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Solving integrals/ (16x+24)sin4xdx

Integral of (16x+24)sin4xdx dx

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 |  (16*x + 24)*sin(4*x) dx
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∫01(16x+24)sin⁡(4x) dx\int\limits_{0}^{1} \left(16 x + 24\right) \sin{\left(4 x \right)}\, dx 
Integral((16*x + 24)*sin(4*x), (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Let u=4xu = 4 x.

      Then let du=4dxdu = 4 dx and substitute dudu:

      ∫(usin⁡(u)+6sin⁡(u)) du\int \left(u \sin{\left(u \right)} + 6 \sin{\left(u \right)}\right)\, du

      1. 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(u)=uu{\left(u \right)} = u and let dv⁡(u)=sin⁡(u)\operatorname{dv}{\left(u \right)} = \sin{\left(u \right)}.

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

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

          1. The integral of sine is negative cosine:

            ∫sin⁡(u) du=−cos⁡(u)\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}

          Now evaluate the sub-integral.

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

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

          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)- \sin{\left(u \right)}

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

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

          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: −6cos⁡(u)- 6 \cos{\left(u \right)}

        The result is: −ucos⁡(u)+sin⁡(u)−6cos⁡(u)- u \cos{\left(u \right)} + \sin{\left(u \right)} - 6 \cos{\left(u \right)}

      Now substitute uu back in:

      −4xcos⁡(4x)+sin⁡(4x)−6cos⁡(4x)- 4 x \cos{\left(4 x \right)} + \sin{\left(4 x \right)} - 6 \cos{\left(4 x \right)}

    Method #2

    1. Rewrite the integrand:

      (16x+24)sin⁡(4x)=16xsin⁡(4x)+24sin⁡(4x)\left(16 x + 24\right) \sin{\left(4 x \right)} = 16 x \sin{\left(4 x \right)} + 24 \sin{\left(4 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:

        ∫16xsin⁡(4x) dx=16∫xsin⁡(4x) dx\int 16 x \sin{\left(4 x \right)}\, dx = 16 \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: −4xcos⁡(4x)+sin⁡(4x)- 4 x \cos{\left(4 x \right)} + \sin{\left(4 x \right)}

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

        ∫24sin⁡(4x) dx=24∫sin⁡(4x) dx\int 24 \sin{\left(4 x \right)}\, dx = 24 \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: −6cos⁡(4x)- 6 \cos{\left(4 x \right)}

      The result is: −4xcos⁡(4x)+sin⁡(4x)−6cos⁡(4x)- 4 x \cos{\left(4 x \right)} + \sin{\left(4 x \right)} - 6 \cos{\left(4 x \right)}

    Method #3

    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)=16x+24u{\left(x \right)} = 16 x + 24 and let dv⁡(x)=sin⁡(4x)\operatorname{dv}{\left(x \right)} = \sin{\left(4 x \right)}.

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

      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:

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

      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)- \sin{\left(4 x \right)}

    Method #4

    1. Rewrite the integrand:

      (16x+24)sin⁡(4x)=16xsin⁡(4x)+24sin⁡(4x)\left(16 x + 24\right) \sin{\left(4 x \right)} = 16 x \sin{\left(4 x \right)} + 24 \sin{\left(4 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:

        ∫16xsin⁡(4x) dx=16∫xsin⁡(4x) dx\int 16 x \sin{\left(4 x \right)}\, dx = 16 \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: −4xcos⁡(4x)+sin⁡(4x)- 4 x \cos{\left(4 x \right)} + \sin{\left(4 x \right)}

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

        ∫24sin⁡(4x) dx=24∫sin⁡(4x) dx\int 24 \sin{\left(4 x \right)}\, dx = 24 \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: −6cos⁡(4x)- 6 \cos{\left(4 x \right)}

      The result is: −4xcos⁡(4x)+sin⁡(4x)−6cos⁡(4x)- 4 x \cos{\left(4 x \right)} + \sin{\left(4 x \right)} - 6 \cos{\left(4 x \right)}

  2. Add the constant of integration:

    −4xcos⁡(4x)+sin⁡(4x)−6cos⁡(4x)+constant- 4 x \cos{\left(4 x \right)} + \sin{\left(4 x \right)} - 6 \cos{\left(4 x \right)}+ \mathrm{constant}

The answer is:

−4xcos⁡(4x)+sin⁡(4x)−6cos⁡(4x)+constant- 4 x \cos{\left(4 x \right)} + \sin{\left(4 x \right)} - 6 \cos{\left(4 x \right)}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                  
 |                                                                   
 | (16*x + 24)*sin(4*x) dx = C - 6*cos(4*x) - 4*x*cos(4*x) + sin(4*x)
 |                                                                   
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∫(16x+24)sin⁡(4x) dx=C−4xcos⁡(4x)+sin⁡(4x)−6cos⁡(4x)\int \left(16 x + 24\right) \sin{\left(4 x \right)}\, dx = C - 4 x \cos{\left(4 x \right)} + \sin{\left(4 x \right)} - 6 \cos{\left(4 x \right)}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.90-5050
Plot the graph f(x)
The answer [src] 
6 - 10*cos(4) + sin(4)
sin⁡(4)+6−10cos⁡(4)\sin{\left(4 \right)} + 6 - 10 \cos{\left(4 \right)} 
=
= 
6 - 10*cos(4) + sin(4)
sin⁡(4)+6−10cos⁡(4)\sin{\left(4 \right)} + 6 - 10 \cos{\left(4 \right)} 
6 - 10*cos(4) + sin(4)
Numerical answer [src] 
11.7796337133282
11.7796337133282

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

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