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(x^2+3x)sin(2x)
Solving integrals/ (x^2+3x)sin(2x)

Integral of (x^2+3x)sin(2x) dx

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

  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

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

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

        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)4 du\int \frac{\sin{\left(u \right)}}{4}\, du

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

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

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

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

      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)=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)4 du\int \frac{\cos{\left(u \right)}}{4}\, du

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

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

      3. 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 \left(- \frac{\sin{\left(2 x \right)}}{2}\right)\, 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)4 du\int \frac{\sin{\left(u \right)}}{4}\, du

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

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

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

        ∫3xsin⁡(2x) dx=3∫xsin⁡(2x) dx\int 3 x \sin{\left(2 x \right)}\, dx = 3 \int x \sin{\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)=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)4 du\int \frac{\sin{\left(u \right)}}{4}\, du

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

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

        2. 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)4 du\int \frac{\cos{\left(u \right)}}{4}\, du

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

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

        So, the result is: −3xcos⁡(2x)2+3sin⁡(2x)4- \frac{3 x \cos{\left(2 x \right)}}{2} + \frac{3 \sin{\left(2 x \right)}}{4}

      The result is: −x2cos⁡(2x)2+xsin⁡(2x)2−3xcos⁡(2x)2+3sin⁡(2x)4+cos⁡(2x)4- \frac{x^{2} \cos{\left(2 x \right)}}{2} + \frac{x \sin{\left(2 x \right)}}{2} - \frac{3 x \cos{\left(2 x \right)}}{2} + \frac{3 \sin{\left(2 x \right)}}{4} + \frac{\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)=x(x+3)u{\left(x \right)} = x \left(x + 3\right) and let dv⁡(x)=sin⁡(2x)\operatorname{dv}{\left(x \right)} = \sin{\left(2 x \right)}.

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

      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)4 du\int \frac{\sin{\left(u \right)}}{4}\, du

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

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

    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−32u{\left(x \right)} = - x - \frac{3}{2} 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)4 du\int \frac{\cos{\left(u \right)}}{4}\, du

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

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

    3. 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 \left(- \frac{\sin{\left(2 x \right)}}{2}\right)\, 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)4 du\int \frac{\sin{\left(u \right)}}{4}\, du

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

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

    Method #3

    1. Rewrite the integrand:

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

          ∫sin⁡(u)4 du\int \frac{\sin{\left(u \right)}}{4}\, du

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

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

      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)=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)4 du\int \frac{\cos{\left(u \right)}}{4}\, du

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

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

      3. 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 \left(- \frac{\sin{\left(2 x \right)}}{2}\right)\, 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)4 du\int \frac{\sin{\left(u \right)}}{4}\, du

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

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

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

        ∫3xsin⁡(2x) dx=3∫xsin⁡(2x) dx\int 3 x \sin{\left(2 x \right)}\, dx = 3 \int x \sin{\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)=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)4 du\int \frac{\sin{\left(u \right)}}{4}\, du

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

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

        2. 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)4 du\int \frac{\cos{\left(u \right)}}{4}\, du

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

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

        So, the result is: −3xcos⁡(2x)2+3sin⁡(2x)4- \frac{3 x \cos{\left(2 x \right)}}{2} + \frac{3 \sin{\left(2 x \right)}}{4}

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

    Method #4

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

      ∫2(x2+3x)sin⁡(x)cos⁡(x) dx=2∫(x2+3x)sin⁡(x)cos⁡(x) dx\int 2 \left(x^{2} + 3 x\right) \sin{\left(x \right)} \cos{\left(x \right)}\, dx = 2 \int \left(x^{2} + 3 x\right) \sin{\left(x \right)} \cos{\left(x \right)}\, dx

      1. Rewrite the integrand:

        (x2+3x)sin⁡(x)cos⁡(x)=x2sin⁡(x)cos⁡(x)+3xsin⁡(x)cos⁡(x)\left(x^{2} + 3 x\right) \sin{\left(x \right)} \cos{\left(x \right)} = x^{2} \sin{\left(x \right)} \cos{\left(x \right)} + 3 x \sin{\left(x \right)} \cos{\left(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⁡(x)cos⁡(x)\operatorname{dv}{\left(x \right)} = \sin{\left(x \right)} \cos{\left(x \right)}.

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

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

          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 u\, du

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

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

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

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

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

          1. Rewrite the integrand:

            cos⁡2(x)=cos⁡(2x)2+12\cos^{2}{\left(x \right)} = \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}

          2. Integrate term-by-term:

            1. 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 \frac{\cos{\left(2 x \right)}}{2}\, 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)4 du\int \frac{\cos{\left(u \right)}}{4}\, du

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

                  ∫cos⁡(u)2 du=∫cos⁡(u) du2\int \frac{\cos{\left(u \right)}}{2}\, 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 is the constant times the variable of integration:

              ∫12 dx=x2\int \frac{1}{2}\, dx = \frac{x}{2}

            The result is: x2+sin⁡(2x)4\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4}

          Now evaluate the sub-integral.

        3. Integrate term-by-term:

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

            ∫(−x2) dx=−∫x dx2\int \left(- \frac{x}{2}\right)\, dx = - \frac{\int x\, dx}{2}

            1. The integral of xnx^{n} is xn+1n+1\frac{x^{n + 1}}{n + 1} when n≠−1n \neq -1:

              ∫x dx=x22\int x\, dx = \frac{x^{2}}{2}

            So, the result is: −x24- \frac{x^{2}}{4}

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

            ∫(−sin⁡(2x)4) dx=−∫sin⁡(2x) dx4\int \left(- \frac{\sin{\left(2 x \right)}}{4}\right)\, dx = - \frac{\int \sin{\left(2 x \right)}\, dx}{4}

            1. Let u=2xu = 2 x.

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

              ∫sin⁡(u)4 du\int \frac{\sin{\left(u \right)}}{4}\, du

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

                ∫sin⁡(u)2 du=∫sin⁡(u) du2\int \frac{\sin{\left(u \right)}}{2}\, 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)8\frac{\cos{\left(2 x \right)}}{8}

          The result is: −x24+cos⁡(2x)8- \frac{x^{2}}{4} + \frac{\cos{\left(2 x \right)}}{8}

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

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

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

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

            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 u\, du

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

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

            Now evaluate the sub-integral.

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

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

            1. Rewrite the integrand:

              cos⁡2(x)=cos⁡(2x)2+12\cos^{2}{\left(x \right)} = \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}

            2. Integrate term-by-term:

              1. 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 \frac{\cos{\left(2 x \right)}}{2}\, 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)4 du\int \frac{\cos{\left(u \right)}}{4}\, du

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

                    ∫cos⁡(u)2 du=∫cos⁡(u) du2\int \frac{\cos{\left(u \right)}}{2}\, 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 is the constant times the variable of integration:

                ∫12 dx=x2\int \frac{1}{2}\, dx = \frac{x}{2}

              The result is: x2+sin⁡(2x)4\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4}

            So, the result is: −x4−sin⁡(2x)8- \frac{x}{4} - \frac{\sin{\left(2 x \right)}}{8}

          So, the result is: −3xcos⁡2(x)2+3x4+3sin⁡(2x)8- \frac{3 x \cos^{2}{\left(x \right)}}{2} + \frac{3 x}{4} + \frac{3 \sin{\left(2 x \right)}}{8}

        The result is: −x2cos⁡2(x)2−x24+x(x2+sin⁡(2x)4)−3xcos⁡2(x)2+3x4+3sin⁡(2x)8+cos⁡(2x)8- \frac{x^{2} \cos^{2}{\left(x \right)}}{2} - \frac{x^{2}}{4} + x \left(\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4}\right) - \frac{3 x \cos^{2}{\left(x \right)}}{2} + \frac{3 x}{4} + \frac{3 \sin{\left(2 x \right)}}{8} + \frac{\cos{\left(2 x \right)}}{8}

      So, the result is: −x2cos⁡2(x)−x22+2x(x2+sin⁡(2x)4)−3xcos⁡2(x)+3x2+3sin⁡(2x)4+cos⁡(2x)4- x^{2} \cos^{2}{\left(x \right)} - \frac{x^{2}}{2} + 2 x \left(\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4}\right) - 3 x \cos^{2}{\left(x \right)} + \frac{3 x}{2} + \frac{3 \sin{\left(2 x \right)}}{4} + \frac{\cos{\left(2 x \right)}}{4}

  2. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
  /                                                                                            
 |                                                                                   2         
 | / 2      \                   cos(2*x)   3*sin(2*x)   x*sin(2*x)   3*x*cos(2*x)   x *cos(2*x)
 | \x  + 3*x/*sin(2*x) dx = C + -------- + ---------- + ---------- - ------------ - -----------
 |                                 4           4            2             2              2     
/
4 x sin⁡(2 x)+(2−4 x2) cos⁡(2 x)4+3 (sin⁡(2 x)−2 x cos⁡(2 x))22{{{{4\,x\,\sin \left(2\,x\right)+\left(2-4\,x^2\right)\,\cos \left( 2\,x\right)}\over{4}}+{{3\,\left(\sin \left(2\,x\right)-2\,x\,\cos \left(2\,x\right)\right)}\over{2}}}\over{2}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.9005
Plot the graph f(x)
The answer [src] 
  1   7*cos(2)   5*sin(2)
- - - -------- + --------
  4      4          4
5 sin⁡2−7 cos⁡24−14{{5\,\sin 2-7\,\cos 2}\over{4}}-{{1}\over{4}} 
=
= 
  1   7*cos(2)   5*sin(2)
- - - -------- + --------
  4      4          4
−14−7cos⁡(2)4+5sin⁡(2)4- \frac{1}{4} - \frac{7 \cos{\left(2 \right)}}{4} + \frac{5 \sin{\left(2 \right)}}{4}
Упростить
Numerical answer [src] 
1.6148787474896
1.6148787474896
The graph
Integral of (x^2+3x)sin(2x) dx

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

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