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(x^2+5x+6)cos2x
Solving integrals/ (x^2+5x+6)cos2x

Integral of (x^2+5x+6)cos2x dx

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∫20((x2+5x)+6)cos⁡(2x) dx\int\limits_{2}^{0} \left(\left(x^{2} + 5 x\right) + 6\right) \cos{\left(2 x \right)}\, dx 
Integral((x^2 + 5*x + 6)*cos(2*x), (x, 2, 0))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      ((x2+5x)+6)cos⁡(2x)=x2cos⁡(2x)+5xcos⁡(2x)+6cos⁡(2x)\left(\left(x^{2} + 5 x\right) + 6\right) \cos{\left(2 x \right)} = x^{2} \cos{\left(2 x \right)} + 5 x \cos{\left(2 x \right)} + 6 \cos{\left(2 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⁡(2x)\operatorname{dv}{\left(x \right)} = \cos{\left(2 x \right)}.

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

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

        1. Let u=2xu = 2 x.

          Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

          ∫cos⁡(u)2 du\int \frac{\cos{\left(u \right)}}{2}\, du

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

            ∫cos⁡(u) du=∫cos⁡(u) du2\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}

            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)2\frac{\sin{\left(u \right)}}{2}

          Now substitute uu back in:

          sin⁡(2x)2\frac{\sin{\left(2 x \right)}}{2}

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

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

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

        1. There are multiple ways to do this integral.

          Method #1

          1. Let u=2xu = 2 x.

            Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

            ∫sin⁡(u)2 du\int \frac{\sin{\left(u \right)}}{2}\, du

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

              ∫sin⁡(u) du=∫sin⁡(u) du2\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}

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

            Now substitute uu back in:

            −cos⁡(2x)2- \frac{\cos{\left(2 x \right)}}{2}

          Method #2

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

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

            1. Let u=cos⁡(x)u = \cos{\left(x \right)}.

              Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute −du- du:

              ∫(−u) du\int \left(- u\right)\, du

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

                ∫u du=−∫u du\int u\, du = - \int u\, du

                1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

                  ∫u du=u22\int u\, du = \frac{u^{2}}{2}

                So, the result is: −u22- \frac{u^{2}}{2}

              Now substitute uu back in:

              −cos⁡2(x)2- \frac{\cos^{2}{\left(x \right)}}{2}

            So, the result is: −cos⁡2(x)- \cos^{2}{\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:

        ∫(−cos⁡(2x)2) dx=−∫cos⁡(2x) dx2\int \left(- \frac{\cos{\left(2 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{2}

        1. Let u=2xu = 2 x.

          Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

          ∫cos⁡(u)2 du\int \frac{\cos{\left(u \right)}}{2}\, du

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

            ∫cos⁡(u) du=∫cos⁡(u) du2\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}

            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)2\frac{\sin{\left(u \right)}}{2}

          Now substitute uu back in:

          sin⁡(2x)2\frac{\sin{\left(2 x \right)}}{2}

        So, the result is: −sin⁡(2x)4- \frac{\sin{\left(2 x \right)}}{4}

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

        ∫5xcos⁡(2x) dx=5∫xcos⁡(2x) dx\int 5 x \cos{\left(2 x \right)}\, dx = 5 \int x \cos{\left(2 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⁡(2x)\operatorname{dv}{\left(x \right)} = \cos{\left(2 x \right)}.

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

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

          1. Let u=2xu = 2 x.

            Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

            ∫cos⁡(u)2 du\int \frac{\cos{\left(u \right)}}{2}\, du

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

              ∫cos⁡(u) du=∫cos⁡(u) du2\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}

              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)2\frac{\sin{\left(u \right)}}{2}

            Now substitute uu back in:

            sin⁡(2x)2\frac{\sin{\left(2 x \right)}}{2}

          Now evaluate the sub-integral.

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

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

          1. Let u=2xu = 2 x.

            Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

            ∫sin⁡(u)2 du\int \frac{\sin{\left(u \right)}}{2}\, du

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

              ∫sin⁡(u) du=∫sin⁡(u) du2\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}

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

            Now substitute uu back in:

            −cos⁡(2x)2- \frac{\cos{\left(2 x \right)}}{2}

          So, the result is: −cos⁡(2x)4- \frac{\cos{\left(2 x \right)}}{4}

        So, the result is: 5xsin⁡(2x)2+5cos⁡(2x)4\frac{5 x \sin{\left(2 x \right)}}{2} + \frac{5 \cos{\left(2 x \right)}}{4}

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

        ∫6cos⁡(2x) dx=6∫cos⁡(2x) dx\int 6 \cos{\left(2 x \right)}\, dx = 6 \int \cos{\left(2 x \right)}\, dx

        1. Let u=2xu = 2 x.

          Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

          ∫cos⁡(u)2 du\int \frac{\cos{\left(u \right)}}{2}\, du

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

            ∫cos⁡(u) du=∫cos⁡(u) du2\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}

            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)2\frac{\sin{\left(u \right)}}{2}

          Now substitute uu back in:

          sin⁡(2x)2\frac{\sin{\left(2 x \right)}}{2}

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

      The result is: x2sin⁡(2x)2+5xsin⁡(2x)2+xcos⁡(2x)2+11sin⁡(2x)4+5cos⁡(2x)4\frac{x^{2} \sin{\left(2 x \right)}}{2} + \frac{5 x \sin{\left(2 x \right)}}{2} + \frac{x \cos{\left(2 x \right)}}{2} + \frac{11 \sin{\left(2 x \right)}}{4} + \frac{5 \cos{\left(2 x \right)}}{4}

    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+5x+6u{\left(x \right)} = x^{2} + 5 x + 6 and let dv⁡(x)=cos⁡(2x)\operatorname{dv}{\left(x \right)} = \cos{\left(2 x \right)}.

      Then du⁡(x)=2x+5\operatorname{du}{\left(x \right)} = 2 x + 5.

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

      1. Let u=2xu = 2 x.

        Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

        ∫cos⁡(u)2 du\int \frac{\cos{\left(u \right)}}{2}\, du

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

          ∫cos⁡(u) du=∫cos⁡(u) du2\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}

          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)2\frac{\sin{\left(u \right)}}{2}

        Now substitute uu back in:

        sin⁡(2x)2\frac{\sin{\left(2 x \right)}}{2}

      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)=x+52u{\left(x \right)} = x + \frac{5}{2} and let dv⁡(x)=sin⁡(2x)\operatorname{dv}{\left(x \right)} = \sin{\left(2 x \right)}.

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

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

      1. Let u=2xu = 2 x.

        Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

        ∫sin⁡(u)2 du\int \frac{\sin{\left(u \right)}}{2}\, du

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

          ∫sin⁡(u) du=∫sin⁡(u) du2\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}

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

        Now substitute uu back in:

        −cos⁡(2x)2- \frac{\cos{\left(2 x \right)}}{2}

      Now evaluate the sub-integral.

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

      ∫(−cos⁡(2x)2) dx=−∫cos⁡(2x) dx2\int \left(- \frac{\cos{\left(2 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{2}

      1. Let u=2xu = 2 x.

        Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

        ∫cos⁡(u)2 du\int \frac{\cos{\left(u \right)}}{2}\, du

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

          ∫cos⁡(u) du=∫cos⁡(u) du2\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}

          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)2\frac{\sin{\left(u \right)}}{2}

        Now substitute uu back in:

        sin⁡(2x)2\frac{\sin{\left(2 x \right)}}{2}

      So, the result is: −sin⁡(2x)4- \frac{\sin{\left(2 x \right)}}{4}

    Method #3

    1. Rewrite the integrand:

      ((x2+5x)+6)cos⁡(2x)=x2cos⁡(2x)+5xcos⁡(2x)+6cos⁡(2x)\left(\left(x^{2} + 5 x\right) + 6\right) \cos{\left(2 x \right)} = x^{2} \cos{\left(2 x \right)} + 5 x \cos{\left(2 x \right)} + 6 \cos{\left(2 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⁡(2x)\operatorname{dv}{\left(x \right)} = \cos{\left(2 x \right)}.

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

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

        1. Let u=2xu = 2 x.

          Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

          ∫cos⁡(u)2 du\int \frac{\cos{\left(u \right)}}{2}\, du

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

            ∫cos⁡(u) du=∫cos⁡(u) du2\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}

            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)2\frac{\sin{\left(u \right)}}{2}

          Now substitute uu back in:

          sin⁡(2x)2\frac{\sin{\left(2 x \right)}}{2}

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

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

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

        1. Let u=2xu = 2 x.

          Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

          ∫sin⁡(u)2 du\int \frac{\sin{\left(u \right)}}{2}\, du

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

            ∫sin⁡(u) du=∫sin⁡(u) du2\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}

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

          Now substitute uu back in:

          −cos⁡(2x)2- \frac{\cos{\left(2 x \right)}}{2}

        Now evaluate the sub-integral.

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

        ∫(−cos⁡(2x)2) dx=−∫cos⁡(2x) dx2\int \left(- \frac{\cos{\left(2 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{2}

        1. Let u=2xu = 2 x.

          Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

          ∫cos⁡(u)2 du\int \frac{\cos{\left(u \right)}}{2}\, du

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

            ∫cos⁡(u) du=∫cos⁡(u) du2\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}

            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)2\frac{\sin{\left(u \right)}}{2}

          Now substitute uu back in:

          sin⁡(2x)2\frac{\sin{\left(2 x \right)}}{2}

        So, the result is: −sin⁡(2x)4- \frac{\sin{\left(2 x \right)}}{4}

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

        ∫5xcos⁡(2x) dx=5∫xcos⁡(2x) dx\int 5 x \cos{\left(2 x \right)}\, dx = 5 \int x \cos{\left(2 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⁡(2x)\operatorname{dv}{\left(x \right)} = \cos{\left(2 x \right)}.

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

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

          1. Let u=2xu = 2 x.

            Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

            ∫cos⁡(u)2 du\int \frac{\cos{\left(u \right)}}{2}\, du

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

              ∫cos⁡(u) du=∫cos⁡(u) du2\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}

              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)2\frac{\sin{\left(u \right)}}{2}

            Now substitute uu back in:

            sin⁡(2x)2\frac{\sin{\left(2 x \right)}}{2}

          Now evaluate the sub-integral.

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

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

          1. Let u=2xu = 2 x.

            Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

            ∫sin⁡(u)2 du\int \frac{\sin{\left(u \right)}}{2}\, du

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

              ∫sin⁡(u) du=∫sin⁡(u) du2\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}

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

            Now substitute uu back in:

            −cos⁡(2x)2- \frac{\cos{\left(2 x \right)}}{2}

          So, the result is: −cos⁡(2x)4- \frac{\cos{\left(2 x \right)}}{4}

        So, the result is: 5xsin⁡(2x)2+5cos⁡(2x)4\frac{5 x \sin{\left(2 x \right)}}{2} + \frac{5 \cos{\left(2 x \right)}}{4}

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

        ∫6cos⁡(2x) dx=6∫cos⁡(2x) dx\int 6 \cos{\left(2 x \right)}\, dx = 6 \int \cos{\left(2 x \right)}\, dx

        1. Let u=2xu = 2 x.

          Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

          ∫cos⁡(u)2 du\int \frac{\cos{\left(u \right)}}{2}\, du

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

            ∫cos⁡(u) du=∫cos⁡(u) du2\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}

            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)2\frac{\sin{\left(u \right)}}{2}

          Now substitute uu back in:

          sin⁡(2x)2\frac{\sin{\left(2 x \right)}}{2}

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

      The result is: x2sin⁡(2x)2+5xsin⁡(2x)2+xcos⁡(2x)2+11sin⁡(2x)4+5cos⁡(2x)4\frac{x^{2} \sin{\left(2 x \right)}}{2} + \frac{5 x \sin{\left(2 x \right)}}{2} + \frac{x \cos{\left(2 x \right)}}{2} + \frac{11 \sin{\left(2 x \right)}}{4} + \frac{5 \cos{\left(2 x \right)}}{4}

  2. Add the constant of integration:

    x2sin⁡(2x)2+5xsin⁡(2x)2+xcos⁡(2x)2+11sin⁡(2x)4+5cos⁡(2x)4+constant\frac{x^{2} \sin{\left(2 x \right)}}{2} + \frac{5 x \sin{\left(2 x \right)}}{2} + \frac{x \cos{\left(2 x \right)}}{2} + \frac{11 \sin{\left(2 x \right)}}{4} + \frac{5 \cos{\left(2 x \right)}}{4}+ \mathrm{constant}

The answer is:

x2sin⁡(2x)2+5xsin⁡(2x)2+xcos⁡(2x)2+11sin⁡(2x)4+5cos⁡(2x)4+constant\frac{x^{2} \sin{\left(2 x \right)}}{2} + \frac{5 x \sin{\left(2 x \right)}}{2} + \frac{x \cos{\left(2 x \right)}}{2} + \frac{11 \sin{\left(2 x \right)}}{4} + \frac{5 \cos{\left(2 x \right)}}{4}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                                                   
 |                                                                           2                        
 | / 2          \                   5*cos(2*x)   11*sin(2*x)   x*cos(2*x)   x *sin(2*x)   5*x*sin(2*x)
 | \x  + 5*x + 6/*cos(2*x) dx = C + ---------- + ----------- + ---------- + ----------- + ------------
 |                                      4             4            2             2             2      
/
∫((x2+5x)+6)cos⁡(2x) dx=C+x2sin⁡(2x)2+5xsin⁡(2x)2+xcos⁡(2x)2+11sin⁡(2x)4+5cos⁡(2x)4\int \left(\left(x^{2} + 5 x\right) + 6\right) \cos{\left(2 x \right)}\, dx = C + \frac{x^{2} \sin{\left(2 x \right)}}{2} + \frac{5 x \sin{\left(2 x \right)}}{2} + \frac{x \cos{\left(2 x \right)}}{2} + \frac{11 \sin{\left(2 x \right)}}{4} + \frac{5 \cos{\left(2 x \right)}}{4}
Simplify
The graph
0.02.00.20.40.60.81.01.21.41.61.8-2525
Plot the graph f(x)
The answer [src] 
5   39*sin(4)   9*cos(4)
- - --------- - --------
4       4          4
54−9cos⁡(4)4−39sin⁡(4)4\frac{5}{4} - \frac{9 \cos{\left(4 \right)}}{4} - \frac{39 \sin{\left(4 \right)}}{4} 
=
= 
5   39*sin(4)   9*cos(4)
- - --------- - --------
4       4          4
54−9cos⁡(4)4−39sin⁡(4)4\frac{5}{4} - \frac{9 \cos{\left(4 \right)}}{4} - \frac{39 \sin{\left(4 \right)}}{4} 
5/4 - 39*sin(4)/4 - 9*cos(4)/4
Numerical answer [src] 
10.0995224761954
10.0995224761954
The graph
Integral of (x^2+5x+6)cos2x dx

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

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