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Solving integrals/ 4xcosxsinx

Integral of 4xcosxsinx dx

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∫014xcos⁡(x)sin⁡(x) dx\int\limits_{0}^{1} 4 x \cos{\left(x \right)} \sin{\left(x \right)}\, dx 
Integral(((4*x)*cos(x))*sin(x), (x, 0, 1))
Detail solution

  1. Use integration by parts:

    ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

    Let u(x)=4xu{\left(x \right)} = 4 x and let dv⁡(x)=sin⁡(x)cos⁡(x)\operatorname{dv}{\left(x \right)} = \sin{\left(x \right)} \cos{\left(x \right)}.

    Then du⁡(x)=4\operatorname{du}{\left(x \right)} = 4.

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

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

      ∫sin⁡(2x)2 dx=∫sin⁡(2x) dx2\int \frac{\sin{\left(2 x \right)}}{2}\, dx = \frac{\int \sin{\left(2 x \right)}\, dx}{2}

      1. There are multiple ways to do this integral.

        Method #1

        1. Let u=2xu = 2 x.

          Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

          ∫sin⁡(u)2 du\int \frac{\sin{\left(u \right)}}{2}\, du

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

            ∫sin⁡(u) du=∫sin⁡(u) du2\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}

            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)2- \frac{\cos{\left(u \right)}}{2}

          Now substitute uu back in:

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

        Method #2

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

          ∫2sin⁡(x)cos⁡(x) dx=2∫sin⁡(x)cos⁡(x) dx\int 2 \sin{\left(x \right)} \cos{\left(x \right)}\, dx = 2 \int \sin{\left(x \right)} \cos{\left(x \right)}\, dx

          1. There are multiple ways to do this integral.

            Method #1

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

              Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute −du- du:

              ∫(−u) du\int \left(- u\right)\, du

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

                ∫u du=−∫u du\int u\, du = - \int u\, du

                1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

                  ∫u du=u22\int u\, du = \frac{u^{2}}{2}

                So, the result is: −u22- \frac{u^{2}}{2}

              Now substitute uu back in:

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

            Method #2

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

              Then let du=cos⁡(x)dxdu = \cos{\left(x \right)} dx and substitute dudu:

              ∫u du\int u\, du

              1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

                ∫u du=u22\int u\, du = \frac{u^{2}}{2}

              Now substitute uu back in:

              sin⁡2(x)2\frac{\sin^{2}{\left(x \right)}}{2}

          So, the result is: −cos⁡2(x)- \cos^{2}{\left(x \right)}

      So, the result is: −cos⁡(2x)4- \frac{\cos{\left(2 x \right)}}{4}

    Now evaluate the sub-integral.

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

    ∫(−cos⁡(2x)) dx=−∫cos⁡(2x) dx\int \left(- \cos{\left(2 x \right)}\right)\, dx = - \int \cos{\left(2 x \right)}\, dx

    1. Let u=2xu = 2 x.

      Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

      ∫cos⁡(u)2 du\int \frac{\cos{\left(u \right)}}{2}\, du

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

        ∫cos⁡(u) du=∫cos⁡(u) du2\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}

        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: sin⁡(u)2\frac{\sin{\left(u \right)}}{2}

      Now substitute uu back in:

      sin⁡(2x)2\frac{\sin{\left(2 x \right)}}{2}

    So, the result is: −sin⁡(2x)2- \frac{\sin{\left(2 x \right)}}{2}

  3. Add the constant of integration:

    −xcos⁡(2x)+sin⁡(2x)2+constant- x \cos{\left(2 x \right)} + \frac{\sin{\left(2 x \right)}}{2}+ \mathrm{constant}

The answer is:

−xcos⁡(2x)+sin⁡(2x)2+constant- x \cos{\left(2 x \right)} + \frac{\sin{\left(2 x \right)}}{2}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                
 |                            sin(2*x)             
 | 4*x*cos(x)*sin(x) dx = C + -------- - x*cos(2*x)
 |                               2                 
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∫4xcos⁡(x)sin⁡(x) dx=C−xcos⁡(2x)+sin⁡(2x)2\int 4 x \cos{\left(x \right)} \sin{\left(x \right)}\, dx = C - x \cos{\left(2 x \right)} + \frac{\sin{\left(2 x \right)}}{2}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.9002
Plot the graph f(x)
The answer [src] 
   2         2                   
sin (1) - cos (1) + cos(1)*sin(1)
−cos⁡2(1)+sin⁡(1)cos⁡(1)+sin⁡2(1)- \cos^{2}{\left(1 \right)} + \sin{\left(1 \right)} \cos{\left(1 \right)} + \sin^{2}{\left(1 \right)} 
=
= 
   2         2                   
sin (1) - cos (1) + cos(1)*sin(1)
−cos⁡2(1)+sin⁡(1)cos⁡(1)+sin⁡2(1)- \cos^{2}{\left(1 \right)} + \sin{\left(1 \right)} \cos{\left(1 \right)} + \sin^{2}{\left(1 \right)} 
sin(1)^2 - cos(1)^2 + cos(1)*sin(1)
Numerical answer [src] 
0.870795549959983
0.870795549959983

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

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