2}$
  Method #2
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 \sin{\left(x \right)} \cos{\left(x \right)}\, dx = 2 \int \sin{\left(x \right)} \cos{\left(x \right)}\, dx$
    1. There are multiple ways to do this integral.
      
      Method #1
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int \left(- u\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u\, du = - \int u\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      So, the result is: $- \frac{u^{2}}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{2}{\left(x \right)}}{2}$
      
      Method #2
      
      Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{2}{\left(x \right)}}{2}$
    So, the result is: $- \cos^{2}{\left(x \right)}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{1}{3 \sin{\left(2 x \right)}}\, dx = \frac{\int \frac{1}{\sin{\left(2 x \right)}}\, dx}{3}$
  1. Don't know the steps in finding this integral.
    
    But the integral is
    
    $\frac{\log{\left(\cos{\left(2 x \right)} - 1 \right)}}{4} - \frac{\log{\left(\cos{\left(2 x \right)} + 1 \right)}}{4}$
  So, the result is: $\frac{\log{\left(\cos{\left(2 x \right)} - 1 \right)}}{12} - \frac{\log{\left(\cos{\left(2 x \right)} + 1 \right)}}{12}$
The result is: $\frac{\log{\left(\cos{\left(2 x \right)} - 1 \right)}}{12} - \frac{\log{\left(\cos{\left(2 x \right)} + 1 \right)}}{12} - \frac{\cos{\left(2 x \right)}}{2}$
Now simplify:

$\frac{\log{\left(- \sin^{2}{\left(x \right)} \right)}}{12} - \frac{\log{\left(\cos^{2}{\left(x \right)} \right)}}{12} - \cos^{2}{\left(x \right)} + \frac{1}{2}$
Add the constant of integration:

$\frac{\log{\left(- \sin^{2}{\left(x \right)} \right)}}{12} - \frac{\log{\left(\cos^{2}{\left(x \right)} \right)}}{12} - \cos^{2}{\left(x \right)} + \frac{1}{2}+ \mathrm{constant}$

The answer is:

$\frac{\log{\left(- \sin^{2}{\left(x \right)} \right)}}{12} - \frac{\log{\left(\cos^{2}{\left(x \right)} \right)}}{12} - \cos^{2}{\left(x \right)} + \frac{1}{2}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                  
 |                                                                                   
 | /               1     \          cos(2*x)   log(1 + cos(2*x))   log(-1 + cos(2*x))
 | |sin(2*x) + ----------| dx = C - -------- - ----------------- + ------------------
 | \           sin(2*x)*3/             2               12                  12        
 |                                                                                   
/

$$\int \left(\sin{\left(2 x \right)} + \frac{1}{3 \sin{\left(2 x \right)}}\right)\, dx = C + \frac{\log{\left(\cos{\left(2 x \right)} - 1 \right)}}{12} - \frac{\log{\left(\cos{\left(2 x \right)} + 1 \right)}}{12} - \frac{\cos{\left(2 x \right)}}{2}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

     pi*I
oo + ----
      12

$$\infty + \frac{i \pi}{12}$$

     pi*I
oo + ----
      12

$$\infty + \frac{i \pi}{12}$$

oo + pi*i/12

Numerical answer [src]

8.13031822795854

8.13031822795854

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

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How to use it?

Identical expressions

Similar expressions

Integral of sin2x+1/sin2*x/3 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #1

Method #2