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Solving integrals/ sin(log2(x))/x

Integral of sin(log2(x))/x dx

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The solution

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  1               
  /               
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 |     /log(x)\   
 |  sin|------|   
 |     \log(2)/   
 |  ----------- dx
 |       x        
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0
∫01sin⁡(log⁡(x)log⁡(2))x dx\int\limits_{0}^{1} \frac{\sin{\left(\frac{\log{\left(x \right)}}{\log{\left(2 \right)}} \right)}}{x}\, dx 
Integral(sin(log(x)/log(2))/x, (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Let u=log⁡(x)log⁡(2)u = \frac{\log{\left(x \right)}}{\log{\left(2 \right)}}.

      Then let du=dxxlog⁡(2)du = \frac{dx}{x \log{\left(2 \right)}} and substitute dulog⁡(2)du \log{\left(2 \right)}:

      ∫log⁡(2)2sin⁡(u) du\int \log{\left(2 \right)}^{2} \sin{\left(u \right)}\, du

      1. The integral of a constant times a function is the constant times the integral of the function:

        ∫log⁡(2)sin⁡(u) du=log⁡(2)∫sin⁡(u) du\int \log{\left(2 \right)} \sin{\left(u \right)}\, du = \log{\left(2 \right)} \int \sin{\left(u \right)}\, du

        1. The integral of sine is negative cosine:

          ∫sin⁡(u) du=−cos⁡(u)\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}

        So, the result is: −log⁡(2)cos⁡(u)- \log{\left(2 \right)} \cos{\left(u \right)}

      Now substitute uu back in:

      −log⁡(2)cos⁡(log⁡(x)log⁡(2))- \log{\left(2 \right)} \cos{\left(\frac{\log{\left(x \right)}}{\log{\left(2 \right)}} \right)}

    Method #2

    1. Let u=1xu = \frac{1}{x}.

      Then let du=−dxx2du = - \frac{dx}{x^{2}} and substitute −du- du:

      ∫sin⁡(log⁡(1u)log⁡(2))u du\int \frac{\sin{\left(\frac{\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:

        ∫(−sin⁡(log⁡(1u)log⁡(2))u) du=−∫sin⁡(log⁡(1u)log⁡(2))u du\int \left(- \frac{\sin{\left(\frac{\log{\left(\frac{1}{u} \right)}}{\log{\left(2 \right)}} \right)}}{u}\right)\, du = - \int \frac{\sin{\left(\frac{\log{\left(\frac{1}{u} \right)}}{\log{\left(2 \right)}} \right)}}{u}\, du

        1. Let u=log⁡(1u)log⁡(2)u = \frac{\log{\left(\frac{1}{u} \right)}}{\log{\left(2 \right)}}.

          Then let du=−duulog⁡(2)du = - \frac{du}{u \log{\left(2 \right)}} and substitute −dulog⁡(2)- du \log{\left(2 \right)}:

          ∫log⁡(2)2sin⁡(u) du\int \log{\left(2 \right)}^{2} \sin{\left(u \right)}\, du

          1. The integral of a constant times a function is the constant times the integral of the function:

            ∫(−log⁡(2)sin⁡(u)) du=−log⁡(2)∫sin⁡(u) du\int \left(- \log{\left(2 \right)} \sin{\left(u \right)}\right)\, du = - \log{\left(2 \right)} \int \sin{\left(u \right)}\, du

            1. The integral of sine is negative cosine:

              ∫sin⁡(u) du=−cos⁡(u)\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}

            So, the result is: log⁡(2)cos⁡(u)\log{\left(2 \right)} \cos{\left(u \right)}

          Now substitute uu back in:

          log⁡(2)cos⁡(log⁡(1u)log⁡(2))\log{\left(2 \right)} \cos{\left(\frac{\log{\left(\frac{1}{u} \right)}}{\log{\left(2 \right)}} \right)}

        So, the result is: −log⁡(2)cos⁡(log⁡(1u)log⁡(2))- \log{\left(2 \right)} \cos{\left(\frac{\log{\left(\frac{1}{u} \right)}}{\log{\left(2 \right)}} \right)}

      Now substitute uu back in:

      −log⁡(2)cos⁡(log⁡(x)log⁡(2))- \log{\left(2 \right)} \cos{\left(\frac{\log{\left(x \right)}}{\log{\left(2 \right)}} \right)}

  2. Now simplify:

    −log⁡(2)cos⁡(log⁡(x)log⁡(2))- \log{\left(2 \right)} \cos{\left(\frac{\log{\left(x \right)}}{\log{\left(2 \right)}} \right)}

  3. Add the constant of integration:

    −log⁡(2)cos⁡(log⁡(x)log⁡(2))+constant- \log{\left(2 \right)} \cos{\left(\frac{\log{\left(x \right)}}{\log{\left(2 \right)}} \right)}+ \mathrm{constant}

The answer is:

−log⁡(2)cos⁡(log⁡(x)log⁡(2))+constant- \log{\left(2 \right)} \cos{\left(\frac{\log{\left(x \right)}}{\log{\left(2 \right)}} \right)}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                       
 |                                        
 |    /log(x)\                            
 | sin|------|                            
 |    \log(2)/             /log(x)\       
 | ----------- dx = C - cos|------|*log(2)
 |      x                  \log(2)/       
 |                                        
/
−log⁡2 cos⁡(log⁡xlog⁡2)-\log 2\,\cos \left({{\log x}\over{\log 2}}\right)
Simplify
The answer [src] 
-log(2) - <-1, 1>*log(2)
∫01sin⁡(log⁡xlog⁡2)x  dx\int_{0}^{1}{{{\sin \left({{\log x}\over{\log 2}}\right)}\over{x}} \;dx} 
=
= 
-log(2) - <-1, 1>*log(2)
−log⁡(2)⟨−1,1⟩−log⁡(2)- \log{\left(2 \right)} \left\langle -1, 1\right\rangle - \log{\left(2 \right)}
Simplify
Numerical answer [src] 
-0.177999323190111
-0.177999323190111

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

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