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Identical expressions
sin(log2x)/x
sinus of ( logarithm of 2x) divide by x
sinlog2x/x
sin(log2x) divide by x
sin(log2x)/xdx

Integral of sin(log2x)/x dx

The solution

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

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

Integral(sin(log(2*x))/x, (x, 0, 1))

Detail solution

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

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

The answer is:

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

The answer (Indefinite) [src]

  /                                    
 |                                     
 | sin(log(2*x))                       
 | ------------- dx = C - cos(log(2*x))
 |       x                             
 |                                     
/

$$-\cos \log \left(2\,x\right)$$

Simplify

The answer [src]

<-1 - cos(log(2)), 1 - cos(log(2))>

$$\int_{0}^{1}{{{\sin \log \left(2\,x\right)}\over{x}}\;dx}$$

<-1 - cos(log(2)), 1 - cos(log(2))>

$$\left\langle -1 - \cos{\left(\log{\left(2 \right)} \right)}, - \cos{\left(\log{\left(2 \right)} \right)} + 1\right\rangle$$

Simplify

Numerical answer [src]

0.0812068795344031

0.0812068795344031

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

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Identical expressions

Integral of sin(log2x)/x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3