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Solving integrals/ (8x-1)sin(5x+3)

Integral of (8x-1)sin(5x+3) dx

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 |  (8*x - 1)*sin(5*x + 3) dx
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∫01(8x−1)sin⁡(5x+3) dx\int\limits_{0}^{1} \left(8 x - 1\right) \sin{\left(5 x + 3 \right)}\, dx 
Integral((8*x - 1)*sin(5*x + 3), (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      (8x−1)sin⁡(5x+3)=8xsin⁡(5x+3)−sin⁡(5x+3)\left(8 x - 1\right) \sin{\left(5 x + 3 \right)} = 8 x \sin{\left(5 x + 3 \right)} - \sin{\left(5 x + 3 \right)}

    2. Integrate term-by-term:

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

        ∫8xsin⁡(5x+3) dx=8∫xsin⁡(5x+3) dx\int 8 x \sin{\left(5 x + 3 \right)}\, dx = 8 \int x \sin{\left(5 x + 3 \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⁡(5x+3)\operatorname{dv}{\left(x \right)} = \sin{\left(5 x + 3 \right)}.

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

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

          1. Let u=5x+3u = 5 x + 3.

            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+3)5- \frac{\cos{\left(5 x + 3 \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:

          ∫(−cos⁡(5x+3)5) dx=−∫cos⁡(5x+3) dx5\int \left(- \frac{\cos{\left(5 x + 3 \right)}}{5}\right)\, dx = - \frac{\int \cos{\left(5 x + 3 \right)}\, dx}{5}

          1. Let u=5x+3u = 5 x + 3.

            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+3)5\frac{\sin{\left(5 x + 3 \right)}}{5}

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

        So, the result is: −8xcos⁡(5x+3)5+8sin⁡(5x+3)25- \frac{8 x \cos{\left(5 x + 3 \right)}}{5} + \frac{8 \sin{\left(5 x + 3 \right)}}{25}

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

        ∫(−sin⁡(5x+3)) dx=−∫sin⁡(5x+3) dx\int \left(- \sin{\left(5 x + 3 \right)}\right)\, dx = - \int \sin{\left(5 x + 3 \right)}\, dx

        1. Let u=5x+3u = 5 x + 3.

          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+3)5- \frac{\cos{\left(5 x + 3 \right)}}{5}

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

      The result is: −8xcos⁡(5x+3)5+8sin⁡(5x+3)25+cos⁡(5x+3)5- \frac{8 x \cos{\left(5 x + 3 \right)}}{5} + \frac{8 \sin{\left(5 x + 3 \right)}}{25} + \frac{\cos{\left(5 x + 3 \right)}}{5}

    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)=8x−1u{\left(x \right)} = 8 x - 1 and let dv⁡(x)=sin⁡(5x+3)\operatorname{dv}{\left(x \right)} = \sin{\left(5 x + 3 \right)}.

      Then du⁡(x)=8\operatorname{du}{\left(x \right)} = 8.

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

      1. Let u=5x+3u = 5 x + 3.

        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+3)5- \frac{\cos{\left(5 x + 3 \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:

      ∫(−8cos⁡(5x+3)5) dx=−8∫cos⁡(5x+3) dx5\int \left(- \frac{8 \cos{\left(5 x + 3 \right)}}{5}\right)\, dx = - \frac{8 \int \cos{\left(5 x + 3 \right)}\, dx}{5}

      1. Let u=5x+3u = 5 x + 3.

        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+3)5\frac{\sin{\left(5 x + 3 \right)}}{5}

      So, the result is: −8sin⁡(5x+3)25- \frac{8 \sin{\left(5 x + 3 \right)}}{25}

    Method #3

    1. Rewrite the integrand:

      (8x−1)sin⁡(5x+3)=8xsin⁡(5x+3)−sin⁡(5x+3)\left(8 x - 1\right) \sin{\left(5 x + 3 \right)} = 8 x \sin{\left(5 x + 3 \right)} - \sin{\left(5 x + 3 \right)}

    2. Integrate term-by-term:

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

        ∫8xsin⁡(5x+3) dx=8∫xsin⁡(5x+3) dx\int 8 x \sin{\left(5 x + 3 \right)}\, dx = 8 \int x \sin{\left(5 x + 3 \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⁡(5x+3)\operatorname{dv}{\left(x \right)} = \sin{\left(5 x + 3 \right)}.

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

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

          1. Let u=5x+3u = 5 x + 3.

            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+3)5- \frac{\cos{\left(5 x + 3 \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:

          ∫(−cos⁡(5x+3)5) dx=−∫cos⁡(5x+3) dx5\int \left(- \frac{\cos{\left(5 x + 3 \right)}}{5}\right)\, dx = - \frac{\int \cos{\left(5 x + 3 \right)}\, dx}{5}

          1. Let u=5x+3u = 5 x + 3.

            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+3)5\frac{\sin{\left(5 x + 3 \right)}}{5}

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

        So, the result is: −8xcos⁡(5x+3)5+8sin⁡(5x+3)25- \frac{8 x \cos{\left(5 x + 3 \right)}}{5} + \frac{8 \sin{\left(5 x + 3 \right)}}{25}

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

        ∫(−sin⁡(5x+3)) dx=−∫sin⁡(5x+3) dx\int \left(- \sin{\left(5 x + 3 \right)}\right)\, dx = - \int \sin{\left(5 x + 3 \right)}\, dx

        1. Let u=5x+3u = 5 x + 3.

          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+3)5- \frac{\cos{\left(5 x + 3 \right)}}{5}

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

      The result is: −8xcos⁡(5x+3)5+8sin⁡(5x+3)25+cos⁡(5x+3)5- \frac{8 x \cos{\left(5 x + 3 \right)}}{5} + \frac{8 \sin{\left(5 x + 3 \right)}}{25} + \frac{\cos{\left(5 x + 3 \right)}}{5}

  2. Add the constant of integration:

    −8xcos⁡(5x+3)5+8sin⁡(5x+3)25+cos⁡(5x+3)5+constant- \frac{8 x \cos{\left(5 x + 3 \right)}}{5} + \frac{8 \sin{\left(5 x + 3 \right)}}{25} + \frac{\cos{\left(5 x + 3 \right)}}{5}+ \mathrm{constant}

The answer is:

−8xcos⁡(5x+3)5+8sin⁡(5x+3)25+cos⁡(5x+3)5+constant- \frac{8 x \cos{\left(5 x + 3 \right)}}{5} + \frac{8 \sin{\left(5 x + 3 \right)}}{25} + \frac{\cos{\left(5 x + 3 \right)}}{5}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                                
 |                                 cos(3 + 5*x)   8*sin(3 + 5*x)   8*x*cos(3 + 5*x)
 | (8*x - 1)*sin(5*x + 3) dx = C + ------------ + -------------- - ----------------
 |                                      5               25                5        
/
∫(8x−1)sin⁡(5x+3) dx=C−8xcos⁡(5x+3)5+8sin⁡(5x+3)25+cos⁡(5x+3)5\int \left(8 x - 1\right) \sin{\left(5 x + 3 \right)}\, dx = C - \frac{8 x \cos{\left(5 x + 3 \right)}}{5} + \frac{8 \sin{\left(5 x + 3 \right)}}{25} + \frac{\cos{\left(5 x + 3 \right)}}{5}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.90-1010
Plot the graph f(x)
The answer [src] 
  8*sin(3)   7*cos(8)   cos(3)   8*sin(8)
- -------- - -------- - ------ + --------
     25         5         5         25
−8sin⁡(3)25−cos⁡(3)5−7cos⁡(8)5+8sin⁡(8)25- \frac{8 \sin{\left(3 \right)}}{25} - \frac{\cos{\left(3 \right)}}{5} - \frac{7 \cos{\left(8 \right)}}{5} + \frac{8 \sin{\left(8 \right)}}{25} 
=
= 
  8*sin(3)   7*cos(8)   cos(3)   8*sin(8)
- -------- - -------- - ------ + --------
     25         5         5         25
−8sin⁡(3)25−cos⁡(3)5−7cos⁡(8)5+8sin⁡(8)25- \frac{8 \sin{\left(3 \right)}}{25} - \frac{\cos{\left(3 \right)}}{5} - \frac{7 \cos{\left(8 \right)}}{5} + \frac{8 \sin{\left(8 \right)}}{25} 
-8*sin(3)/25 - 7*cos(8)/5 - cos(3)/5 + 8*sin(8)/25
Numerical answer [src] 
0.673134782992473
0.673134782992473

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

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