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Solving integrals/ (3x+2)sin4xdx

Integral of (3x+2)sin4xdx dx

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∫01(3x+2)sin⁡(4x) dx\int\limits_{0}^{1} \left(3 x + 2\right) \sin{\left(4 x \right)}\, dx 
Integral((3*x + 2)*sin(4*x), (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      (3x+2)sin⁡(4x)=3xsin⁡(4x)+2sin⁡(4x)\left(3 x + 2\right) \sin{\left(4 x \right)} = 3 x \sin{\left(4 x \right)} + 2 \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:

        ∫3xsin⁡(4x) dx=3∫xsin⁡(4x) dx\int 3 x \sin{\left(4 x \right)}\, dx = 3 \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: −3xcos⁡(4x)4+3sin⁡(4x)16- \frac{3 x \cos{\left(4 x \right)}}{4} + \frac{3 \sin{\left(4 x \right)}}{16}

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

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

      The result is: −3xcos⁡(4x)4+3sin⁡(4x)16−cos⁡(4x)2- \frac{3 x \cos{\left(4 x \right)}}{4} + \frac{3 \sin{\left(4 x \right)}}{16} - \frac{\cos{\left(4 x \right)}}{2}

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

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

      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:

      ∫(−3cos⁡(4x)4) dx=−3∫cos⁡(4x) dx4\int \left(- \frac{3 \cos{\left(4 x \right)}}{4}\right)\, dx = - \frac{3 \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: −3sin⁡(4x)16- \frac{3 \sin{\left(4 x \right)}}{16}

    Method #3

    1. Rewrite the integrand:

      (3x+2)sin⁡(4x)=3xsin⁡(4x)+2sin⁡(4x)\left(3 x + 2\right) \sin{\left(4 x \right)} = 3 x \sin{\left(4 x \right)} + 2 \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:

        ∫3xsin⁡(4x) dx=3∫xsin⁡(4x) dx\int 3 x \sin{\left(4 x \right)}\, dx = 3 \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: −3xcos⁡(4x)4+3sin⁡(4x)16- \frac{3 x \cos{\left(4 x \right)}}{4} + \frac{3 \sin{\left(4 x \right)}}{16}

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

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

      The result is: −3xcos⁡(4x)4+3sin⁡(4x)16−cos⁡(4x)2- \frac{3 x \cos{\left(4 x \right)}}{4} + \frac{3 \sin{\left(4 x \right)}}{16} - \frac{\cos{\left(4 x \right)}}{2}

  2. Add the constant of integration:

    −3xcos⁡(4x)4+3sin⁡(4x)16−cos⁡(4x)2+constant- \frac{3 x \cos{\left(4 x \right)}}{4} + \frac{3 \sin{\left(4 x \right)}}{16} - \frac{\cos{\left(4 x \right)}}{2}+ \mathrm{constant}

The answer is:

−3xcos⁡(4x)4+3sin⁡(4x)16−cos⁡(4x)2+constant- \frac{3 x \cos{\left(4 x \right)}}{4} + \frac{3 \sin{\left(4 x \right)}}{16} - \frac{\cos{\left(4 x \right)}}{2}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                
 |                             cos(4*x)   3*sin(4*x)   3*x*cos(4*x)
 | (3*x + 2)*sin(4*x) dx = C - -------- + ---------- - ------------
 |                                2           16            4      
/
∫(3x+2)sin⁡(4x) dx=C−3xcos⁡(4x)4+3sin⁡(4x)16−cos⁡(4x)2\int \left(3 x + 2\right) \sin{\left(4 x \right)}\, dx = C - \frac{3 x \cos{\left(4 x \right)}}{4} + \frac{3 \sin{\left(4 x \right)}}{16} - \frac{\cos{\left(4 x \right)}}{2}
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   5*cos(4)   3*sin(4)
- - -------- + --------
2      4          16
3sin⁡(4)16+12−5cos⁡(4)4\frac{3 \sin{\left(4 \right)}}{16} + \frac{1}{2} - \frac{5 \cos{\left(4 \right)}}{4} 
=
= 
1   5*cos(4)   3*sin(4)
- - -------- + --------
2      4          16
3sin⁡(4)16+12−5cos⁡(4)4\frac{3 \sin{\left(4 \right)}}{16} + \frac{1}{2} - \frac{5 \cos{\left(4 \right)}}{4} 
1/2 - 5*cos(4)/4 + 3*sin(4)/16
Numerical answer [src] 
1.17515405820928
1.17515405820928

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

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