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Solving integrals/ xcosx+x^2sinx

Integral of xcosx+x^2sinx dx

Limits of integration:

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The graph:

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Piecewise:

The solution

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∫01(x2sin⁡(x)+xcos⁡(x)) dx\int\limits_{0}^{1} \left(x^{2} \sin{\left(x \right)} + x \cos{\left(x \right)}\right)\, dx 
Integral(x*cos(x) + x^2*sin(x), (x, 0, 1))
Detail solution

  1. Integrate term-by-term:

    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)=x2u{\left(x \right)} = x^{2} and let dv⁡(x)=sin⁡(x)\operatorname{dv}{\left(x \right)} = \sin{\left(x \right)}.

      Then du⁡(x)=2x\operatorname{du}{\left(x \right)} = 2 x.

      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. 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)=−2xu{\left(x \right)} = - 2 x and let dv⁡(x)=cos⁡(x)\operatorname{dv}{\left(x \right)} = \cos{\left(x \right)}.

      Then du⁡(x)=−2\operatorname{du}{\left(x \right)} = -2.

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

      1. The integral of cosine is sine:

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

      Now evaluate the sub-integral.

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

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

    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)=xu{\left(x \right)} = x and let dv⁡(x)=cos⁡(x)\operatorname{dv}{\left(x \right)} = \cos{\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 cosine is sine:

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

    The result is: −x2cos⁡(x)+3xsin⁡(x)+3cos⁡(x)- x^{2} \cos{\left(x \right)} + 3 x \sin{\left(x \right)} + 3 \cos{\left(x \right)}

  2. Add the constant of integration:

    −x2cos⁡(x)+3xsin⁡(x)+3cos⁡(x)+constant- x^{2} \cos{\left(x \right)} + 3 x \sin{\left(x \right)} + 3 \cos{\left(x \right)}+ \mathrm{constant}

The answer is:

−x2cos⁡(x)+3xsin⁡(x)+3cos⁡(x)+constant- x^{2} \cos{\left(x \right)} + 3 x \sin{\left(x \right)} + 3 \cos{\left(x \right)}+ \mathrm{constant}

The answer (Indefinite) [src] 
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 | \x*cos(x) + x *sin(x)/ dx = C + 3*cos(x) - x *cos(x) + 3*x*sin(x)
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∫(x2sin⁡(x)+xcos⁡(x)) dx=C−x2cos⁡(x)+3xsin⁡(x)+3cos⁡(x)\int \left(x^{2} \sin{\left(x \right)} + x \cos{\left(x \right)}\right)\, dx = C - x^{2} \cos{\left(x \right)} + 3 x \sin{\left(x \right)} + 3 \cos{\left(x \right)}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.9005
Plot the graph f(x)
The answer [src] 
-3 + 2*cos(1) + 3*sin(1)
−3+2cos⁡(1)+3sin⁡(1)-3 + 2 \cos{\left(1 \right)} + 3 \sin{\left(1 \right)} 
=
= 
-3 + 2*cos(1) + 3*sin(1)
−3+2cos⁡(1)+3sin⁡(1)-3 + 2 \cos{\left(1 \right)} + 3 \sin{\left(1 \right)} 
-3 + 2*cos(1) + 3*sin(1)
Numerical answer [src] 
0.605017566159969
0.605017566159969

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

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