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(4x-2)^2cos3x
Solving integrals/ (4x-2)^2cos3x

Integral of (4x-2)^2cos3x dx

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∫−10(4x−2)2cos⁡(3x) dx\int\limits_{-1}^{0} \left(4 x - 2\right)^{2} \cos{\left(3 x \right)}\, dx 
Integral((4*x - 1*2)^2*cos(3*x), (x, -1, 0))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      (4x−2)2cos⁡(3x)=16x2cos⁡(3x)−16xcos⁡(3x)+4cos⁡(3x)\left(4 x - 2\right)^{2} \cos{\left(3 x \right)} = 16 x^{2} \cos{\left(3 x \right)} - 16 x \cos{\left(3 x \right)} + 4 \cos{\left(3 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:

        ∫16x2cos⁡(3x) dx=16∫x2cos⁡(3x) dx\int 16 x^{2} \cos{\left(3 x \right)}\, dx = 16 \int x^{2} \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)=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}

        So, the result is: 16x2sin⁡(3x)3+32xcos⁡(3x)9−32sin⁡(3x)27\frac{16 x^{2} \sin{\left(3 x \right)}}{3} + \frac{32 x \cos{\left(3 x \right)}}{9} - \frac{32 \sin{\left(3 x \right)}}{27}

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

        ∫(−16xcos⁡(3x)) dx=−16∫xcos⁡(3x) dx\int \left(- 16 x \cos{\left(3 x \right)}\right)\, dx = - 16 \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: −16xsin⁡(3x)3−16cos⁡(3x)9- \frac{16 x \sin{\left(3 x \right)}}{3} - \frac{16 \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: 16x2sin⁡(3x)3−16xsin⁡(3x)3+32xcos⁡(3x)9+4sin⁡(3x)27−16cos⁡(3x)9\frac{16 x^{2} \sin{\left(3 x \right)}}{3} - \frac{16 x \sin{\left(3 x \right)}}{3} + \frac{32 x \cos{\left(3 x \right)}}{9} + \frac{4 \sin{\left(3 x \right)}}{27} - \frac{16 \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)=16x2−16x+4u{\left(x \right)} = 16 x^{2} - 16 x + 4 and let dv⁡(x)=cos⁡(3x)\operatorname{dv}{\left(x \right)} = \cos{\left(3 x \right)}.

      Then du⁡(x)=32x−16\operatorname{du}{\left(x \right)} = 32 x - 16.

      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)=32x3−163u{\left(x \right)} = \frac{32 x}{3} - \frac{16}{3} and let dv⁡(x)=sin⁡(3x)\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}.

      Then du⁡(x)=323\operatorname{du}{\left(x \right)} = \frac{32}{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:

      ∫(−32cos⁡(3x)9) dx=−32∫cos⁡(3x) dx9\int \left(- \frac{32 \cos{\left(3 x \right)}}{9}\right)\, dx = - \frac{32 \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: −32sin⁡(3x)27- \frac{32 \sin{\left(3 x \right)}}{27}

    Method #3

    1. Rewrite the integrand:

      (4x−2)2cos⁡(3x)=16x2cos⁡(3x)−16xcos⁡(3x)+4cos⁡(3x)\left(4 x - 2\right)^{2} \cos{\left(3 x \right)} = 16 x^{2} \cos{\left(3 x \right)} - 16 x \cos{\left(3 x \right)} + 4 \cos{\left(3 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:

        ∫16x2cos⁡(3x) dx=16∫x2cos⁡(3x) dx\int 16 x^{2} \cos{\left(3 x \right)}\, dx = 16 \int x^{2} \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)=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}

        So, the result is: 16x2sin⁡(3x)3+32xcos⁡(3x)9−32sin⁡(3x)27\frac{16 x^{2} \sin{\left(3 x \right)}}{3} + \frac{32 x \cos{\left(3 x \right)}}{9} - \frac{32 \sin{\left(3 x \right)}}{27}

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

        ∫(−16xcos⁡(3x)) dx=−16∫xcos⁡(3x) dx\int \left(- 16 x \cos{\left(3 x \right)}\right)\, dx = - 16 \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: −16xsin⁡(3x)3−16cos⁡(3x)9- \frac{16 x \sin{\left(3 x \right)}}{3} - \frac{16 \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: 16x2sin⁡(3x)3−16xsin⁡(3x)3+32xcos⁡(3x)9+4sin⁡(3x)27−16cos⁡(3x)9\frac{16 x^{2} \sin{\left(3 x \right)}}{3} - \frac{16 x \sin{\left(3 x \right)}}{3} + \frac{32 x \cos{\left(3 x \right)}}{9} + \frac{4 \sin{\left(3 x \right)}}{27} - \frac{16 \cos{\left(3 x \right)}}{9}

  2. Add the constant of integration:

    16x2sin⁡(3x)3−16xsin⁡(3x)3+32xcos⁡(3x)9+4sin⁡(3x)27−16cos⁡(3x)9+constant\frac{16 x^{2} \sin{\left(3 x \right)}}{3} - \frac{16 x \sin{\left(3 x \right)}}{3} + \frac{32 x \cos{\left(3 x \right)}}{9} + \frac{4 \sin{\left(3 x \right)}}{27} - \frac{16 \cos{\left(3 x \right)}}{9}+ \mathrm{constant}

The answer is:

16x2sin⁡(3x)3−16xsin⁡(3x)3+32xcos⁡(3x)9+4sin⁡(3x)27−16cos⁡(3x)9+constant\frac{16 x^{2} \sin{\left(3 x \right)}}{3} - \frac{16 x \sin{\left(3 x \right)}}{3} + \frac{32 x \cos{\left(3 x \right)}}{9} + \frac{4 \sin{\left(3 x \right)}}{27} - \frac{16 \cos{\left(3 x \right)}}{9}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                                                      
 |                                                                             2                         
 |          2                   16*cos(3*x)   4*sin(3*x)   16*x*sin(3*x)   16*x *sin(3*x)   32*x*cos(3*x)
 | (4*x - 2) *cos(3*x) dx = C - ----------- + ---------- - ------------- + -------------- + -------------
 |                                   9            27             3               3                9      
/
16 ((9 x2−2) sin⁡(3 x)+6 x cos⁡(3 x))9−16 (3 x sin⁡(3 x)+cos⁡(3 x))3+4 sin⁡(3 x)3{{{{16\,\left(\left(9\,x^2-2\right)\,\sin \left(3\,x\right)+6\,x\, \cos \left(3\,x\right)\right)}\over{9}}-{{16\,\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
-1.00-0.90-0.80-0.70-0.60-0.50-0.40-0.30-0.20-0.100.00-5050
Plot the graph f(x)
The answer [src] 
  16   16*cos(3)   292*sin(3)
- -- + --------- + ----------
  9        3           27
292 sin⁡3+144 cos⁡327−169{{292\,\sin 3+144\,\cos 3}\over{27}}-{{16}\over{9}} 
=
= 
  16   16*cos(3)   292*sin(3)
- -- + --------- + ----------
  9        3           27
16cos⁡(3)3−169+292sin⁡(3)27\frac{16 \cos{\left(3 \right)}}{3} - \frac{16}{9} + \frac{292 \sin{\left(3 \right)}}{27}
Simplify
Numerical answer [src] 
-5.53155100581418
-5.53155100581418
The graph
Integral of (4x-2)^2cos3x dx

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

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