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(-2x^2-12x+6)cos6x
Solving integrals/ (-2x^2-12x+6)cos6x

Integral of (-2x^2-12x+6)cos6x dx

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∫01(−2x2−12x+6)cos⁡(6x) dx\int\limits_{0}^{1} \left(- 2 x^{2} - 12 x + 6\right) \cos{\left(6 x \right)}\, dx 
Integral((-2*x^2 - 12*x + 6)*cos(6*x), (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      (−2x2−12x+6)cos⁡(6x)=−2x2cos⁡(6x)−12xcos⁡(6x)+6cos⁡(6x)\left(- 2 x^{2} - 12 x + 6\right) \cos{\left(6 x \right)} = - 2 x^{2} \cos{\left(6 x \right)} - 12 x \cos{\left(6 x \right)} + 6 \cos{\left(6 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:

        ∫(−2x2cos⁡(6x)) dx=−2∫x2cos⁡(6x) dx\int \left(- 2 x^{2} \cos{\left(6 x \right)}\right)\, dx = - 2 \int x^{2} \cos{\left(6 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⁡(6x)\operatorname{dv}{\left(x \right)} = \cos{\left(6 x \right)}.

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

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

          1. Let u=6xu = 6 x.

            Then let du=6dxdu = 6 dx and substitute du6\frac{du}{6}:

            ∫cos⁡(u)36 du\int \frac{\cos{\left(u \right)}}{36}\, du

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

              ∫cos⁡(u)6 du=∫cos⁡(u) du6\int \frac{\cos{\left(u \right)}}{6}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}

              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)6\frac{\sin{\left(u \right)}}{6}

            Now substitute uu back in:

            sin⁡(6x)6\frac{\sin{\left(6 x \right)}}{6}

          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)=x3u{\left(x \right)} = \frac{x}{3} and let dv⁡(x)=sin⁡(6x)\operatorname{dv}{\left(x \right)} = \sin{\left(6 x \right)}.

          Then du⁡(x)=13\operatorname{du}{\left(x \right)} = \frac{1}{3}.

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

          1. Let u=6xu = 6 x.

            Then let du=6dxdu = 6 dx and substitute du6\frac{du}{6}:

            ∫sin⁡(u)36 du\int \frac{\sin{\left(u \right)}}{36}\, du

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

              ∫sin⁡(u)6 du=∫sin⁡(u) du6\int \frac{\sin{\left(u \right)}}{6}\, du = \frac{\int \sin{\left(u \right)}\, du}{6}

              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)6- \frac{\cos{\left(u \right)}}{6}

            Now substitute uu back in:

            −cos⁡(6x)6- \frac{\cos{\left(6 x \right)}}{6}

          Now evaluate the sub-integral.

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

          ∫(−cos⁡(6x)18) dx=−∫cos⁡(6x) dx18\int \left(- \frac{\cos{\left(6 x \right)}}{18}\right)\, dx = - \frac{\int \cos{\left(6 x \right)}\, dx}{18}

          1. Let u=6xu = 6 x.

            Then let du=6dxdu = 6 dx and substitute du6\frac{du}{6}:

            ∫cos⁡(u)36 du\int \frac{\cos{\left(u \right)}}{36}\, du

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

              ∫cos⁡(u)6 du=∫cos⁡(u) du6\int \frac{\cos{\left(u \right)}}{6}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}

              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)6\frac{\sin{\left(u \right)}}{6}

            Now substitute uu back in:

            sin⁡(6x)6\frac{\sin{\left(6 x \right)}}{6}

          So, the result is: −sin⁡(6x)108- \frac{\sin{\left(6 x \right)}}{108}

        So, the result is: −x2sin⁡(6x)3−xcos⁡(6x)9+sin⁡(6x)54- \frac{x^{2} \sin{\left(6 x \right)}}{3} - \frac{x \cos{\left(6 x \right)}}{9} + \frac{\sin{\left(6 x \right)}}{54}

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

        ∫(−12xcos⁡(6x)) dx=−12∫xcos⁡(6x) dx\int \left(- 12 x \cos{\left(6 x \right)}\right)\, dx = - 12 \int x \cos{\left(6 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⁡(6x)\operatorname{dv}{\left(x \right)} = \cos{\left(6 x \right)}.

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

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

          1. Let u=6xu = 6 x.

            Then let du=6dxdu = 6 dx and substitute du6\frac{du}{6}:

            ∫cos⁡(u)36 du\int \frac{\cos{\left(u \right)}}{36}\, du

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

              ∫cos⁡(u)6 du=∫cos⁡(u) du6\int \frac{\cos{\left(u \right)}}{6}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}

              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)6\frac{\sin{\left(u \right)}}{6}

            Now substitute uu back in:

            sin⁡(6x)6\frac{\sin{\left(6 x \right)}}{6}

          Now evaluate the sub-integral.

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

          ∫sin⁡(6x)6 dx=∫sin⁡(6x) dx6\int \frac{\sin{\left(6 x \right)}}{6}\, dx = \frac{\int \sin{\left(6 x \right)}\, dx}{6}

          1. Let u=6xu = 6 x.

            Then let du=6dxdu = 6 dx and substitute du6\frac{du}{6}:

            ∫sin⁡(u)36 du\int \frac{\sin{\left(u \right)}}{36}\, du

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

              ∫sin⁡(u)6 du=∫sin⁡(u) du6\int \frac{\sin{\left(u \right)}}{6}\, du = \frac{\int \sin{\left(u \right)}\, du}{6}

              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)6- \frac{\cos{\left(u \right)}}{6}

            Now substitute uu back in:

            −cos⁡(6x)6- \frac{\cos{\left(6 x \right)}}{6}

          So, the result is: −cos⁡(6x)36- \frac{\cos{\left(6 x \right)}}{36}

        So, the result is: −2xsin⁡(6x)−cos⁡(6x)3- 2 x \sin{\left(6 x \right)} - \frac{\cos{\left(6 x \right)}}{3}

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

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

        1. Let u=6xu = 6 x.

          Then let du=6dxdu = 6 dx and substitute du6\frac{du}{6}:

          ∫cos⁡(u)36 du\int \frac{\cos{\left(u \right)}}{36}\, du

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

            ∫cos⁡(u)6 du=∫cos⁡(u) du6\int \frac{\cos{\left(u \right)}}{6}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}

            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)6\frac{\sin{\left(u \right)}}{6}

          Now substitute uu back in:

          sin⁡(6x)6\frac{\sin{\left(6 x \right)}}{6}

        So, the result is: sin⁡(6x)\sin{\left(6 x \right)}

      The result is: −x2sin⁡(6x)3−2xsin⁡(6x)−xcos⁡(6x)9+55sin⁡(6x)54−cos⁡(6x)3- \frac{x^{2} \sin{\left(6 x \right)}}{3} - 2 x \sin{\left(6 x \right)} - \frac{x \cos{\left(6 x \right)}}{9} + \frac{55 \sin{\left(6 x \right)}}{54} - \frac{\cos{\left(6 x \right)}}{3}

    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)=−2x2−12x+6u{\left(x \right)} = - 2 x^{2} - 12 x + 6 and let dv⁡(x)=cos⁡(6x)\operatorname{dv}{\left(x \right)} = \cos{\left(6 x \right)}.

      Then du⁡(x)=−4x−12\operatorname{du}{\left(x \right)} = - 4 x - 12.

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

      1. Let u=6xu = 6 x.

        Then let du=6dxdu = 6 dx and substitute du6\frac{du}{6}:

        ∫cos⁡(u)36 du\int \frac{\cos{\left(u \right)}}{36}\, du

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

          ∫cos⁡(u)6 du=∫cos⁡(u) du6\int \frac{\cos{\left(u \right)}}{6}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}

          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)6\frac{\sin{\left(u \right)}}{6}

        Now substitute uu back in:

        sin⁡(6x)6\frac{\sin{\left(6 x \right)}}{6}

      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)=−2x3−2u{\left(x \right)} = - \frac{2 x}{3} - 2 and let dv⁡(x)=sin⁡(6x)\operatorname{dv}{\left(x \right)} = \sin{\left(6 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=6xu = 6 x.

        Then let du=6dxdu = 6 dx and substitute du6\frac{du}{6}:

        ∫sin⁡(u)36 du\int \frac{\sin{\left(u \right)}}{36}\, du

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

          ∫sin⁡(u)6 du=∫sin⁡(u) du6\int \frac{\sin{\left(u \right)}}{6}\, du = \frac{\int \sin{\left(u \right)}\, du}{6}

          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)6- \frac{\cos{\left(u \right)}}{6}

        Now substitute uu back in:

        −cos⁡(6x)6- \frac{\cos{\left(6 x \right)}}{6}

      Now evaluate the sub-integral.

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

      ∫cos⁡(6x)9 dx=∫cos⁡(6x) dx9\int \frac{\cos{\left(6 x \right)}}{9}\, dx = \frac{\int \cos{\left(6 x \right)}\, dx}{9}

      1. Let u=6xu = 6 x.

        Then let du=6dxdu = 6 dx and substitute du6\frac{du}{6}:

        ∫cos⁡(u)36 du\int \frac{\cos{\left(u \right)}}{36}\, du

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

          ∫cos⁡(u)6 du=∫cos⁡(u) du6\int \frac{\cos{\left(u \right)}}{6}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}

          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)6\frac{\sin{\left(u \right)}}{6}

        Now substitute uu back in:

        sin⁡(6x)6\frac{\sin{\left(6 x \right)}}{6}

      So, the result is: sin⁡(6x)54\frac{\sin{\left(6 x \right)}}{54}

    Method #3

    1. Rewrite the integrand:

      (−2x2−12x+6)cos⁡(6x)=−2x2cos⁡(6x)−12xcos⁡(6x)+6cos⁡(6x)\left(- 2 x^{2} - 12 x + 6\right) \cos{\left(6 x \right)} = - 2 x^{2} \cos{\left(6 x \right)} - 12 x \cos{\left(6 x \right)} + 6 \cos{\left(6 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:

        ∫(−2x2cos⁡(6x)) dx=−2∫x2cos⁡(6x) dx\int \left(- 2 x^{2} \cos{\left(6 x \right)}\right)\, dx = - 2 \int x^{2} \cos{\left(6 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⁡(6x)\operatorname{dv}{\left(x \right)} = \cos{\left(6 x \right)}.

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

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

          1. Let u=6xu = 6 x.

            Then let du=6dxdu = 6 dx and substitute du6\frac{du}{6}:

            ∫cos⁡(u)36 du\int \frac{\cos{\left(u \right)}}{36}\, du

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

              ∫cos⁡(u)6 du=∫cos⁡(u) du6\int \frac{\cos{\left(u \right)}}{6}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}

              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)6\frac{\sin{\left(u \right)}}{6}

            Now substitute uu back in:

            sin⁡(6x)6\frac{\sin{\left(6 x \right)}}{6}

          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)=x3u{\left(x \right)} = \frac{x}{3} and let dv⁡(x)=sin⁡(6x)\operatorname{dv}{\left(x \right)} = \sin{\left(6 x \right)}.

          Then du⁡(x)=13\operatorname{du}{\left(x \right)} = \frac{1}{3}.

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

          1. Let u=6xu = 6 x.

            Then let du=6dxdu = 6 dx and substitute du6\frac{du}{6}:

            ∫sin⁡(u)36 du\int \frac{\sin{\left(u \right)}}{36}\, du

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

              ∫sin⁡(u)6 du=∫sin⁡(u) du6\int \frac{\sin{\left(u \right)}}{6}\, du = \frac{\int \sin{\left(u \right)}\, du}{6}

              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)6- \frac{\cos{\left(u \right)}}{6}

            Now substitute uu back in:

            −cos⁡(6x)6- \frac{\cos{\left(6 x \right)}}{6}

          Now evaluate the sub-integral.

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

          ∫(−cos⁡(6x)18) dx=−∫cos⁡(6x) dx18\int \left(- \frac{\cos{\left(6 x \right)}}{18}\right)\, dx = - \frac{\int \cos{\left(6 x \right)}\, dx}{18}

          1. Let u=6xu = 6 x.

            Then let du=6dxdu = 6 dx and substitute du6\frac{du}{6}:

            ∫cos⁡(u)36 du\int \frac{\cos{\left(u \right)}}{36}\, du

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

              ∫cos⁡(u)6 du=∫cos⁡(u) du6\int \frac{\cos{\left(u \right)}}{6}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}

              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)6\frac{\sin{\left(u \right)}}{6}

            Now substitute uu back in:

            sin⁡(6x)6\frac{\sin{\left(6 x \right)}}{6}

          So, the result is: −sin⁡(6x)108- \frac{\sin{\left(6 x \right)}}{108}

        So, the result is: −x2sin⁡(6x)3−xcos⁡(6x)9+sin⁡(6x)54- \frac{x^{2} \sin{\left(6 x \right)}}{3} - \frac{x \cos{\left(6 x \right)}}{9} + \frac{\sin{\left(6 x \right)}}{54}

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

        ∫(−12xcos⁡(6x)) dx=−12∫xcos⁡(6x) dx\int \left(- 12 x \cos{\left(6 x \right)}\right)\, dx = - 12 \int x \cos{\left(6 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⁡(6x)\operatorname{dv}{\left(x \right)} = \cos{\left(6 x \right)}.

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

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

          1. Let u=6xu = 6 x.

            Then let du=6dxdu = 6 dx and substitute du6\frac{du}{6}:

            ∫cos⁡(u)36 du\int \frac{\cos{\left(u \right)}}{36}\, du

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

              ∫cos⁡(u)6 du=∫cos⁡(u) du6\int \frac{\cos{\left(u \right)}}{6}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}

              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)6\frac{\sin{\left(u \right)}}{6}

            Now substitute uu back in:

            sin⁡(6x)6\frac{\sin{\left(6 x \right)}}{6}

          Now evaluate the sub-integral.

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

          ∫sin⁡(6x)6 dx=∫sin⁡(6x) dx6\int \frac{\sin{\left(6 x \right)}}{6}\, dx = \frac{\int \sin{\left(6 x \right)}\, dx}{6}

          1. Let u=6xu = 6 x.

            Then let du=6dxdu = 6 dx and substitute du6\frac{du}{6}:

            ∫sin⁡(u)36 du\int \frac{\sin{\left(u \right)}}{36}\, du

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

              ∫sin⁡(u)6 du=∫sin⁡(u) du6\int \frac{\sin{\left(u \right)}}{6}\, du = \frac{\int \sin{\left(u \right)}\, du}{6}

              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)6- \frac{\cos{\left(u \right)}}{6}

            Now substitute uu back in:

            −cos⁡(6x)6- \frac{\cos{\left(6 x \right)}}{6}

          So, the result is: −cos⁡(6x)36- \frac{\cos{\left(6 x \right)}}{36}

        So, the result is: −2xsin⁡(6x)−cos⁡(6x)3- 2 x \sin{\left(6 x \right)} - \frac{\cos{\left(6 x \right)}}{3}

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

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

        1. Let u=6xu = 6 x.

          Then let du=6dxdu = 6 dx and substitute du6\frac{du}{6}:

          ∫cos⁡(u)36 du\int \frac{\cos{\left(u \right)}}{36}\, du

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

            ∫cos⁡(u)6 du=∫cos⁡(u) du6\int \frac{\cos{\left(u \right)}}{6}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}

            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)6\frac{\sin{\left(u \right)}}{6}

          Now substitute uu back in:

          sin⁡(6x)6\frac{\sin{\left(6 x \right)}}{6}

        So, the result is: sin⁡(6x)\sin{\left(6 x \right)}

      The result is: −x2sin⁡(6x)3−2xsin⁡(6x)−xcos⁡(6x)9+55sin⁡(6x)54−cos⁡(6x)3- \frac{x^{2} \sin{\left(6 x \right)}}{3} - 2 x \sin{\left(6 x \right)} - \frac{x \cos{\left(6 x \right)}}{9} + \frac{55 \sin{\left(6 x \right)}}{54} - \frac{\cos{\left(6 x \right)}}{3}

  2. Add the constant of integration:

    −x2sin⁡(6x)3−2xsin⁡(6x)−xcos⁡(6x)9+55sin⁡(6x)54−cos⁡(6x)3+constant- \frac{x^{2} \sin{\left(6 x \right)}}{3} - 2 x \sin{\left(6 x \right)} - \frac{x \cos{\left(6 x \right)}}{9} + \frac{55 \sin{\left(6 x \right)}}{54} - \frac{\cos{\left(6 x \right)}}{3}+ \mathrm{constant}

The answer is:

−x2sin⁡(6x)3−2xsin⁡(6x)−xcos⁡(6x)9+55sin⁡(6x)54−cos⁡(6x)3+constant- \frac{x^{2} \sin{\left(6 x \right)}}{3} - 2 x \sin{\left(6 x \right)} - \frac{x \cos{\left(6 x \right)}}{9} + \frac{55 \sin{\left(6 x \right)}}{54} - \frac{\cos{\left(6 x \right)}}{3}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                                                      
 |                                                                                2                      
 | /     2           \                   cos(6*x)   55*sin(6*x)                  x *sin(6*x)   x*cos(6*x)
 | \- 2*x  - 12*x + 6/*cos(6*x) dx = C - -------- + ----------- - 2*x*sin(6*x) - ----------- - ----------
 |                                          3            54                           3            9     
/
−(36 x2−2) sin⁡(6 x)+12 x cos⁡(6 x)18−2 (6 x sin⁡(6 x)+cos⁡(6 x))+6 sin⁡(6 x)6{{-{{\left(36\,x^2-2\right)\,\sin \left(6\,x\right)+12\,x\,\cos \left(6\,x\right)}\over{18}}-2\,\left(6\,x\,\sin \left(6\,x\right)+ \cos \left(6\,x\right)\right)+6\,\sin \left(6\,x\right)}\over{6}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.90-1010
Plot the graph f(x)
The answer [src] 
1   71*sin(6)   4*cos(6)
- - --------- - --------
3       54         9
13−71 sin⁡6+24 cos⁡654{{1}\over{3}}-{{71\,\sin 6+24\,\cos 6}\over{54}} 
=
= 
1   71*sin(6)   4*cos(6)
- - --------- - --------
3       54         9
−4cos⁡(6)9+13−71sin⁡(6)54- \frac{4 \cos{\left(6 \right)}}{9} + \frac{1}{3} - \frac{71 \sin{\left(6 \right)}}{54}
Упростить
Numerical answer [src] 
0.273970620231758
0.273970620231758
The graph
Integral of (-2x^2-12x+6)cos6x dx

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

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