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(x+2)^2*cos(3x)
Solving integrals/ (x+2)^2*cos(3x)

Integral of (x+2)^2*cos(3x) dx

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∫−20(x+2)2cos⁡(3x) dx\int\limits_{-2}^{0} \left(x + 2\right)^{2} \cos{\left(3 x \right)}\, dx 
Integral((x + 2)^2*cos(3*x), (x, -2, 0))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      (x+2)2cos⁡(3x)=x2cos⁡(3x)+4xcos⁡(3x)+4cos⁡(3x)\left(x + 2\right)^{2} \cos{\left(3 x \right)} = x^{2} \cos{\left(3 x \right)} + 4 x \cos{\left(3 x \right)} + 4 \cos{\left(3 x \right)}

    2. 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)=cos⁡(3x)\operatorname{dv}{\left(x \right)} = \cos{\left(3 x \right)}.

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

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

        1. Let u=3xu = 3 x.

          Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

          ∫cos⁡(u)9 du\int \frac{\cos{\left(u \right)}}{9}\, du

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

            ∫cos⁡(u)3 du=∫cos⁡(u) du3\int \frac{\cos{\left(u \right)}}{3}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}

            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)3\frac{\sin{\left(u \right)}}{3}

          Now substitute uu back in:

          sin⁡(3x)3\frac{\sin{\left(3 x \right)}}{3}

        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)=2x3u{\left(x \right)} = \frac{2 x}{3} and let dv⁡(x)=sin⁡(3x)\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}.

        Then du⁡(x)=23\operatorname{du}{\left(x \right)} = \frac{2}{3}.

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

        1. Let u=3xu = 3 x.

          Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

          ∫sin⁡(u)9 du\int \frac{\sin{\left(u \right)}}{9}\, du

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

            ∫sin⁡(u)3 du=∫sin⁡(u) du3\int \frac{\sin{\left(u \right)}}{3}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}

            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)3- \frac{\cos{\left(u \right)}}{3}

          Now substitute uu back in:

          −cos⁡(3x)3- \frac{\cos{\left(3 x \right)}}{3}

        Now evaluate the sub-integral.

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

        ∫(−2cos⁡(3x)9) dx=−2∫cos⁡(3x) dx9\int \left(- \frac{2 \cos{\left(3 x \right)}}{9}\right)\, dx = - \frac{2 \int \cos{\left(3 x \right)}\, dx}{9}

        1. Let u=3xu = 3 x.

          Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

          ∫cos⁡(u)9 du\int \frac{\cos{\left(u \right)}}{9}\, du

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

            ∫cos⁡(u)3 du=∫cos⁡(u) du3\int \frac{\cos{\left(u \right)}}{3}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}

            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)3\frac{\sin{\left(u \right)}}{3}

          Now substitute uu back in:

          sin⁡(3x)3\frac{\sin{\left(3 x \right)}}{3}

        So, the result is: −2sin⁡(3x)27- \frac{2 \sin{\left(3 x \right)}}{27}

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

        ∫4xcos⁡(3x) dx=4∫xcos⁡(3x) dx\int 4 x \cos{\left(3 x \right)}\, dx = 4 \int x \cos{\left(3 x \right)}\, dx

        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⁡(3x)\operatorname{dv}{\left(x \right)} = \cos{\left(3 x \right)}.

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

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

          1. Let u=3xu = 3 x.

            Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

            ∫cos⁡(u)9 du\int \frac{\cos{\left(u \right)}}{9}\, du

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

              ∫cos⁡(u)3 du=∫cos⁡(u) du3\int \frac{\cos{\left(u \right)}}{3}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}

              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)3\frac{\sin{\left(u \right)}}{3}

            Now substitute uu back in:

            sin⁡(3x)3\frac{\sin{\left(3 x \right)}}{3}

          Now evaluate the sub-integral.

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

          ∫sin⁡(3x)3 dx=∫sin⁡(3x) dx3\int \frac{\sin{\left(3 x \right)}}{3}\, dx = \frac{\int \sin{\left(3 x \right)}\, dx}{3}

          1. Let u=3xu = 3 x.

            Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

            ∫sin⁡(u)9 du\int \frac{\sin{\left(u \right)}}{9}\, du

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

              ∫sin⁡(u)3 du=∫sin⁡(u) du3\int \frac{\sin{\left(u \right)}}{3}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}

              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)3- \frac{\cos{\left(u \right)}}{3}

            Now substitute uu back in:

            −cos⁡(3x)3- \frac{\cos{\left(3 x \right)}}{3}

          So, the result is: −cos⁡(3x)9- \frac{\cos{\left(3 x \right)}}{9}

        So, the result is: 4xsin⁡(3x)3+4cos⁡(3x)9\frac{4 x \sin{\left(3 x \right)}}{3} + \frac{4 \cos{\left(3 x \right)}}{9}

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

        ∫4cos⁡(3x) dx=4∫cos⁡(3x) dx\int 4 \cos{\left(3 x \right)}\, dx = 4 \int \cos{\left(3 x \right)}\, dx

        1. Let u=3xu = 3 x.

          Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

          ∫cos⁡(u)9 du\int \frac{\cos{\left(u \right)}}{9}\, du

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

            ∫cos⁡(u)3 du=∫cos⁡(u) du3\int \frac{\cos{\left(u \right)}}{3}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}

            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)3\frac{\sin{\left(u \right)}}{3}

          Now substitute uu back in:

          sin⁡(3x)3\frac{\sin{\left(3 x \right)}}{3}

        So, the result is: 4sin⁡(3x)3\frac{4 \sin{\left(3 x \right)}}{3}

      The result is: x2sin⁡(3x)3+4xsin⁡(3x)3+2xcos⁡(3x)9+34sin⁡(3x)27+4cos⁡(3x)9\frac{x^{2} \sin{\left(3 x \right)}}{3} + \frac{4 x \sin{\left(3 x \right)}}{3} + \frac{2 x \cos{\left(3 x \right)}}{9} + \frac{34 \sin{\left(3 x \right)}}{27} + \frac{4 \cos{\left(3 x \right)}}{9}

    Method #2

    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)=x2+4x+4u{\left(x \right)} = x^{2} + 4 x + 4 and let dv⁡(x)=cos⁡(3x)\operatorname{dv}{\left(x \right)} = \cos{\left(3 x \right)}.

      Then du⁡(x)=2x+4\operatorname{du}{\left(x \right)} = 2 x + 4.

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

      1. Let u=3xu = 3 x.

        Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

        ∫cos⁡(u)9 du\int \frac{\cos{\left(u \right)}}{9}\, du

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

          ∫cos⁡(u)3 du=∫cos⁡(u) du3\int \frac{\cos{\left(u \right)}}{3}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}

          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)3\frac{\sin{\left(u \right)}}{3}

        Now substitute uu back in:

        sin⁡(3x)3\frac{\sin{\left(3 x \right)}}{3}

      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)=2x3+43u{\left(x \right)} = \frac{2 x}{3} + \frac{4}{3} and let dv⁡(x)=sin⁡(3x)\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}.

      Then du⁡(x)=23\operatorname{du}{\left(x \right)} = \frac{2}{3}.

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

      1. Let u=3xu = 3 x.

        Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

        ∫sin⁡(u)9 du\int \frac{\sin{\left(u \right)}}{9}\, du

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

          ∫sin⁡(u)3 du=∫sin⁡(u) du3\int \frac{\sin{\left(u \right)}}{3}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}

          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)3- \frac{\cos{\left(u \right)}}{3}

        Now substitute uu back in:

        −cos⁡(3x)3- \frac{\cos{\left(3 x \right)}}{3}

      Now evaluate the sub-integral.

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

      ∫(−2cos⁡(3x)9) dx=−2∫cos⁡(3x) dx9\int \left(- \frac{2 \cos{\left(3 x \right)}}{9}\right)\, dx = - \frac{2 \int \cos{\left(3 x \right)}\, dx}{9}

      1. Let u=3xu = 3 x.

        Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

        ∫cos⁡(u)9 du\int \frac{\cos{\left(u \right)}}{9}\, du

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

          ∫cos⁡(u)3 du=∫cos⁡(u) du3\int \frac{\cos{\left(u \right)}}{3}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}

          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)3\frac{\sin{\left(u \right)}}{3}

        Now substitute uu back in:

        sin⁡(3x)3\frac{\sin{\left(3 x \right)}}{3}

      So, the result is: −2sin⁡(3x)27- \frac{2 \sin{\left(3 x \right)}}{27}

    Method #3

    1. Rewrite the integrand:

      (x+2)2cos⁡(3x)=x2cos⁡(3x)+4xcos⁡(3x)+4cos⁡(3x)\left(x + 2\right)^{2} \cos{\left(3 x \right)} = x^{2} \cos{\left(3 x \right)} + 4 x \cos{\left(3 x \right)} + 4 \cos{\left(3 x \right)}

    2. 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)=cos⁡(3x)\operatorname{dv}{\left(x \right)} = \cos{\left(3 x \right)}.

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

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

        1. Let u=3xu = 3 x.

          Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

          ∫cos⁡(u)9 du\int \frac{\cos{\left(u \right)}}{9}\, du

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

            ∫cos⁡(u)3 du=∫cos⁡(u) du3\int \frac{\cos{\left(u \right)}}{3}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}

            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)3\frac{\sin{\left(u \right)}}{3}

          Now substitute uu back in:

          sin⁡(3x)3\frac{\sin{\left(3 x \right)}}{3}

        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)=2x3u{\left(x \right)} = \frac{2 x}{3} and let dv⁡(x)=sin⁡(3x)\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}.

        Then du⁡(x)=23\operatorname{du}{\left(x \right)} = \frac{2}{3}.

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

        1. Let u=3xu = 3 x.

          Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

          ∫sin⁡(u)9 du\int \frac{\sin{\left(u \right)}}{9}\, du

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

            ∫sin⁡(u)3 du=∫sin⁡(u) du3\int \frac{\sin{\left(u \right)}}{3}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}

            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)3- \frac{\cos{\left(u \right)}}{3}

          Now substitute uu back in:

          −cos⁡(3x)3- \frac{\cos{\left(3 x \right)}}{3}

        Now evaluate the sub-integral.

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

        ∫(−2cos⁡(3x)9) dx=−2∫cos⁡(3x) dx9\int \left(- \frac{2 \cos{\left(3 x \right)}}{9}\right)\, dx = - \frac{2 \int \cos{\left(3 x \right)}\, dx}{9}

        1. Let u=3xu = 3 x.

          Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

          ∫cos⁡(u)9 du\int \frac{\cos{\left(u \right)}}{9}\, du

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

            ∫cos⁡(u)3 du=∫cos⁡(u) du3\int \frac{\cos{\left(u \right)}}{3}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}

            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)3\frac{\sin{\left(u \right)}}{3}

          Now substitute uu back in:

          sin⁡(3x)3\frac{\sin{\left(3 x \right)}}{3}

        So, the result is: −2sin⁡(3x)27- \frac{2 \sin{\left(3 x \right)}}{27}

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

        ∫4xcos⁡(3x) dx=4∫xcos⁡(3x) dx\int 4 x \cos{\left(3 x \right)}\, dx = 4 \int x \cos{\left(3 x \right)}\, dx

        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⁡(3x)\operatorname{dv}{\left(x \right)} = \cos{\left(3 x \right)}.

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

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

          1. Let u=3xu = 3 x.

            Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

            ∫cos⁡(u)9 du\int \frac{\cos{\left(u \right)}}{9}\, du

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

              ∫cos⁡(u)3 du=∫cos⁡(u) du3\int \frac{\cos{\left(u \right)}}{3}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}

              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)3\frac{\sin{\left(u \right)}}{3}

            Now substitute uu back in:

            sin⁡(3x)3\frac{\sin{\left(3 x \right)}}{3}

          Now evaluate the sub-integral.

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

          ∫sin⁡(3x)3 dx=∫sin⁡(3x) dx3\int \frac{\sin{\left(3 x \right)}}{3}\, dx = \frac{\int \sin{\left(3 x \right)}\, dx}{3}

          1. Let u=3xu = 3 x.

            Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

            ∫sin⁡(u)9 du\int \frac{\sin{\left(u \right)}}{9}\, du

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

              ∫sin⁡(u)3 du=∫sin⁡(u) du3\int \frac{\sin{\left(u \right)}}{3}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}

              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)3- \frac{\cos{\left(u \right)}}{3}

            Now substitute uu back in:

            −cos⁡(3x)3- \frac{\cos{\left(3 x \right)}}{3}

          So, the result is: −cos⁡(3x)9- \frac{\cos{\left(3 x \right)}}{9}

        So, the result is: 4xsin⁡(3x)3+4cos⁡(3x)9\frac{4 x \sin{\left(3 x \right)}}{3} + \frac{4 \cos{\left(3 x \right)}}{9}

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

        ∫4cos⁡(3x) dx=4∫cos⁡(3x) dx\int 4 \cos{\left(3 x \right)}\, dx = 4 \int \cos{\left(3 x \right)}\, dx

        1. Let u=3xu = 3 x.

          Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

          ∫cos⁡(u)9 du\int \frac{\cos{\left(u \right)}}{9}\, du

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

            ∫cos⁡(u)3 du=∫cos⁡(u) du3\int \frac{\cos{\left(u \right)}}{3}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}

            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)3\frac{\sin{\left(u \right)}}{3}

          Now substitute uu back in:

          sin⁡(3x)3\frac{\sin{\left(3 x \right)}}{3}

        So, the result is: 4sin⁡(3x)3\frac{4 \sin{\left(3 x \right)}}{3}

      The result is: x2sin⁡(3x)3+4xsin⁡(3x)3+2xcos⁡(3x)9+34sin⁡(3x)27+4cos⁡(3x)9\frac{x^{2} \sin{\left(3 x \right)}}{3} + \frac{4 x \sin{\left(3 x \right)}}{3} + \frac{2 x \cos{\left(3 x \right)}}{9} + \frac{34 \sin{\left(3 x \right)}}{27} + \frac{4 \cos{\left(3 x \right)}}{9}

  2. Add the constant of integration:

    x2sin⁡(3x)3+4xsin⁡(3x)3+2xcos⁡(3x)9+34sin⁡(3x)27+4cos⁡(3x)9+constant\frac{x^{2} \sin{\left(3 x \right)}}{3} + \frac{4 x \sin{\left(3 x \right)}}{3} + \frac{2 x \cos{\left(3 x \right)}}{9} + \frac{34 \sin{\left(3 x \right)}}{27} + \frac{4 \cos{\left(3 x \right)}}{9}+ \mathrm{constant}

The answer is:

x2sin⁡(3x)3+4xsin⁡(3x)3+2xcos⁡(3x)9+34sin⁡(3x)27+4cos⁡(3x)9+constant\frac{x^{2} \sin{\left(3 x \right)}}{3} + \frac{4 x \sin{\left(3 x \right)}}{3} + \frac{2 x \cos{\left(3 x \right)}}{9} + \frac{34 \sin{\left(3 x \right)}}{27} + \frac{4 \cos{\left(3 x \right)}}{9}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                                               
 |                                                        2                                       
 |        2                   4*cos(3*x)   34*sin(3*x)   x *sin(3*x)   2*x*cos(3*x)   4*x*sin(3*x)
 | (x + 2) *cos(3*x) dx = C + ---------- + ----------- + ----------- + ------------ + ------------
 |                                9             27            3             9              3      
/
(9 x2−2) sin⁡(3 x)+6 x cos⁡(3 x)9+4 (3 x sin⁡(3 x)+cos⁡(3 x))3+4 sin⁡(3 x)3{{{{\left(9\,x^2-2\right)\,\sin \left(3\,x\right)+6\,x\,\cos \left( 3\,x\right)}\over{9}}+{{4\,\left(3\,x\,\sin \left(3\,x\right)+\cos \left(3\,x\right)\right)}\over{3}}+4\,\sin \left(3\,x\right)}\over{3 }}
Simplify
The graph
-2.0-1.8-1.6-1.4-1.2-1.0-0.8-0.6-0.4-0.20.05-5
Plot the graph f(x)
The answer [src] 
4   2*sin(6)
- - --------
9      27
49−2 sin⁡627{{4}\over{9}}-{{2\,\sin 6}\over{27}} 
=
= 
4   2*sin(6)
- - --------
9      27
−2sin⁡(6)27+49- \frac{2 \sin{\left(6 \right)}}{27} + \frac{4}{9}
Упростить
Numerical answer [src] 
0.465141888755476
0.465141888755476
The graph
Integral of (x+2)^2*cos(3x) dx

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

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