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Solving integrals/ 8x^3sin(4x^4+2)

Integral of 8x^3sin(4x^4+2) dx

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The solution

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∫018x3sin⁡(4x4+2) dx\int\limits_{0}^{1} 8 x^{3} \sin{\left(4 x^{4} + 2 \right)}\, dx 
Integral((8*x^3)*sin(4*x^4 + 2), (x, 0, 1))
Detail solution

  1. Let u=4x4+2u = 4 x^{4} + 2.

    Then let du=16x3dxdu = 16 x^{3} dx and substitute du2\frac{du}{2}:

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

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

      ∫sin⁡(u) du=∫sin⁡(u) du2\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}

      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)2- \frac{\cos{\left(u \right)}}{2}

    Now substitute uu back in:

    −cos⁡(4x4+2)2- \frac{\cos{\left(4 x^{4} + 2 \right)}}{2}

  2. Now simplify:

    −cos⁡(4x4+2)2- \frac{\cos{\left(4 x^{4} + 2 \right)}}{2}

  3. Add the constant of integration:

    −cos⁡(4x4+2)2+constant- \frac{\cos{\left(4 x^{4} + 2 \right)}}{2}+ \mathrm{constant}

The answer is:

−cos⁡(4x4+2)2+constant- \frac{\cos{\left(4 x^{4} + 2 \right)}}{2}+ \mathrm{constant}

The answer (Indefinite) [src] 
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 |    3    /   4    \          cos\4*x  + 2/
 | 8*x *sin\4*x  + 2/ dx = C - -------------
 |                                   2      
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∫8x3sin⁡(4x4+2) dx=C−cos⁡(4x4+2)2\int 8 x^{3} \sin{\left(4 x^{4} + 2 \right)}\, dx = C - \frac{\cos{\left(4 x^{4} + 2 \right)}}{2}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.90-1010
Plot the graph f(x)
The answer [src] 
cos(2)   cos(6)
------ - ------
  2        2
−cos⁡(6)2+cos⁡(2)2- \frac{\cos{\left(6 \right)}}{2} + \frac{\cos{\left(2 \right)}}{2} 
=
= 
cos(2)   cos(6)
------ - ------
  2        2
−cos⁡(6)2+cos⁡(2)2- \frac{\cos{\left(6 \right)}}{2} + \frac{\cos{\left(2 \right)}}{2} 
cos(2)/2 - cos(6)/2
Numerical answer [src] 
-0.688158561598754
-0.688158561598754

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

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