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Integral of cos(lnx)/x^2 dx

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  1               
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 |  cos(log(x))   
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01cos(log(x))x2dx\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:

    eucos(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:

              2eucos(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)2eucos(u)2- \frac{e^{u} \sin{\left(u \right)}}{2} - \frac{e^{u} \cos{\left(u \right)}}{2}

        Now substitute uu back in:

        eusin(u)2eucos(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 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=eusin(u)dueucos(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 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=(eucos(u))du+eusin(u)eucos(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:

          2eucos(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)2eucos(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))2xcos(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))x2dx=C+sin(log(x))2xcos(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}
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.