Mister Exam

Other calculators

Solving integrals/ ((x^2+4x+3))*sin4x

Integral of ((x^2+4x+3))*sin4x dx

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src] 
  0                           
  /                           
 |                            
 |  / 2          \            
 |  \x  + 4*x + 3/*sin(4*x) dx
 |                            
/                             
-1
∫−10((x2+4x)+3)sin⁡(4x) dx\int\limits_{-1}^{0} \left(\left(x^{2} + 4 x\right) + 3\right) \sin{\left(4 x \right)}\, dx 
Integral((x^2 + 4*x + 3)*sin(4*x), (x, -1, 0))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

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

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

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

        1. Let u=4xu = 4 x.

          Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

          ∫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) du=∫sin⁡(u) du4\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}

            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)4- \frac{\cos{\left(u \right)}}{4}

          Now substitute uu back in:

          −cos⁡(4x)4- \frac{\cos{\left(4 x \right)}}{4}

        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)=−x2u{\left(x \right)} = - \frac{x}{2} and let dv⁡(x)=cos⁡(4x)\operatorname{dv}{\left(x \right)} = \cos{\left(4 x \right)}.

        Then du⁡(x)=−12\operatorname{du}{\left(x \right)} = - \frac{1}{2}.

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

        1. Let u=4xu = 4 x.

          Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

          ∫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) du=∫cos⁡(u) du4\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}

            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)4\frac{\sin{\left(u \right)}}{4}

          Now substitute uu back in:

          sin⁡(4x)4\frac{\sin{\left(4 x \right)}}{4}

        Now evaluate the sub-integral.

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

        ∫(−sin⁡(4x)8) dx=−∫sin⁡(4x) dx8\int \left(- \frac{\sin{\left(4 x \right)}}{8}\right)\, dx = - \frac{\int \sin{\left(4 x \right)}\, dx}{8}

        1. Let u=4xu = 4 x.

          Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

          ∫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) du=∫sin⁡(u) du4\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}

            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)4- \frac{\cos{\left(u \right)}}{4}

          Now substitute uu back in:

          −cos⁡(4x)4- \frac{\cos{\left(4 x \right)}}{4}

        So, the result is: cos⁡(4x)32\frac{\cos{\left(4 x \right)}}{32}

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

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

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

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

          1. Let u=4xu = 4 x.

            Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

            ∫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) du=∫sin⁡(u) du4\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}

              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)4- \frac{\cos{\left(u \right)}}{4}

            Now substitute uu back in:

            −cos⁡(4x)4- \frac{\cos{\left(4 x \right)}}{4}

          Now evaluate the sub-integral.

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

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

          1. Let u=4xu = 4 x.

            Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

            ∫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) du=∫cos⁡(u) du4\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}

              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)4\frac{\sin{\left(u \right)}}{4}

            Now substitute uu back in:

            sin⁡(4x)4\frac{\sin{\left(4 x \right)}}{4}

          So, the result is: −sin⁡(4x)16- \frac{\sin{\left(4 x \right)}}{16}

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

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

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

        1. Let u=4xu = 4 x.

          Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

          ∫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) du=∫sin⁡(u) du4\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}

            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)4- \frac{\cos{\left(u \right)}}{4}

          Now substitute uu back in:

          −cos⁡(4x)4- \frac{\cos{\left(4 x \right)}}{4}

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

      The result is: −x2cos⁡(4x)4+xsin⁡(4x)8−xcos⁡(4x)+sin⁡(4x)4−23cos⁡(4x)32- \frac{x^{2} \cos{\left(4 x \right)}}{4} + \frac{x \sin{\left(4 x \right)}}{8} - x \cos{\left(4 x \right)} + \frac{\sin{\left(4 x \right)}}{4} - \frac{23 \cos{\left(4 x \right)}}{32}

    Method #2

    1. Use integration by parts:

      ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

      Let u(x)=x2+4x+3u{\left(x \right)} = x^{2} + 4 x + 3 and let dv⁡(x)=sin⁡(4x)\operatorname{dv}{\left(x \right)} = \sin{\left(4 x \right)}.

      Then du⁡(x)=2x+4\operatorname{du}{\left(x \right)} = 2 x + 4.

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

      1. Let u=4xu = 4 x.

        Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

        ∫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) du=∫sin⁡(u) du4\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}

          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)4- \frac{\cos{\left(u \right)}}{4}

        Now substitute uu back in:

        −cos⁡(4x)4- \frac{\cos{\left(4 x \right)}}{4}

      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)=−x2−1u{\left(x \right)} = - \frac{x}{2} - 1 and let dv⁡(x)=cos⁡(4x)\operatorname{dv}{\left(x \right)} = \cos{\left(4 x \right)}.

      Then du⁡(x)=−12\operatorname{du}{\left(x \right)} = - \frac{1}{2}.

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

      1. Let u=4xu = 4 x.

        Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

        ∫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) du=∫cos⁡(u) du4\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}

          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)4\frac{\sin{\left(u \right)}}{4}

        Now substitute uu back in:

        sin⁡(4x)4\frac{\sin{\left(4 x \right)}}{4}

      Now evaluate the sub-integral.

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

      ∫(−sin⁡(4x)8) dx=−∫sin⁡(4x) dx8\int \left(- \frac{\sin{\left(4 x \right)}}{8}\right)\, dx = - \frac{\int \sin{\left(4 x \right)}\, dx}{8}

      1. Let u=4xu = 4 x.

        Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

        ∫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) du=∫sin⁡(u) du4\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}

          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)4- \frac{\cos{\left(u \right)}}{4}

        Now substitute uu back in:

        −cos⁡(4x)4- \frac{\cos{\left(4 x \right)}}{4}

      So, the result is: cos⁡(4x)32\frac{\cos{\left(4 x \right)}}{32}

    Method #3

    1. Rewrite the integrand:

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

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

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

        1. Let u=4xu = 4 x.

          Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

          ∫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) du=∫sin⁡(u) du4\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}

            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)4- \frac{\cos{\left(u \right)}}{4}

          Now substitute uu back in:

          −cos⁡(4x)4- \frac{\cos{\left(4 x \right)}}{4}

        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)=−x2u{\left(x \right)} = - \frac{x}{2} and let dv⁡(x)=cos⁡(4x)\operatorname{dv}{\left(x \right)} = \cos{\left(4 x \right)}.

        Then du⁡(x)=−12\operatorname{du}{\left(x \right)} = - \frac{1}{2}.

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

        1. Let u=4xu = 4 x.

          Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

          ∫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) du=∫cos⁡(u) du4\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}

            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)4\frac{\sin{\left(u \right)}}{4}

          Now substitute uu back in:

          sin⁡(4x)4\frac{\sin{\left(4 x \right)}}{4}

        Now evaluate the sub-integral.

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

        ∫(−sin⁡(4x)8) dx=−∫sin⁡(4x) dx8\int \left(- \frac{\sin{\left(4 x \right)}}{8}\right)\, dx = - \frac{\int \sin{\left(4 x \right)}\, dx}{8}

        1. Let u=4xu = 4 x.

          Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

          ∫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) du=∫sin⁡(u) du4\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}

            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)4- \frac{\cos{\left(u \right)}}{4}

          Now substitute uu back in:

          −cos⁡(4x)4- \frac{\cos{\left(4 x \right)}}{4}

        So, the result is: cos⁡(4x)32\frac{\cos{\left(4 x \right)}}{32}

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

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

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

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

          1. Let u=4xu = 4 x.

            Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

            ∫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) du=∫sin⁡(u) du4\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}

              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)4- \frac{\cos{\left(u \right)}}{4}

            Now substitute uu back in:

            −cos⁡(4x)4- \frac{\cos{\left(4 x \right)}}{4}

          Now evaluate the sub-integral.

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

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

          1. Let u=4xu = 4 x.

            Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

            ∫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) du=∫cos⁡(u) du4\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}

              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)4\frac{\sin{\left(u \right)}}{4}

            Now substitute uu back in:

            sin⁡(4x)4\frac{\sin{\left(4 x \right)}}{4}

          So, the result is: −sin⁡(4x)16- \frac{\sin{\left(4 x \right)}}{16}

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

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

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

        1. Let u=4xu = 4 x.

          Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

          ∫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) du=∫sin⁡(u) du4\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}

            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)4- \frac{\cos{\left(u \right)}}{4}

          Now substitute uu back in:

          −cos⁡(4x)4- \frac{\cos{\left(4 x \right)}}{4}

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

      The result is: −x2cos⁡(4x)4+xsin⁡(4x)8−xcos⁡(4x)+sin⁡(4x)4−23cos⁡(4x)32- \frac{x^{2} \cos{\left(4 x \right)}}{4} + \frac{x \sin{\left(4 x \right)}}{8} - x \cos{\left(4 x \right)} + \frac{\sin{\left(4 x \right)}}{4} - \frac{23 \cos{\left(4 x \right)}}{32}

  2. Add the constant of integration:

    −x2cos⁡(4x)4+xsin⁡(4x)8−xcos⁡(4x)+sin⁡(4x)4−23cos⁡(4x)32+constant- \frac{x^{2} \cos{\left(4 x \right)}}{4} + \frac{x \sin{\left(4 x \right)}}{8} - x \cos{\left(4 x \right)} + \frac{\sin{\left(4 x \right)}}{4} - \frac{23 \cos{\left(4 x \right)}}{32}+ \mathrm{constant}

The answer is:

−x2cos⁡(4x)4+xsin⁡(4x)8−xcos⁡(4x)+sin⁡(4x)4−23cos⁡(4x)32+constant- \frac{x^{2} \cos{\left(4 x \right)}}{4} + \frac{x \sin{\left(4 x \right)}}{8} - x \cos{\left(4 x \right)} + \frac{\sin{\left(4 x \right)}}{4} - \frac{23 \cos{\left(4 x \right)}}{32}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                                               
 |                                                                         2                      
 | / 2          \                   23*cos(4*x)   sin(4*x)                x *cos(4*x)   x*sin(4*x)
 | \x  + 4*x + 3/*sin(4*x) dx = C - ----------- + -------- - x*cos(4*x) - ----------- + ----------
 |                                       32          4                         4            8     
/
∫((x2+4x)+3)sin⁡(4x) dx=C−x2cos⁡(4x)4+xsin⁡(4x)8−xcos⁡(4x)+sin⁡(4x)4−23cos⁡(4x)32\int \left(\left(x^{2} + 4 x\right) + 3\right) \sin{\left(4 x \right)}\, dx = C - \frac{x^{2} \cos{\left(4 x \right)}}{4} + \frac{x \sin{\left(4 x \right)}}{8} - x \cos{\left(4 x \right)} + \frac{\sin{\left(4 x \right)}}{4} - \frac{23 \cos{\left(4 x \right)}}{32}
Simplify
The graph
-1.00-0.90-0.80-0.70-0.60-0.50-0.40-0.30-0.20-0.100.002-2
Plot the graph f(x)
The answer [src] 
  23   cos(4)   sin(4)
- -- - ------ + ------
  32     32       8
−2332+sin⁡(4)8−cos⁡(4)32- \frac{23}{32} + \frac{\sin{\left(4 \right)}}{8} - \frac{\cos{\left(4 \right)}}{32} 
=
= 
  23   cos(4)   sin(4)
- -- - ------ + ------
  32     32       8
−2332+sin⁡(4)8−cos⁡(4)32- \frac{23}{32} + \frac{\sin{\left(4 \right)}}{8} - \frac{\cos{\left(4 \right)}}{32} 
-23/32 - cos(4)/32 + sin(4)/8
Numerical answer [src] 
-0.792923948761503
-0.792923948761503

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

    © Mister Exam – Calculator