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(x-3)sin(x/3)dx
Solving integrals/ (x-3)sin(x/3)dx

Integral of (x-3)sin(x/3)dx dx

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∫01(x−3)sin⁡(x3) dx\int\limits_{0}^{1} \left(x - 3\right) \sin{\left(\frac{x}{3} \right)}\, dx 
Integral((x - 3)*sin(x/3), (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      (x−3)sin⁡(x3)=xsin⁡(x3)−3sin⁡(x3)\left(x - 3\right) \sin{\left(\frac{x}{3} \right)} = x \sin{\left(\frac{x}{3} \right)} - 3 \sin{\left(\frac{x}{3} \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⁡(x3)\operatorname{dv}{\left(x \right)} = \sin{\left(\frac{x}{3} \right)}.

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

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

        1. Let u=x3u = \frac{x}{3}.

          Then let du=dx3du = \frac{dx}{3} and substitute 3du3 du:

          ∫3sin⁡(u) du\int 3 \sin{\left(u \right)}\, du

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

            ∫sin⁡(u) du=3∫sin⁡(u) du\int \sin{\left(u \right)}\, du = 3 \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: −3cos⁡(u)- 3 \cos{\left(u \right)}

          Now substitute uu back in:

          −3cos⁡(x3)- 3 \cos{\left(\frac{x}{3} \right)}

        Now evaluate the sub-integral.

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

        ∫(−3cos⁡(x3)) dx=−3∫cos⁡(x3) dx\int \left(- 3 \cos{\left(\frac{x}{3} \right)}\right)\, dx = - 3 \int \cos{\left(\frac{x}{3} \right)}\, dx

        1. Let u=x3u = \frac{x}{3}.

          Then let du=dx3du = \frac{dx}{3} and substitute 3du3 du:

          ∫3cos⁡(u) du\int 3 \cos{\left(u \right)}\, du

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

            ∫cos⁡(u) du=3∫cos⁡(u) du\int \cos{\left(u \right)}\, du = 3 \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: 3sin⁡(u)3 \sin{\left(u \right)}

          Now substitute uu back in:

          3sin⁡(x3)3 \sin{\left(\frac{x}{3} \right)}

        So, the result is: −9sin⁡(x3)- 9 \sin{\left(\frac{x}{3} \right)}

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

        ∫(−3sin⁡(x3)) dx=−3∫sin⁡(x3) dx\int \left(- 3 \sin{\left(\frac{x}{3} \right)}\right)\, dx = - 3 \int \sin{\left(\frac{x}{3} \right)}\, dx

        1. Let u=x3u = \frac{x}{3}.

          Then let du=dx3du = \frac{dx}{3} and substitute 3du3 du:

          ∫3sin⁡(u) du\int 3 \sin{\left(u \right)}\, du

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

            ∫sin⁡(u) du=3∫sin⁡(u) du\int \sin{\left(u \right)}\, du = 3 \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: −3cos⁡(u)- 3 \cos{\left(u \right)}

          Now substitute uu back in:

          −3cos⁡(x3)- 3 \cos{\left(\frac{x}{3} \right)}

        So, the result is: 9cos⁡(x3)9 \cos{\left(\frac{x}{3} \right)}

      The result is: −3xcos⁡(x3)+9sin⁡(x3)+9cos⁡(x3)- 3 x \cos{\left(\frac{x}{3} \right)} + 9 \sin{\left(\frac{x}{3} \right)} + 9 \cos{\left(\frac{x}{3} \right)}

    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⁡(x3)\operatorname{dv}{\left(x \right)} = \sin{\left(\frac{x}{3} \right)}.

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

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

      1. Let u=x3u = \frac{x}{3}.

        Then let du=dx3du = \frac{dx}{3} and substitute 3du3 du:

        ∫3sin⁡(u) du\int 3 \sin{\left(u \right)}\, du

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

          ∫sin⁡(u) du=3∫sin⁡(u) du\int \sin{\left(u \right)}\, du = 3 \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: −3cos⁡(u)- 3 \cos{\left(u \right)}

        Now substitute uu back in:

        −3cos⁡(x3)- 3 \cos{\left(\frac{x}{3} \right)}

      Now evaluate the sub-integral.

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

      ∫(−3cos⁡(x3)) dx=−3∫cos⁡(x3) dx\int \left(- 3 \cos{\left(\frac{x}{3} \right)}\right)\, dx = - 3 \int \cos{\left(\frac{x}{3} \right)}\, dx

      1. Let u=x3u = \frac{x}{3}.

        Then let du=dx3du = \frac{dx}{3} and substitute 3du3 du:

        ∫3cos⁡(u) du\int 3 \cos{\left(u \right)}\, du

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

          ∫cos⁡(u) du=3∫cos⁡(u) du\int \cos{\left(u \right)}\, du = 3 \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: 3sin⁡(u)3 \sin{\left(u \right)}

        Now substitute uu back in:

        3sin⁡(x3)3 \sin{\left(\frac{x}{3} \right)}

      So, the result is: −9sin⁡(x3)- 9 \sin{\left(\frac{x}{3} \right)}

    Method #3

    1. Rewrite the integrand:

      (x−3)sin⁡(x3)=xsin⁡(x3)−3sin⁡(x3)\left(x - 3\right) \sin{\left(\frac{x}{3} \right)} = x \sin{\left(\frac{x}{3} \right)} - 3 \sin{\left(\frac{x}{3} \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⁡(x3)\operatorname{dv}{\left(x \right)} = \sin{\left(\frac{x}{3} \right)}.

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

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

        1. Let u=x3u = \frac{x}{3}.

          Then let du=dx3du = \frac{dx}{3} and substitute 3du3 du:

          ∫3sin⁡(u) du\int 3 \sin{\left(u \right)}\, du

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

            ∫sin⁡(u) du=3∫sin⁡(u) du\int \sin{\left(u \right)}\, du = 3 \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: −3cos⁡(u)- 3 \cos{\left(u \right)}

          Now substitute uu back in:

          −3cos⁡(x3)- 3 \cos{\left(\frac{x}{3} \right)}

        Now evaluate the sub-integral.

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

        ∫(−3cos⁡(x3)) dx=−3∫cos⁡(x3) dx\int \left(- 3 \cos{\left(\frac{x}{3} \right)}\right)\, dx = - 3 \int \cos{\left(\frac{x}{3} \right)}\, dx

        1. Let u=x3u = \frac{x}{3}.

          Then let du=dx3du = \frac{dx}{3} and substitute 3du3 du:

          ∫3cos⁡(u) du\int 3 \cos{\left(u \right)}\, du

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

            ∫cos⁡(u) du=3∫cos⁡(u) du\int \cos{\left(u \right)}\, du = 3 \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: 3sin⁡(u)3 \sin{\left(u \right)}

          Now substitute uu back in:

          3sin⁡(x3)3 \sin{\left(\frac{x}{3} \right)}

        So, the result is: −9sin⁡(x3)- 9 \sin{\left(\frac{x}{3} \right)}

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

        ∫(−3sin⁡(x3)) dx=−3∫sin⁡(x3) dx\int \left(- 3 \sin{\left(\frac{x}{3} \right)}\right)\, dx = - 3 \int \sin{\left(\frac{x}{3} \right)}\, dx

        1. Let u=x3u = \frac{x}{3}.

          Then let du=dx3du = \frac{dx}{3} and substitute 3du3 du:

          ∫3sin⁡(u) du\int 3 \sin{\left(u \right)}\, du

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

            ∫sin⁡(u) du=3∫sin⁡(u) du\int \sin{\left(u \right)}\, du = 3 \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: −3cos⁡(u)- 3 \cos{\left(u \right)}

          Now substitute uu back in:

          −3cos⁡(x3)- 3 \cos{\left(\frac{x}{3} \right)}

        So, the result is: 9cos⁡(x3)9 \cos{\left(\frac{x}{3} \right)}

      The result is: −3xcos⁡(x3)+9sin⁡(x3)+9cos⁡(x3)- 3 x \cos{\left(\frac{x}{3} \right)} + 9 \sin{\left(\frac{x}{3} \right)} + 9 \cos{\left(\frac{x}{3} \right)}

  2. Now simplify:

    −3xcos⁡(x3)+92sin⁡(x3+π4)- 3 x \cos{\left(\frac{x}{3} \right)} + 9 \sqrt{2} \sin{\left(\frac{x}{3} + \frac{\pi}{4} \right)}

  3. Add the constant of integration:

    −3xcos⁡(x3)+92sin⁡(x3+π4)+constant- 3 x \cos{\left(\frac{x}{3} \right)} + 9 \sqrt{2} \sin{\left(\frac{x}{3} + \frac{\pi}{4} \right)}+ \mathrm{constant}

The answer is:

−3xcos⁡(x3)+92sin⁡(x3+π4)+constant- 3 x \cos{\left(\frac{x}{3} \right)} + 9 \sqrt{2} \sin{\left(\frac{x}{3} + \frac{\pi}{4} \right)}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                        
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 |            /x\               /x\        /x\          /x\
 | (x - 3)*sin|-| dx = C + 9*cos|-| + 9*sin|-| - 3*x*cos|-|
 |            \3/               \3/        \3/          \3/
 |                                                         
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∫(x−3)sin⁡(x3) dx=C−3xcos⁡(x3)+9sin⁡(x3)+9cos⁡(x3)\int \left(x - 3\right) \sin{\left(\frac{x}{3} \right)}\, dx = C - 3 x \cos{\left(\frac{x}{3} \right)} + 9 \sin{\left(\frac{x}{3} \right)} + 9 \cos{\left(\frac{x}{3} \right)}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.90-1010
Plot the graph f(x)
The answer [src] 
-9 + 6*cos(1/3) + 9*sin(1/3)
−9+9sin⁡(13)+6cos⁡(13)-9 + 9 \sin{\left(\frac{1}{3} \right)} + 6 \cos{\left(\frac{1}{3} \right)} 
=
= 
-9 + 6*cos(1/3) + 9*sin(1/3)
−9+9sin⁡(13)+6cos⁡(13)-9 + 9 \sin{\left(\frac{1}{3} \right)} + 6 \cos{\left(\frac{1}{3} \right)} 
-9 + 6*cos(1/3) + 9*sin(1/3)
Numerical answer [src] 
-0.385506050946204
-0.385506050946204
The graph
Integral of (x-3)sin(x/3)dx dx

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

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