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(2-x)sinx/2

Integral of (2-x)sinx/2 dx

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 |  (2 - x)*sin(x)   
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01(2x)sin(x)2dx\int\limits_{0}^{1} \frac{\left(2 - x\right) \sin{\left(x \right)}}{2}\, dx
Integral(2 - x*sin(x)/2, (x, 0, 1))
Detail solution
  1. The integral of a constant times a function is the constant times the integral of the function:

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

    1. There are multiple ways to do this integral.

      Method #1

      1. Let u=xu = - x.

        Then let du=dxdu = - dx and substitute dudu:

        (usin(u)+2sin(u))du\int \left(u \sin{\left(u \right)} + 2 \sin{\left(u \right)}\right)\, du

        1. Integrate term-by-term:

          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)=sin(u)\operatorname{dv}{\left(u \right)} = \sin{\left(u \right)}.

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

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

            1. The integral of sine is negative cosine:

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

            Now evaluate the sub-integral.

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

            (cos(u))du=cos(u)du\int \left(- \cos{\left(u \right)}\right)\, 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)}

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

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

            2sin(u)du=2sin(u)du\int 2 \sin{\left(u \right)}\, du = 2 \int \sin{\left(u \right)}\, du

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

          The result is: ucos(u)+sin(u)2cos(u)- u \cos{\left(u \right)} + \sin{\left(u \right)} - 2 \cos{\left(u \right)}

        Now substitute uu back in:

        xcos(x)sin(x)2cos(x)x \cos{\left(x \right)} - \sin{\left(x \right)} - 2 \cos{\left(x \right)}

      Method #2

      1. Rewrite the integrand:

        (2x)sin(x)=xsin(x)+2sin(x)\left(2 - x\right) \sin{\left(x \right)} = - x \sin{\left(x \right)} + 2 \sin{\left(x \right)}

      2. Integrate term-by-term:

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

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

          1. Use integration by parts:

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

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

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

            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 a constant times a function is the constant times the integral of the function:

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

            1. The integral of cosine is sine:

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

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

          So, the result is: xcos(x)sin(x)x \cos{\left(x \right)} - \sin{\left(x \right)}

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

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

          1. The integral of sine is negative cosine:

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

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

        The result is: xcos(x)sin(x)2cos(x)x \cos{\left(x \right)} - \sin{\left(x \right)} - 2 \cos{\left(x \right)}

      Method #3

      1. Use integration by parts:

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

        Let u(x)=2xu{\left(x \right)} = 2 - x and let dv(x)=sin(x)\operatorname{dv}{\left(x \right)} = \sin{\left(x \right)}.

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

        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 cosine is sine:

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

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

  2. Add the constant of integration:

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


The answer is:

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

The answer (Indefinite) [src]
  /                                                  
 |                                                   
 | (2 - x)*sin(x)                   sin(x)   x*cos(x)
 | -------------- dx = C - cos(x) - ------ + --------
 |       2                            2         2    
 |                                                   
/                                                    
(2x)sin(x)2dx=C+xcos(x)2sin(x)2cos(x)\int \frac{\left(2 - x\right) \sin{\left(x \right)}}{2}\, dx = C + \frac{x \cos{\left(x \right)}}{2} - \frac{\sin{\left(x \right)}}{2} - \cos{\left(x \right)}
The graph
0.001.000.100.200.300.400.500.600.700.800.902-2
The answer [src]
    cos(1)   sin(1)
1 - ------ - ------
      2        2   
sin(1)2cos(1)2+1- \frac{\sin{\left(1 \right)}}{2} - \frac{\cos{\left(1 \right)}}{2} + 1
=
=
    cos(1)   sin(1)
1 - ------ - ------
      2        2   
sin(1)2cos(1)2+1- \frac{\sin{\left(1 \right)}}{2} - \frac{\cos{\left(1 \right)}}{2} + 1
Numerical answer [src]
0.309113354661982
0.309113354661982
The graph
Integral of (2-x)sinx/2 dx

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