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(3x+2)sin3xdx
Solving integrals/ (3x+2)sin3xdx

Integral of (3x+2)sin3xdx dx

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

  1. There are multiple ways to do this integral.

    Method #1

    1. Let u=3xu = 3 x.

      Then let du=3dxdu = 3 dx and substitute dudu:

      ∫(usin⁡(u)3+2sin⁡(u)3) du\int \left(\frac{u \sin{\left(u \right)}}{3} + \frac{2 \sin{\left(u \right)}}{3}\right)\, du

      1. Integrate term-by-term:

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

          ∫usin⁡(u)3 du=∫usin⁡(u) du3\int \frac{u \sin{\left(u \right)}}{3}\, du = \frac{\int u \sin{\left(u \right)}\, du}{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(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)}

          So, the result is: −ucos⁡(u)3+sin⁡(u)3- \frac{u \cos{\left(u \right)}}{3} + \frac{\sin{\left(u \right)}}{3}

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

          ∫2sin⁡(u)3 du=2∫sin⁡(u) du3\int \frac{2 \sin{\left(u \right)}}{3}\, du = \frac{2 \int \sin{\left(u \right)}\, du}{3}

          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: −2cos⁡(u)3- \frac{2 \cos{\left(u \right)}}{3}

        The result is: −ucos⁡(u)3+sin⁡(u)3−2cos⁡(u)3- \frac{u \cos{\left(u \right)}}{3} + \frac{\sin{\left(u \right)}}{3} - \frac{2 \cos{\left(u \right)}}{3}

      Now substitute uu back in:

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

    Method #2

    1. Rewrite the integrand:

      (3x+2)sin⁡(3x)1=3xsin⁡(3x)+2sin⁡(3x)\left(3 x + 2\right) \sin{\left(3 x \right)} 1 = 3 x \sin{\left(3 x \right)} + 2 \sin{\left(3 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⁡(3x) dx=3∫xsin⁡(3x) dx\int 3 x \sin{\left(3 x \right)}\, dx = 3 \int x \sin{\left(3 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⁡(3x)\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}.

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

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

          1. Let u=3xu = 3 x.

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

            ∫sin⁡(u)9 du\int \frac{\sin{\left(u \right)}}{9}\, du

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

              ∫sin⁡(u)3 du=∫sin⁡(u) du3\int \frac{\sin{\left(u \right)}}{3}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}

              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)3- \frac{\cos{\left(u \right)}}{3}

            Now substitute uu back in:

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

          Now evaluate the sub-integral.

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

          ∫(−cos⁡(3x)3) dx=−∫cos⁡(3x) dx3\int \left(- \frac{\cos{\left(3 x \right)}}{3}\right)\, dx = - \frac{\int \cos{\left(3 x \right)}\, dx}{3}

          1. Let u=3xu = 3 x.

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

            ∫cos⁡(u)9 du\int \frac{\cos{\left(u \right)}}{9}\, du

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

              ∫cos⁡(u)3 du=∫cos⁡(u) du3\int \frac{\cos{\left(u \right)}}{3}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}

              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)3\frac{\sin{\left(u \right)}}{3}

            Now substitute uu back in:

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

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

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

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

        ∫2sin⁡(3x) dx=2∫sin⁡(3x) dx\int 2 \sin{\left(3 x \right)}\, dx = 2 \int \sin{\left(3 x \right)}\, dx

        1. Let u=3xu = 3 x.

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

          ∫sin⁡(u)9 du\int \frac{\sin{\left(u \right)}}{9}\, du

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

            ∫sin⁡(u)3 du=∫sin⁡(u) du3\int \frac{\sin{\left(u \right)}}{3}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}

            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)3- \frac{\cos{\left(u \right)}}{3}

          Now substitute uu back in:

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

        So, the result is: −2cos⁡(3x)3- \frac{2 \cos{\left(3 x \right)}}{3}

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

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

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

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

      1. Let u=3xu = 3 x.

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

        ∫sin⁡(u)9 du\int \frac{\sin{\left(u \right)}}{9}\, du

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

          ∫sin⁡(u)3 du=∫sin⁡(u) du3\int \frac{\sin{\left(u \right)}}{3}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}

          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)3- \frac{\cos{\left(u \right)}}{3}

        Now substitute uu back in:

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

      Now evaluate the sub-integral.

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

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

      1. Let u=3xu = 3 x.

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

        ∫cos⁡(u)9 du\int \frac{\cos{\left(u \right)}}{9}\, du

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

          ∫cos⁡(u)3 du=∫cos⁡(u) du3\int \frac{\cos{\left(u \right)}}{3}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}

          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)3\frac{\sin{\left(u \right)}}{3}

        Now substitute uu back in:

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

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

    Method #4

    1. Rewrite the integrand:

      (3x+2)sin⁡(3x)1=3xsin⁡(3x)+2sin⁡(3x)\left(3 x + 2\right) \sin{\left(3 x \right)} 1 = 3 x \sin{\left(3 x \right)} + 2 \sin{\left(3 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⁡(3x) dx=3∫xsin⁡(3x) dx\int 3 x \sin{\left(3 x \right)}\, dx = 3 \int x \sin{\left(3 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⁡(3x)\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}.

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

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

          1. Let u=3xu = 3 x.

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

            ∫sin⁡(u)9 du\int \frac{\sin{\left(u \right)}}{9}\, du

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

              ∫sin⁡(u)3 du=∫sin⁡(u) du3\int \frac{\sin{\left(u \right)}}{3}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}

              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)3- \frac{\cos{\left(u \right)}}{3}

            Now substitute uu back in:

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

          Now evaluate the sub-integral.

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

          ∫(−cos⁡(3x)3) dx=−∫cos⁡(3x) dx3\int \left(- \frac{\cos{\left(3 x \right)}}{3}\right)\, dx = - \frac{\int \cos{\left(3 x \right)}\, dx}{3}

          1. Let u=3xu = 3 x.

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

            ∫cos⁡(u)9 du\int \frac{\cos{\left(u \right)}}{9}\, du

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

              ∫cos⁡(u)3 du=∫cos⁡(u) du3\int \frac{\cos{\left(u \right)}}{3}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}

              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)3\frac{\sin{\left(u \right)}}{3}

            Now substitute uu back in:

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

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

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

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

        ∫2sin⁡(3x) dx=2∫sin⁡(3x) dx\int 2 \sin{\left(3 x \right)}\, dx = 2 \int \sin{\left(3 x \right)}\, dx

        1. Let u=3xu = 3 x.

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

          ∫sin⁡(u)9 du\int \frac{\sin{\left(u \right)}}{9}\, du

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

            ∫sin⁡(u)3 du=∫sin⁡(u) du3\int \frac{\sin{\left(u \right)}}{3}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}

            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)3- \frac{\cos{\left(u \right)}}{3}

          Now substitute uu back in:

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

        So, the result is: −2cos⁡(3x)3- \frac{2 \cos{\left(3 x \right)}}{3}

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

  2. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
  /                                                                
 |                               2*cos(3*x)   sin(3*x)             
 | (3*x + 2)*sin(3*x)*1 dx = C - ---------- + -------- - x*cos(3*x)
 |                                   3           3                 
/
sin⁡(3 x)−3 x cos⁡(3 x)−2 cos⁡(3 x)3{{\sin \left(3\,x\right)-3\,x\,\cos \left(3\,x\right)-2\,\cos \left(3\,x\right)}\over{3}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.905-5
Plot the graph f(x)
The answer [src] 
2   5*cos(3)   sin(3)
- - -------- + ------
3      3         3
sin⁡3−5 cos⁡33+23{{\sin 3-5\,\cos 3}\over{3}}+{{2}\over{3}} 
=
= 
2   5*cos(3)   sin(3)
- - -------- + ------
3      3         3
sin⁡(3)3+23−5cos⁡(3)3\frac{\sin{\left(3 \right)}}{3} + \frac{2}{3} - \frac{5 \cos{\left(3 \right)}}{3}
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Numerical answer [src] 
2.36369416368736
2.36369416368736
The graph
Integral of (3x+2)sin3xdx dx

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

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