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(x^2+4x-2)sin3xdx
Solving integrals/ (x^2+4x-2)sin3xdx

Integral of (x^2+4x-2)sin3xdx dx

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∫01((x2+4x)−2)sin⁡(3x) dx\int\limits_{0}^{1} \left(\left(x^{2} + 4 x\right) - 2\right) \sin{\left(3 x \right)}\, dx 
Integral((x^2 + 4*x - 2)*sin(3*x), (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      ((x2+4x)−2)sin⁡(3x)=x2sin⁡(3x)+4xsin⁡(3x)−2sin⁡(3x)\left(\left(x^{2} + 4 x\right) - 2\right) \sin{\left(3 x \right)} = x^{2} \sin{\left(3 x \right)} + 4 x \sin{\left(3 x \right)} - 2 \sin{\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)=sin⁡(3x)\operatorname{dv}{\left(x \right)} = \sin{\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}:

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

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

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

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

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

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

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

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

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

        1. Let u=3xu = 3 x.

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

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

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

            ∫sin⁡(u) du=∫sin⁡(u) du3\int \sin{\left(u \right)}\, 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: 2cos⁡(3x)27\frac{2 \cos{\left(3 x \right)}}{27}

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

        ∫4xsin⁡(3x) dx=4∫xsin⁡(3x) dx\int 4 x \sin{\left(3 x \right)}\, dx = 4 \int x \sin{\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)=sin⁡(3x)\operatorname{dv}{\left(x \right)} = \sin{\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}:

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

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

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

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

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

          1. Let u=3xu = 3 x.

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

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

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

              ∫cos⁡(u) du=∫cos⁡(u) du3\int \cos{\left(u \right)}\, 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: −sin⁡(3x)9- \frac{\sin{\left(3 x \right)}}{9}

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

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

        ∫(−2sin⁡(3x)) dx=−2∫sin⁡(3x) dx\int \left(- 2 \sin{\left(3 x \right)}\right)\, dx = - 2 \int \sin{\left(3 x \right)}\, dx

        1. Let u=3xu = 3 x.

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

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

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

            ∫sin⁡(u) du=∫sin⁡(u) du3\int \sin{\left(u \right)}\, 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: 2cos⁡(3x)3\frac{2 \cos{\left(3 x \right)}}{3}

      The result is: −x2cos⁡(3x)3+2xsin⁡(3x)9−4xcos⁡(3x)3+4sin⁡(3x)9+20cos⁡(3x)27- \frac{x^{2} \cos{\left(3 x \right)}}{3} + \frac{2 x \sin{\left(3 x \right)}}{9} - \frac{4 x \cos{\left(3 x \right)}}{3} + \frac{4 \sin{\left(3 x \right)}}{9} + \frac{20 \cos{\left(3 x \right)}}{27}

    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−2u{\left(x \right)} = x^{2} + 4 x - 2 and let dv⁡(x)=sin⁡(3x)\operatorname{dv}{\left(x \right)} = \sin{\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}:

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

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

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

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

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

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

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

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

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

      1. Let u=3xu = 3 x.

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

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

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

          ∫sin⁡(u) du=∫sin⁡(u) du3\int \sin{\left(u \right)}\, 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: 2cos⁡(3x)27\frac{2 \cos{\left(3 x \right)}}{27}

    Method #3

    1. Rewrite the integrand:

      ((x2+4x)−2)sin⁡(3x)=x2sin⁡(3x)+4xsin⁡(3x)−2sin⁡(3x)\left(\left(x^{2} + 4 x\right) - 2\right) \sin{\left(3 x \right)} = x^{2} \sin{\left(3 x \right)} + 4 x \sin{\left(3 x \right)} - 2 \sin{\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)=sin⁡(3x)\operatorname{dv}{\left(x \right)} = \sin{\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}:

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

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

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

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

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

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

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

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

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

        1. Let u=3xu = 3 x.

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

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

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

            ∫sin⁡(u) du=∫sin⁡(u) du3\int \sin{\left(u \right)}\, 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: 2cos⁡(3x)27\frac{2 \cos{\left(3 x \right)}}{27}

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

        ∫4xsin⁡(3x) dx=4∫xsin⁡(3x) dx\int 4 x \sin{\left(3 x \right)}\, dx = 4 \int x \sin{\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)=sin⁡(3x)\operatorname{dv}{\left(x \right)} = \sin{\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}:

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

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

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

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

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

          1. Let u=3xu = 3 x.

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

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

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

              ∫cos⁡(u) du=∫cos⁡(u) du3\int \cos{\left(u \right)}\, 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: −sin⁡(3x)9- \frac{\sin{\left(3 x \right)}}{9}

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

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

        ∫(−2sin⁡(3x)) dx=−2∫sin⁡(3x) dx\int \left(- 2 \sin{\left(3 x \right)}\right)\, dx = - 2 \int \sin{\left(3 x \right)}\, dx

        1. Let u=3xu = 3 x.

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

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

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

            ∫sin⁡(u) du=∫sin⁡(u) du3\int \sin{\left(u \right)}\, 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: 2cos⁡(3x)3\frac{2 \cos{\left(3 x \right)}}{3}

      The result is: −x2cos⁡(3x)3+2xsin⁡(3x)9−4xcos⁡(3x)3+4sin⁡(3x)9+20cos⁡(3x)27- \frac{x^{2} \cos{\left(3 x \right)}}{3} + \frac{2 x \sin{\left(3 x \right)}}{9} - \frac{4 x \cos{\left(3 x \right)}}{3} + \frac{4 \sin{\left(3 x \right)}}{9} + \frac{20 \cos{\left(3 x \right)}}{27}

  2. Add the constant of integration:

    −x2cos⁡(3x)3+2xsin⁡(3x)9−4xcos⁡(3x)3+4sin⁡(3x)9+20cos⁡(3x)27+constant- \frac{x^{2} \cos{\left(3 x \right)}}{3} + \frac{2 x \sin{\left(3 x \right)}}{9} - \frac{4 x \cos{\left(3 x \right)}}{3} + \frac{4 \sin{\left(3 x \right)}}{9} + \frac{20 \cos{\left(3 x \right)}}{27}+ \mathrm{constant}

The answer is:

−x2cos⁡(3x)3+2xsin⁡(3x)9−4xcos⁡(3x)3+4sin⁡(3x)9+20cos⁡(3x)27+constant- \frac{x^{2} \cos{\left(3 x \right)}}{3} + \frac{2 x \sin{\left(3 x \right)}}{9} - \frac{4 x \cos{\left(3 x \right)}}{3} + \frac{4 \sin{\left(3 x \right)}}{9} + \frac{20 \cos{\left(3 x \right)}}{27}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                                                     
 |                                                                             2                        
 | / 2          \                   4*sin(3*x)   20*cos(3*x)   4*x*cos(3*x)   x *cos(3*x)   2*x*sin(3*x)
 | \x  + 4*x - 2/*sin(3*x) dx = C + ---------- + ----------- - ------------ - ----------- + ------------
 |                                      9             27            3              3             9      
/
∫((x2+4x)−2)sin⁡(3x) dx=C−x2cos⁡(3x)3+2xsin⁡(3x)9−4xcos⁡(3x)3+4sin⁡(3x)9+20cos⁡(3x)27\int \left(\left(x^{2} + 4 x\right) - 2\right) \sin{\left(3 x \right)}\, dx = C - \frac{x^{2} \cos{\left(3 x \right)}}{3} + \frac{2 x \sin{\left(3 x \right)}}{9} - \frac{4 x \cos{\left(3 x \right)}}{3} + \frac{4 \sin{\left(3 x \right)}}{9} + \frac{20 \cos{\left(3 x \right)}}{27}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.902-2
Plot the graph f(x)
The answer [src] 
  20   25*cos(3)   2*sin(3)
- -- - --------- + --------
  27       27         3
−2027+2sin⁡(3)3−25cos⁡(3)27- \frac{20}{27} + \frac{2 \sin{\left(3 \right)}}{3} - \frac{25 \cos{\left(3 \right)}}{27} 
=
= 
  20   25*cos(3)   2*sin(3)
- -- - --------- + --------
  27       27         3
−2027+2sin⁡(3)3−25cos⁡(3)27- \frac{20}{27} + \frac{2 \sin{\left(3 \right)}}{3} - \frac{25 \cos{\left(3 \right)}}{27} 
-20/27 - 25*cos(3)/27 + 2*sin(3)/3
Numerical answer [src] 
0.269998983706991
0.269998983706991
The graph
Integral of (x^2+4x-2)sin3xdx dx

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

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