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Solving integrals/ (e*cos(ln(x)))/x

Integral of (e*cos(ln(x)))/x dx

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The solution

You have entered [src] 
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 |  E*cos(log(x))   
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∫01ecos⁡(log⁡(x))x dx\int\limits_{0}^{1} \frac{e \cos{\left(\log{\left(x \right)} \right)}}{x}\, dx 
Integral((E*cos(log(x)))/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 edue du:

      ∫ecos⁡(u) du\int e \cos{\left(u \right)}\, du

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

        ∫cos⁡(u) du=e∫cos⁡(u) du\int \cos{\left(u \right)}\, du = e \int \cos{\left(u \right)}\, du

        1. The integral of cosine is sine:

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

        So, the result is: esin⁡(u)e \sin{\left(u \right)}

      Now substitute uu back in:

      esin⁡(log⁡(x))e \sin{\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 −edu- e du:

      ∫(−ecos⁡(log⁡(1u))u) du\int \left(- \frac{e \cos{\left(\log{\left(\frac{1}{u} \right)} \right)}}{u}\right)\, du

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

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

        1. Let u=1uu = \frac{1}{u}.

          Then let du=−duu2du = - \frac{du}{u^{2}} and substitute −du- du:

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

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

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

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

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

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

              1. The integral of cosine is sine:

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

              Now substitute uu back in:

              sin⁡(log⁡(u))\sin{\left(\log{\left(u \right)} \right)}

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

          Now substitute uu back in:

          sin⁡(log⁡(u))\sin{\left(\log{\left(u \right)} \right)}

        So, the result is: −esin⁡(log⁡(u))- e \sin{\left(\log{\left(u \right)} \right)}

      Now substitute uu back in:

      esin⁡(log⁡(x))e \sin{\left(\log{\left(x \right)} \right)}

  2. Add the constant of integration:

    esin⁡(log⁡(x))+constante \sin{\left(\log{\left(x \right)} \right)}+ \mathrm{constant}

The answer is:

esin⁡(log⁡(x))+constante \sin{\left(\log{\left(x \right)} \right)}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                    
 |                                     
 | E*cos(log(x))                       
 | ------------- dx = C + E*sin(log(x))
 |       x                             
 |                                     
/
∫ecos⁡(log⁡(x))x dx=C+esin⁡(log⁡(x))\int \frac{e \cos{\left(\log{\left(x \right)} \right)}}{x}\, dx = C + e \sin{\left(\log{\left(x \right)} \right)}
Simplify
The answer [src] 
-<-1, 1>*E
−⟨−1,1⟩e- \left\langle -1, 1\right\rangle e 
=
= 
-<-1, 1>*E
−⟨−1,1⟩e- \left\langle -1, 1\right\rangle e 
-AccumBounds(-1, 1)*E
Numerical answer [src] 
0.299165685009485
0.299165685009485

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

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