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Solving integrals/ (1-8x^2)cos4xdx

Integral of (1-8x^2)cos4xdx dx

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The solution

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 2*pi                      
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  |  /       2\            
  |  \1 - 8*x /*cos(4*x) dx
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 0
∫02π(1−8x2)cos⁡(4x) dx\int\limits_{0}^{2 \pi} \left(1 - 8 x^{2}\right) \cos{\left(4 x \right)}\, dx 
Integral((1 - 8*x^2)*cos(4*x), (x, 0, 2*pi))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      (1−8x2)cos⁡(4x)=−8x2cos⁡(4x)+cos⁡(4x)\left(1 - 8 x^{2}\right) \cos{\left(4 x \right)} = - 8 x^{2} \cos{\left(4 x \right)} + \cos{\left(4 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:

        ∫(−8x2cos⁡(4x)) dx=−8∫x2cos⁡(4x) dx\int \left(- 8 x^{2} \cos{\left(4 x \right)}\right)\, dx = - 8 \int x^{2} \cos{\left(4 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)=x2u{\left(x \right)} = x^{2} and let dv⁡(x)=cos⁡(4x)\operatorname{dv}{\left(x \right)} = \cos{\left(4 x \right)}.

          Then du⁡(x)=2x\operatorname{du}{\left(x \right)} = 2 x.

          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}:

            ∫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}

          Now evaluate the sub-integral.

        2. 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)=x2u{\left(x \right)} = \frac{x}{2} and let dv⁡(x)=sin⁡(4x)\operatorname{dv}{\left(x \right)} = \sin{\left(4 x \right)}.

          Then du⁡(x)=12\operatorname{du}{\left(x \right)} = \frac{1}{2}.

          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.

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

          ∫(−cos⁡(4x)8) dx=−∫cos⁡(4x) dx8\int \left(- \frac{\cos{\left(4 x \right)}}{8}\right)\, dx = - \frac{\int \cos{\left(4 x \right)}\, dx}{8}

          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)32- \frac{\sin{\left(4 x \right)}}{32}

        So, the result is: −2x2sin⁡(4x)−xcos⁡(4x)+sin⁡(4x)4- 2 x^{2} \sin{\left(4 x \right)} - x \cos{\left(4 x \right)} + \frac{\sin{\left(4 x \right)}}{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}

      The result is: −2x2sin⁡(4x)−xcos⁡(4x)+sin⁡(4x)2- 2 x^{2} \sin{\left(4 x \right)} - x \cos{\left(4 x \right)} + \frac{\sin{\left(4 x \right)}}{2}

    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)=1−8x2u{\left(x \right)} = 1 - 8 x^{2} and let dv⁡(x)=cos⁡(4x)\operatorname{dv}{\left(x \right)} = \cos{\left(4 x \right)}.

      Then du⁡(x)=−16x\operatorname{du}{\left(x \right)} = - 16 x.

      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}:

        ∫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}

      Now evaluate the sub-integral.

    2. 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)=−4xu{\left(x \right)} = - 4 x and let dv⁡(x)=sin⁡(4x)\operatorname{dv}{\left(x \right)} = \sin{\left(4 x \right)}.

      Then du⁡(x)=−4\operatorname{du}{\left(x \right)} = -4.

      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.

    3. 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}

    Method #3

    1. Rewrite the integrand:

      (1−8x2)cos⁡(4x)=−8x2cos⁡(4x)+cos⁡(4x)\left(1 - 8 x^{2}\right) \cos{\left(4 x \right)} = - 8 x^{2} \cos{\left(4 x \right)} + \cos{\left(4 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:

        ∫(−8x2cos⁡(4x)) dx=−8∫x2cos⁡(4x) dx\int \left(- 8 x^{2} \cos{\left(4 x \right)}\right)\, dx = - 8 \int x^{2} \cos{\left(4 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)=x2u{\left(x \right)} = x^{2} and let dv⁡(x)=cos⁡(4x)\operatorname{dv}{\left(x \right)} = \cos{\left(4 x \right)}.

          Then du⁡(x)=2x\operatorname{du}{\left(x \right)} = 2 x.

          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}:

            ∫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}

          Now evaluate the sub-integral.

        2. 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)=x2u{\left(x \right)} = \frac{x}{2} and let dv⁡(x)=sin⁡(4x)\operatorname{dv}{\left(x \right)} = \sin{\left(4 x \right)}.

          Then du⁡(x)=12\operatorname{du}{\left(x \right)} = \frac{1}{2}.

          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.

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

          ∫(−cos⁡(4x)8) dx=−∫cos⁡(4x) dx8\int \left(- \frac{\cos{\left(4 x \right)}}{8}\right)\, dx = - \frac{\int \cos{\left(4 x \right)}\, dx}{8}

          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)32- \frac{\sin{\left(4 x \right)}}{32}

        So, the result is: −2x2sin⁡(4x)−xcos⁡(4x)+sin⁡(4x)4- 2 x^{2} \sin{\left(4 x \right)} - x \cos{\left(4 x \right)} + \frac{\sin{\left(4 x \right)}}{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}

      The result is: −2x2sin⁡(4x)−xcos⁡(4x)+sin⁡(4x)2- 2 x^{2} \sin{\left(4 x \right)} - x \cos{\left(4 x \right)} + \frac{\sin{\left(4 x \right)}}{2}

  2. Add the constant of integration:

    −2x2sin⁡(4x)−xcos⁡(4x)+sin⁡(4x)2+constant- 2 x^{2} \sin{\left(4 x \right)} - x \cos{\left(4 x \right)} + \frac{\sin{\left(4 x \right)}}{2}+ \mathrm{constant}

The answer is:

−2x2sin⁡(4x)−xcos⁡(4x)+sin⁡(4x)2+constant- 2 x^{2} \sin{\left(4 x \right)} - x \cos{\left(4 x \right)} + \frac{\sin{\left(4 x \right)}}{2}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                  
 |                                                                   
 | /       2\                   sin(4*x)                   2         
 | \1 - 8*x /*cos(4*x) dx = C + -------- - x*cos(4*x) - 2*x *sin(4*x)
 |                                 2                                 
/
∫(1−8x2)cos⁡(4x) dx=C−2x2sin⁡(4x)−xcos⁡(4x)+sin⁡(4x)2\int \left(1 - 8 x^{2}\right) \cos{\left(4 x \right)}\, dx = C - 2 x^{2} \sin{\left(4 x \right)} - x \cos{\left(4 x \right)} + \frac{\sin{\left(4 x \right)}}{2}
Simplify
The graph
0.00.51.01.52.02.53.03.54.04.55.05.56.0-500500
Plot the graph f(x)
The answer [src] 
-2*pi
−2π- 2 \pi 
=
= 
-2*pi
−2π- 2 \pi 
-2*pi
Numerical answer [src] 
-6.28318530717951
-6.28318530717951

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

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