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Solving integrals/ (sin(lnx)/x)dx

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

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The solution

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∫01sin⁡(log⁡(x))1x1 dx\int\limits_{0}^{1} \sin{\left(\log{\left(x \right)} \right)} \frac{1}{x} 1\, dx 
Integral(sin(log(x))*1/x, (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

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

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

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

      Now substitute uu back in:

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

  2. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
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∫sin⁡(log⁡(x))1x1 dx=C−cos⁡(log⁡(x))\int \sin{\left(\log{\left(x \right)} \right)} \frac{1}{x} 1\, dx = C - \cos{\left(\log{\left(x \right)} \right)}
Simplify
The answer [src] 
<-2, 0>
⟨−2,0⟩\left\langle -2, 0\right\rangle 
=
= 
<-2, 0>
⟨−2,0⟩\left\langle -2, 0\right\rangle
Упростить
Numerical answer [src] 
0.0141500631560091
0.0141500631560091

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

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