Mister Exam

Integral of sinxlncosx dx

Limits of integration:

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The solution

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 |  sin(x)*log(cos(x)) dx
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01log(cos(x))sin(x)dx\int\limits_{0}^{1} \log{\left(\cos{\left(x \right)} \right)} \sin{\left(x \right)}\, dx
Integral(sin(x)*log(cos(x)), (x, 0, 1))
Detail solution
  1. There are multiple ways to do this integral.

    Method #1

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

      Then let du=sin(x)dxcos(x)du = - \frac{\sin{\left(x \right)} dx}{\cos{\left(x \right)}} and substitute du- du:

      ueudu\int u e^{u}\, du

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

        (ueu)du=ueudu\int \left(- u e^{u}\right)\, du = - \int u e^{u}\, du

        1. Use integration by parts:

          udv=uvvdu\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

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

          Then du(u)=1\operatorname{du}{\left(u \right)} = 1.

          To find v(u)v{\left(u \right)}:

          1. The integral of the exponential function is itself.

            eudu=eu\int e^{u}\, du = e^{u}

          Now evaluate the sub-integral.

        2. The integral of the exponential function is itself.

          eudu=eu\int e^{u}\, du = e^{u}

        So, the result is: ueu+eu- u e^{u} + e^{u}

      Now substitute uu back in:

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

    Method #2

    1. Use integration by parts:

      udv=uvvdu\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

      Let u(x)=log(cos(x))u{\left(x \right)} = \log{\left(\cos{\left(x \right)} \right)} and let dv(x)=sin(x)\operatorname{dv}{\left(x \right)} = \sin{\left(x \right)}.

      Then du(x)=sin(x)cos(x)\operatorname{du}{\left(x \right)} = - \frac{\sin{\left(x \right)}}{\cos{\left(x \right)}}.

      To find v(x)v{\left(x \right)}:

      1. The integral of sine is negative cosine:

        sin(x)dx=cos(x)\int \sin{\left(x \right)}\, dx = - \cos{\left(x \right)}

      Now evaluate the sub-integral.

    2. The integral of sine is negative cosine:

      sin(x)dx=cos(x)\int \sin{\left(x \right)}\, dx = - \cos{\left(x \right)}

  2. Now simplify:

    (1log(cos(x)))cos(x)\left(1 - \log{\left(\cos{\left(x \right)} \right)}\right) \cos{\left(x \right)}

  3. Add the constant of integration:

    (1log(cos(x)))cos(x)+constant\left(1 - \log{\left(\cos{\left(x \right)} \right)}\right) \cos{\left(x \right)}+ \mathrm{constant}


The answer is:

(1log(cos(x)))cos(x)+constant\left(1 - \log{\left(\cos{\left(x \right)} \right)}\right) \cos{\left(x \right)}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                                       
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 | sin(x)*log(cos(x)) dx = C - cos(x)*log(cos(x)) + cos(x)
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cosxcosxlogcosx\cos x-\cos x\,\log \cos x
The graph
0.001.000.100.200.300.400.500.600.700.800.902-2
The answer [src]
-1 - cos(1)*log(cos(1)) + cos(1)
cos1logcos1+cos11-\cos 1\,\log \cos 1+\cos 1-1
=
=
-1 - cos(1)*log(cos(1)) + cos(1)
1log(cos(1))cos(1)+cos(1)-1 - \log{\left(\cos{\left(1 \right)} \right)} \cos{\left(1 \right)} + \cos{\left(1 \right)}
Numerical answer [src]
-0.127073292628833
-0.127073292628833
The graph
Integral of sinxlncosx dx

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