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Solving integrals/ sin(log2x)/x

Integral of sin(log2x)/x dx

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The solution

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 |  sin(log(2*x))   
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∫01sin⁡(log⁡(2x))x dx\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

  1. There are multiple ways to do this integral.

    Method #1

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

      Then let du=dxxdu = \frac{dx}{x} and substitute dudu:

      ∫sin⁡(u) du\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)}

      Now substitute uu back in:

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

    Method #2

    1. Rewrite the integrand:

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

          Then let du=−duudu = - \frac{du}{u} and substitute −du- du:

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

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

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

          Now substitute uu back in:

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

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

      Now substitute uu back in:

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

    Method #3

    1. Rewrite the integrand:

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

          Then let du=−duudu = - \frac{du}{u} and substitute −du- du:

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

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

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

          Now substitute uu back in:

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

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

      Now substitute uu back in:

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

  2. Add the constant of integration:

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

The answer is:

−cos⁡(log⁡(2x))+constant- \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⁡(2 x)-\cos \log \left(2\,x\right)
Simplify
The answer [src] 
<-1 - cos(log(2)), 1 - cos(log(2))>
∫01sin⁡log⁡(2 x)x  dx\int_{0}^{1}{{{\sin \log \left(2\,x\right)}\over{x}}\;dx} 
=
= 
<-1 - cos(log(2)), 1 - cos(log(2))>
⟨−1−cos⁡(log⁡(2)),−cos⁡(log⁡(2))+1⟩\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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