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Solving integrals/ (x-3)*sin4x

Integral of (x-3)*sin4x dx

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

  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      (x−3)sin⁡(4x)=xsin⁡(4x)−3sin⁡(4x)\left(x - 3\right) \sin{\left(4 x \right)} = 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)=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}

      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 \left(- 3 \sin{\left(4 x \right)}\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: −xcos⁡(4x)4+sin⁡(4x)16+3cos⁡(4x)4- \frac{x \cos{\left(4 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{16} + \frac{3 \cos{\left(4 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−3u{\left(x \right)} = x - 3 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}

    Method #3

    1. Rewrite the integrand:

      (x−3)sin⁡(4x)=xsin⁡(4x)−3sin⁡(4x)\left(x - 3\right) \sin{\left(4 x \right)} = 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)=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}

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

  2. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
  /                                                            
 |                           sin(4*x)   3*cos(4*x)   x*cos(4*x)
 | (x - 3)*sin(4*x) dx = C + -------- + ---------- - ----------
 |                              16          4            4     
/
∫(x−3)sin⁡(4x) dx=C−xcos⁡(4x)4+sin⁡(4x)16+3cos⁡(4x)4\int \left(x - 3\right) \sin{\left(4 x \right)}\, dx = C - \frac{x \cos{\left(4 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{16} + \frac{3 \cos{\left(4 x \right)}}{4}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.905-5
Plot the graph f(x)
The answer [src] 
  3   cos(4)   sin(4)
- - + ------ + ------
  4     2        16
−34+cos⁡(4)2+sin⁡(4)16- \frac{3}{4} + \frac{\cos{\left(4 \right)}}{2} + \frac{\sin{\left(4 \right)}}{16} 
=
= 
  3   cos(4)   sin(4)
- - + ------ + ------
  4     2        16
−34+cos⁡(4)2+sin⁡(4)16- \frac{3}{4} + \frac{\cos{\left(4 \right)}}{2} + \frac{\sin{\left(4 \right)}}{16} 
-3/4 + cos(4)/2 + sin(4)/16
Numerical answer [src] 
-1.12412196638855
-1.12412196638855

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

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