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Solving integrals/ (3*x-2)*cos(5*x)

Integral of (3*x-2)*cos(5*x) dx

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∫01(3x−2)cos⁡(5x) dx\int\limits_{0}^{1} \left(3 x - 2\right) \cos{\left(5 x \right)}\, dx 
Integral((3*x - 2)*cos(5*x), (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      (3x−2)cos⁡(5x)=3xcos⁡(5x)−2cos⁡(5x)\left(3 x - 2\right) \cos{\left(5 x \right)} = 3 x \cos{\left(5 x \right)} - 2 \cos{\left(5 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:

        ∫3xcos⁡(5x) dx=3∫xcos⁡(5x) dx\int 3 x \cos{\left(5 x \right)}\, dx = 3 \int x \cos{\left(5 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)=cos⁡(5x)\operatorname{dv}{\left(x \right)} = \cos{\left(5 x \right)}.

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

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

          1. Let u=5xu = 5 x.

            Then let du=5dxdu = 5 dx and substitute du5\frac{du}{5}:

            ∫cos⁡(u)5 du\int \frac{\cos{\left(u \right)}}{5}\, du

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

              ∫cos⁡(u) du=∫cos⁡(u) du5\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{5}

              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)5\frac{\sin{\left(u \right)}}{5}

            Now substitute uu back in:

            sin⁡(5x)5\frac{\sin{\left(5 x \right)}}{5}

          Now evaluate the sub-integral.

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

          ∫sin⁡(5x)5 dx=∫sin⁡(5x) dx5\int \frac{\sin{\left(5 x \right)}}{5}\, dx = \frac{\int \sin{\left(5 x \right)}\, dx}{5}

          1. Let u=5xu = 5 x.

            Then let du=5dxdu = 5 dx and substitute du5\frac{du}{5}:

            ∫sin⁡(u)5 du\int \frac{\sin{\left(u \right)}}{5}\, du

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

              ∫sin⁡(u) du=∫sin⁡(u) du5\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{5}

              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)5- \frac{\cos{\left(u \right)}}{5}

            Now substitute uu back in:

            −cos⁡(5x)5- \frac{\cos{\left(5 x \right)}}{5}

          So, the result is: −cos⁡(5x)25- \frac{\cos{\left(5 x \right)}}{25}

        So, the result is: 3xsin⁡(5x)5+3cos⁡(5x)25\frac{3 x \sin{\left(5 x \right)}}{5} + \frac{3 \cos{\left(5 x \right)}}{25}

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

        ∫(−2cos⁡(5x)) dx=−2∫cos⁡(5x) dx\int \left(- 2 \cos{\left(5 x \right)}\right)\, dx = - 2 \int \cos{\left(5 x \right)}\, dx

        1. Let u=5xu = 5 x.

          Then let du=5dxdu = 5 dx and substitute du5\frac{du}{5}:

          ∫cos⁡(u)5 du\int \frac{\cos{\left(u \right)}}{5}\, du

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

            ∫cos⁡(u) du=∫cos⁡(u) du5\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{5}

            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)5\frac{\sin{\left(u \right)}}{5}

          Now substitute uu back in:

          sin⁡(5x)5\frac{\sin{\left(5 x \right)}}{5}

        So, the result is: −2sin⁡(5x)5- \frac{2 \sin{\left(5 x \right)}}{5}

      The result is: 3xsin⁡(5x)5−2sin⁡(5x)5+3cos⁡(5x)25\frac{3 x \sin{\left(5 x \right)}}{5} - \frac{2 \sin{\left(5 x \right)}}{5} + \frac{3 \cos{\left(5 x \right)}}{25}

    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)=cos⁡(5x)\operatorname{dv}{\left(x \right)} = \cos{\left(5 x \right)}.

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

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

      1. Let u=5xu = 5 x.

        Then let du=5dxdu = 5 dx and substitute du5\frac{du}{5}:

        ∫cos⁡(u)5 du\int \frac{\cos{\left(u \right)}}{5}\, du

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

          ∫cos⁡(u) du=∫cos⁡(u) du5\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{5}

          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)5\frac{\sin{\left(u \right)}}{5}

        Now substitute uu back in:

        sin⁡(5x)5\frac{\sin{\left(5 x \right)}}{5}

      Now evaluate the sub-integral.

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

      ∫3sin⁡(5x)5 dx=3∫sin⁡(5x) dx5\int \frac{3 \sin{\left(5 x \right)}}{5}\, dx = \frac{3 \int \sin{\left(5 x \right)}\, dx}{5}

      1. Let u=5xu = 5 x.

        Then let du=5dxdu = 5 dx and substitute du5\frac{du}{5}:

        ∫sin⁡(u)5 du\int \frac{\sin{\left(u \right)}}{5}\, du

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

          ∫sin⁡(u) du=∫sin⁡(u) du5\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{5}

          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)5- \frac{\cos{\left(u \right)}}{5}

        Now substitute uu back in:

        −cos⁡(5x)5- \frac{\cos{\left(5 x \right)}}{5}

      So, the result is: −3cos⁡(5x)25- \frac{3 \cos{\left(5 x \right)}}{25}

    Method #3

    1. Rewrite the integrand:

      (3x−2)cos⁡(5x)=3xcos⁡(5x)−2cos⁡(5x)\left(3 x - 2\right) \cos{\left(5 x \right)} = 3 x \cos{\left(5 x \right)} - 2 \cos{\left(5 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:

        ∫3xcos⁡(5x) dx=3∫xcos⁡(5x) dx\int 3 x \cos{\left(5 x \right)}\, dx = 3 \int x \cos{\left(5 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)=cos⁡(5x)\operatorname{dv}{\left(x \right)} = \cos{\left(5 x \right)}.

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

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

          1. Let u=5xu = 5 x.

            Then let du=5dxdu = 5 dx and substitute du5\frac{du}{5}:

            ∫cos⁡(u)5 du\int \frac{\cos{\left(u \right)}}{5}\, du

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

              ∫cos⁡(u) du=∫cos⁡(u) du5\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{5}

              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)5\frac{\sin{\left(u \right)}}{5}

            Now substitute uu back in:

            sin⁡(5x)5\frac{\sin{\left(5 x \right)}}{5}

          Now evaluate the sub-integral.

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

          ∫sin⁡(5x)5 dx=∫sin⁡(5x) dx5\int \frac{\sin{\left(5 x \right)}}{5}\, dx = \frac{\int \sin{\left(5 x \right)}\, dx}{5}

          1. Let u=5xu = 5 x.

            Then let du=5dxdu = 5 dx and substitute du5\frac{du}{5}:

            ∫sin⁡(u)5 du\int \frac{\sin{\left(u \right)}}{5}\, du

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

              ∫sin⁡(u) du=∫sin⁡(u) du5\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{5}

              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)5- \frac{\cos{\left(u \right)}}{5}

            Now substitute uu back in:

            −cos⁡(5x)5- \frac{\cos{\left(5 x \right)}}{5}

          So, the result is: −cos⁡(5x)25- \frac{\cos{\left(5 x \right)}}{25}

        So, the result is: 3xsin⁡(5x)5+3cos⁡(5x)25\frac{3 x \sin{\left(5 x \right)}}{5} + \frac{3 \cos{\left(5 x \right)}}{25}

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

        ∫(−2cos⁡(5x)) dx=−2∫cos⁡(5x) dx\int \left(- 2 \cos{\left(5 x \right)}\right)\, dx = - 2 \int \cos{\left(5 x \right)}\, dx

        1. Let u=5xu = 5 x.

          Then let du=5dxdu = 5 dx and substitute du5\frac{du}{5}:

          ∫cos⁡(u)5 du\int \frac{\cos{\left(u \right)}}{5}\, du

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

            ∫cos⁡(u) du=∫cos⁡(u) du5\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{5}

            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)5\frac{\sin{\left(u \right)}}{5}

          Now substitute uu back in:

          sin⁡(5x)5\frac{\sin{\left(5 x \right)}}{5}

        So, the result is: −2sin⁡(5x)5- \frac{2 \sin{\left(5 x \right)}}{5}

      The result is: 3xsin⁡(5x)5−2sin⁡(5x)5+3cos⁡(5x)25\frac{3 x \sin{\left(5 x \right)}}{5} - \frac{2 \sin{\left(5 x \right)}}{5} + \frac{3 \cos{\left(5 x \right)}}{25}

  2. Add the constant of integration:

    3xsin⁡(5x)5−2sin⁡(5x)5+3cos⁡(5x)25+constant\frac{3 x \sin{\left(5 x \right)}}{5} - \frac{2 \sin{\left(5 x \right)}}{5} + \frac{3 \cos{\left(5 x \right)}}{25}+ \mathrm{constant}

The answer is:

3xsin⁡(5x)5−2sin⁡(5x)5+3cos⁡(5x)25+constant\frac{3 x \sin{\left(5 x \right)}}{5} - \frac{2 \sin{\left(5 x \right)}}{5} + \frac{3 \cos{\left(5 x \right)}}{25}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                  
 |                             2*sin(5*x)   3*cos(5*x)   3*x*sin(5*x)
 | (3*x - 2)*cos(5*x) dx = C - ---------- + ---------- + ------------
 |                                 5            25            5      
/
∫(3x−2)cos⁡(5x) dx=C+3xsin⁡(5x)5−2sin⁡(5x)5+3cos⁡(5x)25\int \left(3 x - 2\right) \cos{\left(5 x \right)}\, dx = C + \frac{3 x \sin{\left(5 x \right)}}{5} - \frac{2 \sin{\left(5 x \right)}}{5} + \frac{3 \cos{\left(5 x \right)}}{25}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.902.5-2.5
Plot the graph f(x)
The answer [src] 
  3    sin(5)   3*cos(5)
- -- + ------ + --------
  25     5         25
sin⁡(5)5−325+3cos⁡(5)25\frac{\sin{\left(5 \right)}}{5} - \frac{3}{25} + \frac{3 \cos{\left(5 \right)}}{25} 
=
= 
  3    sin(5)   3*cos(5)
- -- + ------ + --------
  25     5         25
sin⁡(5)5−325+3cos⁡(5)25\frac{\sin{\left(5 \right)}}{5} - \frac{3}{25} + \frac{3 \cos{\left(5 \right)}}{25} 
-3/25 + sin(5)/5 + 3*cos(5)/25
Numerical answer [src] 
-0.277745392677041
-0.277745392677041

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

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