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(x^2)sin4xdx
Solving integrals/ (x^2)sin4xdx

Integral of (x^2)sin4xdx dx

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∫01x2sin⁡(4x)1 dx\int\limits_{0}^{1} x^{2} \sin{\left(4 x \right)} 1\, dx 
Integral(x^2*sin(4*x)*1, (x, 0, 1))
Detail solution

  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)=sin⁡(4x)\operatorname{dv}{\left(x \right)} = \sin{\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}:

      ∫sin⁡(u)16 du\int \frac{\sin{\left(u \right)}}{16}\, du

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

        ∫sin⁡(u)4 du=∫sin⁡(u) du4\int \frac{\sin{\left(u \right)}}{4}\, 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.

  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)=cos⁡(4x)\operatorname{dv}{\left(x \right)} = \cos{\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}:

      ∫cos⁡(u)16 du\int \frac{\cos{\left(u \right)}}{16}\, du

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

        ∫cos⁡(u)4 du=∫cos⁡(u) du4\int \frac{\cos{\left(u \right)}}{4}\, 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.

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

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

    1. Let u=4xu = 4 x.

      Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

      ∫sin⁡(u)16 du\int \frac{\sin{\left(u \right)}}{16}\, du

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

        ∫sin⁡(u)4 du=∫sin⁡(u) du4\int \frac{\sin{\left(u \right)}}{4}\, 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}

    So, the result is: cos⁡(4x)32\frac{\cos{\left(4 x \right)}}{32}

  4. Add the constant of integration:

    −x2cos⁡(4x)4+xsin⁡(4x)8+cos⁡(4x)32+constant- \frac{x^{2} \cos{\left(4 x \right)}}{4} + \frac{x \sin{\left(4 x \right)}}{8} + \frac{\cos{\left(4 x \right)}}{32}+ \mathrm{constant}

The answer is:

−x2cos⁡(4x)4+xsin⁡(4x)8+cos⁡(4x)32+constant- \frac{x^{2} \cos{\left(4 x \right)}}{4} + \frac{x \sin{\left(4 x \right)}}{8} + \frac{\cos{\left(4 x \right)}}{32}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                          
 |                                    2                      
 |  2                     cos(4*x)   x *cos(4*x)   x*sin(4*x)
 | x *sin(4*x)*1 dx = C + -------- - ----------- + ----------
 |                           32           4            8     
/
8 x sin⁡(4 x)+(2−16 x2) cos⁡(4 x)64{{8\,x\,\sin \left(4\,x\right)+\left(2-16\,x^2\right)\,\cos \left(4 \,x\right)}\over{64}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.901-1
Plot the graph f(x)
The answer [src] 
  1    7*cos(4)   sin(4)
- -- - -------- + ------
  32      32        8
4 sin⁡4−7 cos⁡432−132{{4\,\sin 4-7\,\cos 4}\over{32}}-{{1}\over{32}} 
=
= 
  1    7*cos(4)   sin(4)
- -- - -------- + ------
  32      32        8
sin⁡(4)8−132−7cos⁡(4)32\frac{\sin{\left(4 \right)}}{8} - \frac{1}{32} - \frac{7 \cos{\left(4 \right)}}{32}
Упростить
Numerical answer [src] 
0.0171342301504241
0.0171342301504241
The graph
Integral of (x^2)sin4xdx dx

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

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