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Solving integrals/ cos(lnx)/x^2

Integral of cos(lnx)/x^2 dx

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The solution

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∫01cos⁡(log⁡(x))x2 dx\int\limits_{0}^{1} \frac{\cos{\left(\log{\left(x \right)} \right)}}{x^{2}}\, dx 
Integral(cos(log(x))/x^2, (x, 0, 1))
Detail solution

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

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

    ∫e−ucos⁡(u) du\int e^{- u} \cos{\left(u \right)}\, du

    1. There are multiple ways to do this integral.

      Method #1

      1. Let u=−uu = - u.

        Then let du=−dudu = - du and substitute −du- du:

        ∫(−eucos⁡(u)) du\int \left(- e^{u} \cos{\left(u \right)}\right)\, du

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

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

          1. Use integration by parts, noting that the integrand eventually repeats itself.

            1. For the integrand eucos⁡(u)e^{u} \cos{\left(u \right)}:

              Let u(u)=cos⁡(u)u{\left(u \right)} = \cos{\left(u \right)} and let dv⁡(u)=eu\operatorname{dv}{\left(u \right)} = e^{u}.

              Then ∫eucos⁡(u) du=eucos⁡(u)−∫(−eusin⁡(u)) du\int e^{u} \cos{\left(u \right)}\, du = e^{u} \cos{\left(u \right)} - \int \left(- e^{u} \sin{\left(u \right)}\right)\, du.

            2. For the integrand −eusin⁡(u)- e^{u} \sin{\left(u \right)}:

              Let u(u)=−sin⁡(u)u{\left(u \right)} = - \sin{\left(u \right)} and let dv⁡(u)=eu\operatorname{dv}{\left(u \right)} = e^{u}.

              Then ∫eucos⁡(u) du=eusin⁡(u)+eucos⁡(u)+∫(−eucos⁡(u)) du\int e^{u} \cos{\left(u \right)}\, du = e^{u} \sin{\left(u \right)} + e^{u} \cos{\left(u \right)} + \int \left(- e^{u} \cos{\left(u \right)}\right)\, du.

            3. Notice that the integrand has repeated itself, so move it to one side:

              2∫eucos⁡(u) du=eusin⁡(u)+eucos⁡(u)2 \int e^{u} \cos{\left(u \right)}\, du = e^{u} \sin{\left(u \right)} + e^{u} \cos{\left(u \right)}

              Therefore,

              ∫eucos⁡(u) du=eusin⁡(u)2+eucos⁡(u)2\int e^{u} \cos{\left(u \right)}\, du = \frac{e^{u} \sin{\left(u \right)}}{2} + \frac{e^{u} \cos{\left(u \right)}}{2}

          So, the result is: −eusin⁡(u)2−eucos⁡(u)2- \frac{e^{u} \sin{\left(u \right)}}{2} - \frac{e^{u} \cos{\left(u \right)}}{2}

        Now substitute uu back in:

        e−usin⁡(u)2−e−ucos⁡(u)2\frac{e^{- u} \sin{\left(u \right)}}{2} - \frac{e^{- u} \cos{\left(u \right)}}{2}

      Method #2

      1. Use integration by parts, noting that the integrand eventually repeats itself.

        1. For the integrand e−ucos⁡(u)e^{- u} \cos{\left(u \right)}:

          Let u(u)=cos⁡(u)u{\left(u \right)} = \cos{\left(u \right)} and let dv⁡(u)=e−u\operatorname{dv}{\left(u \right)} = e^{- u}.

          Then ∫e−ucos⁡(u) du=−∫e−usin⁡(u) du−e−ucos⁡(u)\int e^{- u} \cos{\left(u \right)}\, du = - \int e^{- u} \sin{\left(u \right)}\, du - e^{- u} \cos{\left(u \right)}.

        2. For the integrand e−usin⁡(u)e^{- u} \sin{\left(u \right)}:

          Let u(u)=sin⁡(u)u{\left(u \right)} = \sin{\left(u \right)} and let dv⁡(u)=e−u\operatorname{dv}{\left(u \right)} = e^{- u}.

          Then ∫e−ucos⁡(u) du=∫(−e−ucos⁡(u)) du+e−usin⁡(u)−e−ucos⁡(u)\int e^{- u} \cos{\left(u \right)}\, du = \int \left(- e^{- u} \cos{\left(u \right)}\right)\, du + e^{- u} \sin{\left(u \right)} - e^{- u} \cos{\left(u \right)}.

        3. Notice that the integrand has repeated itself, so move it to one side:

          2∫e−ucos⁡(u) du=e−usin⁡(u)−e−ucos⁡(u)2 \int e^{- u} \cos{\left(u \right)}\, du = e^{- u} \sin{\left(u \right)} - e^{- u} \cos{\left(u \right)}

          Therefore,

          ∫e−ucos⁡(u) du=e−usin⁡(u)2−e−ucos⁡(u)2\int e^{- u} \cos{\left(u \right)}\, du = \frac{e^{- u} \sin{\left(u \right)}}{2} - \frac{e^{- u} \cos{\left(u \right)}}{2}

    Now substitute uu back in:

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

  2. Now simplify:

    −2cos⁡(log⁡(x)+π4)2x- \frac{\sqrt{2} \cos{\left(\log{\left(x \right)} + \frac{\pi}{4} \right)}}{2 x}

  3. Add the constant of integration:

    −2cos⁡(log⁡(x)+π4)2x+constant- \frac{\sqrt{2} \cos{\left(\log{\left(x \right)} + \frac{\pi}{4} \right)}}{2 x}+ \mathrm{constant}

The answer is:

−2cos⁡(log⁡(x)+π4)2x+constant- \frac{\sqrt{2} \cos{\left(\log{\left(x \right)} + \frac{\pi}{4} \right)}}{2 x}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                              
 |                                               
 | cos(log(x))          sin(log(x))   cos(log(x))
 | ----------- dx = C + ----------- - -----------
 |       2                  2*x           2*x    
 |      x                                        
 |                                               
/
∫cos⁡(log⁡(x))x2 dx=C+sin⁡(log⁡(x))2x−cos⁡(log⁡(x))2x\int \frac{\cos{\left(\log{\left(x \right)} \right)}}{x^{2}}\, dx = C + \frac{\sin{\left(\log{\left(x \right)} \right)}}{2 x} - \frac{\cos{\left(\log{\left(x \right)} \right)}}{2 x}
Simplify
The answer [src] 
<-oo, oo>
⟨−∞,∞⟩\left\langle -\infty, \infty\right\rangle 
=
= 
<-oo, oo>
⟨−∞,∞⟩\left\langle -\infty, \infty\right\rangle 
AccumBounds(-oo, oo)
Numerical answer [src] 
7.49656888054513e+18
7.49656888054513e+18

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

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